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Anurag Mishra Mechanics 1 with www.puucho.com

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·' !II Published by:

SHRI BALAJI PUBLICATIONS (EDUCATIONAL PUBLISHERS & DISTRIBUTORS)

6, Gulshan Vihar, Gali No. 1, Opp. Mahalaxmi Enclave, Jansath Road, Muzaffarnagar (U.P.) Phone: 0131-2660440 (0), 2600503 (R) website : www.shribalajibooks.com email : [email protected]

!II First edition

2009

!II Fourth edition !II Fifth edition !II Reprint

2012 2013 2014

!II © All rights reserved with author

!II Price : {

428.00

!II Typeset by : Sun Creation Muzaffarnagar

!II _printed at : Dayal Offset Printers Meerut (U.P.)

!II All the rights reserved .. No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the author and publisher. Any violation/breach shall be taken into legal action.

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Preface

I have been involved in teaching Physics for last 16 years. This book is an opportunity to present my experiences. During my interaction with UT-JEE aspirants. I realised that most feared topic is mechanics. Some of the reasons put forward by students behind this thought were: @

No spontaneous thoughts.appear after reading a problem. Mind goes blank. Can not . proceed in a problem.

@

How to ~roceed in a problem? Which law is applicable; that is a given problem will involve conservation of energy or momentum or both.

@

If some one says solve the problem in non-inertial reference frame, horrible thoughts appear inmind.

@

Total confusion about CM frame.

@

Proper understanding of constraints.

@

S)lort cut approach in relative motion.

@

No single book available that gives large no. of solved examples with elaboration of concepts in asolµtion.

This book will help the students in building analytical and quantitative skills, addressing keyl,Jlisconceptions and developing.confidence in problem solving. I sincerely wish that this book will fulfill all the aspirations of the readers. Although utmost full care has been taken to make the book free from error but some errors. . ina_dvertently may creep-in. Author and Publisher shall be highly obliged if suggestions. regarding improvement and errors are pointed out by readers. I am indebted Neeraj Ji for providing me an opportunity to write a book of this magnitude. -

.

I am indebted to my father Sh. Bhavesh Mishra; my mother Smt. Priyamvada Mishra, my wife Manjari, my sister Parul, my little kids Vrishank and Ira for giving their valuable time which I utilized during the writing of this book and people of Morada bad, who supported. . ·me throughout my career. In the last, !also pay my sincere thanks to all the esteemed members ofM/s. ShriBalaji Publications in bringing out this book in"the present form.

Anurag Mishra.

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Our Other Publications for

(Engineering Entrance Examination€)

Er. Anurag Mishra

Er. Anurag Mishra

Mechanics !•JEE

• Simple Harmonic Motion • Wave Motion

Electricity& Magnetism

forJEE

• Solid and Fluids • Gravitation

Vol. II

• Electrostatics , •, Electric Current

j•

yapacitors

I

l• 1•

The Magnetic Field

/

and A.G. Circuits

'

Electromagnetic Induction

'I-----~

Er. Anurag Mishra

)feat& Thermodynamics

/•JEE

I'

.

1 •

Temperature, Heat & tl)e. equation of State. Heat Transfer

Optics

• Thermodynamics

.I ,

'

.

: : ·,-:' \ ~

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. foJ EE

.

Geometrical Optics'

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Wave Opti~

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How to face the challenge ? Following are some doubts which arise in the mind of almost all the students but may face them by taking some care. 1.

I can not solve numerical because my concepts are not clear. In fact numerical solvingitselfis an exercise t~ learn concepts.

·z.

1 /can not study because I am in depression, I fell into it because I was not ' studying! Depression is escape mechanism of people afraid of facing failures. Failure is integral partoflearning.

3.

I understand everything in class but can not solve on my own.WRITING work is · vital. It is a multiple activity, initially idea comes in mind then we put into language to express it, we are focussed in hand eye coordination, eyes create visual impression on brain which isrecorded there. WRITING WORKS ARE EMBOSSED ON BRAIN LIKE CARVINGS OF AJANTACAVES.

4.

In exams my brain goes blank, but I can crack them at home. Home attempt is your second attempt! you are contemplating about it while home back. You do not behave differently in exam you replicate your instincts. Once a fast bowler was bowling no balls. His coach placed a-stump on crease, in fear of injury he got it right. CONCEPTUALIZATION, WRITING EQUATION, SOLVING, THEN PROBLEM GETS TO CONCLUSION!

5.

I am an average student. It is a rationalization used by people afraid of hard work. In their reference frame Newton's first law applies "if I have a misconception I will continue with it unless pushed by an external agent even I will surround him in my web of misconception yielding zero resultant:' AVERAGE IS NOT DUE TO CAPACITY LACUANEBUT DUE TO LACK OF DETERMINATION TO SHED INERTNESS.

6.

A famous cliche "/ do not have luck in my favour' PRINCIPLE OF CAUSALITY: CAUSE OF AN EVENT OCCURS IN TIME BEFORE OCCURRENCE OF THAT EVENT i.e., cause occurs first then event occurs. SHINING OF LUCK IS NOT AN iNSTANTANEOUS EVENT IT IS PRECEDED BY RELENTLESS HARD WORK. Sow a seed ofaspiration in mind, water it with passion, dedication it will bear fruit, luckcan give you sweeter fruit.

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Do not take study as a burden actually )ts a skill like singing and dancing. It has to be honed by proper devotion and dedication. ·

2.

Withou't strong sen_se of achievement you can't excel. Before entering the cmppeti.tive field strong counselling by parents is must. Majority do not · know what for they are here. No strategic planning, they behave like a tail : ende_r batting in frontofSteyn's bouncers.

3. - 'science is not a subje,ct based on well laid dowh procedures or· based on learning sonie facts, it ipvolves very intuitive and exploratory approach. Unless their is desire and passion to learn you can not discover new ideas. It requires p'atience and hard work, whose fruits may be tangible later on.

4.

Some students realize very late that they are studying for acquiring skills and , .honingthem. Their is a feeling that they can ride at the back of instructor and · , achieve ~xcellence. Study comes as tqrtu·rous exercise enforced on them and their is some mechanism that can take this burden-of them. -

5.

Science is not about gaining good marks, up toXth by reading key points good marks are achieved but beyond that only those survive who have genuine interest in learning and exp to ring. Selfstudy habit is must.

6. ·

IF YOU WANT TO GAIN LEAD START EARLY. Majority of successful students try to finish .major portion-elementary part of syllabus before they enter Coaching Institute. Due to this their maturity level as comparecl to others is · more tliey get ample time to adj4st with th~ fast pace. They are less· traumatized by the scientific matter handed over. For those who enter fresh must be counselled to not get bullied by ·early starters but work harder initially within first two months initial edge is neutralized.·

7.

Once a sl:l,ldent lags behind due to scime forced or unforced errors his mind begins to play rationalization remarks like I am an average student, my mind is not sharp enough, I have low IQ etc: These words are mechanisms _used to · a\Toid hard work. These words a,re relative terms a person who has .early start may be intelligentr~lative t9y6u.' . . .

,

'

' :i.

0

lntelligence means _cu~ulative i-~sult of h_ard _work of previous years, that hard work has eventuallY. led to a developm~nt of instinct tci crack 'things easily. _., · /

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CONTENTS

--

(j UNIT AND DIMENSIONS (j I. DESCRIPTION OF MOTION Subject of kinematics (19), Vector notation (20), Displacement

I-

vector (20),Parallelogram law of vector addition (21)., Component ofvector(23), Unitvector(24), Expressing a vector in unit vector notati_on (25), Position vector (27), Rectangular resolution of a vector in three dimensions (28), Vector multiplication (31), The scalar product of two vectors (32), The vector product of two vectors (35), Rectilinear motion of a particle (37), Calculus supplementary (41), Rectilinear motion (44), Instantaneous velocity ( 45), Integration ( 48), Interpretation of graphs (53), Average velocity & Average speed (58), Two-dimensional motion with constant acceleration (69), Projection on an inclined plane (82), Relative motion (87), Application of advanced concept of relative motion (88), Equation of motion for relative motion (98), Projection of a particle in an accelerated elevator (101), Projection of a ball in horizontally moving trolley (101 ), Closest distance of approach between two moving bodies (102), Problems: Level-1 (106), Level-2(113), Level-3 (117),Answers(123), Solutions (124).

Q

2. FORCE ANALYSIS The concept of force (138), Reference frame (139), Ideal string

~~~~~p=--=~-=::;~

(143), ideal pulley (143), Contact force (143), Concept of external and internal force (144), Pulley system (145), Tension in a hanging rope (145), Constrained motion (148), Pulley constraint (154), Normal constraint (155), Elastic force of spring (158), Parallel combination (159), Friction (165), The laws of sliding friction (165), Direction of kinetic Friction (171), FBD when arm is in_ deceleration (185), Circular motion (186), Angular velodty vector (187), Concept of , pseudo force (192), Non-inertial reference frame (193), Whirling rope (195), Lift Force on an airplane (196), Non-uniform circular motion on horizontal plane {196), Problems: Level-1 (210), Level-2 (221),Level-3 (230),Answers (238), Solutions (240).

1

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-CJ 3. WORK AND ENERGY Work done (265), Unit of work (266), Conservative and non-conservative force (268), Concept of potential energy (269), Classical work energy theorem (270), Conservation of mechanical energy (271), Work done by friction (271), Work done by spring force (271), Work depends on the frame of reference (272), Work due to internal force (friction) (272); Work energy theorem in a non-inertial reference frame (273), How to apply coriservation of energy_ equation (273), Vertical circ_ular motion (283), Power (293), lntemai energy so!-lrces & work (296), Problems Level-1 (299), Level-2 (306), Level-3 (310),Answers (316), Solutions (318).

CJ 4. IMPULSE AND MOMENTUM

Impulse (328), Conservation of momentum (328), Conservation momentum for a two particle system (329), Relative velocity and the conservation of momentum (330), Recoil disintegration, explosions (335), Impulsive force (336), Centripetal acceleration revisite~ (338), Centre of mass (340), Position of COM of two particles (340), Centre of gravity (341 ), Motion of the centre of mass (34,1), Kinetic energy of a system of particles (342), Most important concept (343), Finding the centre of mass by integration (353), Collisions (361), Models for elastic & inelastic collisions (362), Oblique impact (365), The velocity of the centre of mass for collisions (370), Elastic collisions in the CM reference frame (371), Inelastic collisions in CM reference frame (372), System of variable mass; Rocket propulsion (380), Problems Level-1 (383), Level-2 (395), Level-3 (399), Answers (406), Solutions (408). ·

(J 5. RIGID BODY MOTION What is rigid body(421), General rigid body motion (421 ), Rotation about centre of mass (422), Kinematics of fixed axis rotation (422), Vector representation of rotational quantities (425), .Torque (427), Newton's second law for rotation (429), Rotational kinetic energy and moment of inertia (430), Rotational kinetic energy of a collection of particles (431), Perpendicular axis theorem (437), Dynamics of a rigid body(441), Angular momentum (447), The ladder (450), Work done due to torque (457), Angular ,, momentum of a projectile (462), Angular momentum of an i[]verted conical pendulum (462), Angular '=--~='-~=-~=~-impulse0angular momentum theorem (464), Two bodies rotatory system (466), Kinematics ofrigidbody rotation (475), Total kinetic energy of body (484), Dynamics of rigid body in plane motion (486), Torque on the rotating skew rod (506), Problem: Level-1 (511); Level-2 (519), Level,3 (525), Answers (531), Solutions (533), Exercisg_,_advanced problems (544), Comprehension based pro blems (553), Assertion and reason type problems (562).

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UNIT AND DIMENSIONS .

Physics is that branch of science in which we observe, measure and describe natural phenomena related to matter and energy. Like all the science, physics is ultimately based on observation. To ~ssemble the relevant observations into a coherent picture by, constructing a logical framework is called theory. Theory enables the physicist to account for past observations and to decide how new ones should be made. Nearly all physical observation are quantitative; they require measurement.

(a) Magnitude of Physical Quantity= Numerical value x Unit th us for a given physical quantity when the unit will change, numerical value will also change, e.g. density of water =lg- cc- 1 = 10 3 kg- m-3 and not lkg- m-3 .

Every measurement is a comparison of a quantity with a standard quantity that is, an agreed upon quantity of the same kind. To measure a

Criteria for Standards The choice of the standard is arbitrary. However, several criteria must be met if a standard is to be as useful as possible. 1. Stability : The standard should not vary with time. If this criterion is satisfied, measurements made at different time, using the same standard, can be meaningfully compared. 2. Reproducibility: The standard should be accurately reproducible so that copies, ideally identical with the standard itself can be used elsewhere. If this criterion is satisfied, measurements made at different places can be compared. 3. Acceptability : The standard should be universally accepted so as to eliminate clumsy and possibly inaccurate comparisons among measurement made with separate standard. 4. Accessibility : As nearly as possible, the standard should be readily accessible to everyone ·vho needs to use it. 5. Precision : It should be possible to measure the standard itself with a precision at least as great as the precision with which any comparable measurement can be made.

length for example, you adopt as your standard a convenient measuring rod, whose length you use as the unit of length. You count the number of times that the rod fits into the length to be measured. This number given the length in terms of the chosen unit. Physical Quantities The quantities by means of which describe the laws of physics are called Physical Quantities. A physical quantity is complete specified if it has ~---e>(A) Numerical value only ratio e.g., refractive index, dielectric constant, etc.

or Magnitude only Scalar e.g., mass, charge,current etc.

or Magnitude and direction Vector e.g., displacement, torque, etc.

In expressing the magnitude of a physical quantity we choose a unit and then find physical quantity how many times that unit is contained in the given physical quantity, i.e.

(b) Larger the unit smaller will be the magnitude and vice-versa, e.g., 1 kg= 1000 gm then as 1000 is greater than 1, gm is smaller unit than kg of mass.

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6. Security : The standard should be as safe as possible from and preferably' immune from all possible· causes of damage. · . . · When a' standard meets these criteria as nearly as. possible, it can. be·taken as the-primary standard. The 1 primary standard Ca.I\ 'the!J be,yse_d to produce secondary· standards'calibrateo in terms of the primary, and so on. · A set of base units, toget:li,(!r with the ~es required to express·all other units fn tenns·of.them, constitutes a system of units. The systein,in general use throughout the world is called _the Interil.ati(!nal System, of Units. The, short form SI (from the French System International) is used in ' all languages.

Systems of Units'. ·. ,. A complete ·set ~f tmits; both.fundamental anc\ derived for all kinds of phys.ical quantities, is called ·a system of units. There are· several systems of units· which liave been· employed for describing measurements. A few common ' .. systems are give'n ):,elo.;,. : ' A. CGS system . _ , The systeni is' also c+y+2zT-2x-y-zl

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·-~!U_Nf_TAN_D_Df_M_EN_SI_ON_S_ _ _ _ _ _ _ _ _ _ _ _ _ _-'--_

Equating the exponents of sl~ilar quantities on both Sides Of the equation, ' . , 'i I , -x+z =·o, 3x·+ y + 2z·=·l 'and -ix·-y-z = 0 Solving ·these for x, y and z, we get 1 ·, .:.3 1 and Z=x=2; Y =2 0

2

So eqn .. (1) becomes, QL

=KG112c-,12h v2

If the constant K is assumed, to be unity . QL =~Gh/c 3,'

_;__-'----'·-:,.,.~,:~·c,;..c\"----~----·''·i_j]

if dimensions· are g!ven. ·For example,._if.)he di111ensional formula of a physical quantity is'[ML2T-2l, it may be"work or ' . energy or torque. - ,., . , , . (2) Numerical constant [Kl having no dimensions such as (1/2), 1 or 21t, etc., cannot be expressed by,.the niethods .,.of dimensions. · · (3) The method of dimensions cannot used. to derive relations involving produci:,of physical qua~tities.,'it-cannot be used to' derive relations other )hari product of power functions. For example, · ' . ,, , s = ut + (1/2) at. 2 , Qr·... y =.asincot ' Cannot be derived by USlng- thi~, th~Ory.: , ' , I ,

.

0

•

L.!=~"A-t;ll\RJ~,.@> fij ::e~city,. f~;c~ ~d- time are tizl~n to be fundamental tquantities

find dimensional formula for (a) Mass, and

·

~(l,)_Efl~r~--_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.,

Solution Let the quantity be Q, then · Q = f (v,F,T) Assuming that the function is the product of power functions of v, F and T, · Q =K vxpyyz .... (1)

where K is a dimensionless constant of proportionality. The above equation dimensionally becomes. [Q] = [LT- 1 Y[MLT-2]Y[T]'

i.e.

.. .. (2)

We can check the dimensional correctness of these relations. , . , (4) Dimensioni,J analysis is 'not ~seful for· deriving formula for a physical quantity that depends·on more than 3 physical quantities as then ,there will be less number of equations than the unknowns. However,. still we can check correctness of the given equation dimensionally. For example, T =21t~I I mgL cannot ·b~ derived _by theory of dimensions but its dimensional correctness can be checked. (5) Even if a physical quantity depends on 3 physical quantities, out of which two have same dimensions, the formula cannot be deriv~d liy theory of dimensions, e.g.' · formula for the frequency of a tuning fork f =.(d/L2 )v cannot be derived by· dimensional analysis. '

Now i.e., [Ql Q = Mass So eqn. (2) becomes [Ml = MyLx+yT-x- 2y+zl (a)

= [Ml

its dimensional correctness requires. y = 1,x+y = Oand-x-2y +z = 0 which on solving yields x=-1,y=landz=l Substituting it in eqn. (1), we get Q=Kv-1Ff, (b)

'

PHYSICAL QUANTITIES FROM ., . . HEAT AN!;> TH~,R~ODYNAMICS

Q = Energy i.e. [Q] = [ML2T-2l

So eqn. (2) becomes [ML2T-2l = [MYLx+yrx-2y+zl Which· is the light of principle of homogeneity yields y = l;x+y = 2and-x-2y +z = -2 which on solving yields

1. Temperature: It is _a 'fundamental quantity with dimensions [0l and unit kelvin '[Kl. 2. Heat: By definition," it is energy transferred due to energy transferred, so its dimensions' ate [ML2T-2l arid SI unit joule (J). ~racticai unit of heat .is calorie (cal} and 1 calorie = 4.18 joule. · . · . 3. Coefficient of lin~ar-Expansion a '

So

Le.

Q=KvFf

Limitations of Theory of Dimensions (1) If dimensions ate given, many physical quantities have same dimensions. Physical quantity may not be unique

Af.

4. Specific heat c As

X=y=z=l

So eqn. (1) becomes

'

It is defined as a= - ' . L,i0 Le., '[al =[B-1 l . ' So its unit is (C 0 i-1 or K~J' Q= mc,iB; Q'

c=--mi\0·. -: . [ML2 F 2 l [cl= c......,~~ ' "[M][0l

[cl= [L2 r 20-1 l So its SI unit -will be . J/kg:~. "".hile· practical unit

i.e.

cal/g-co.

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[p +

5. Latent Heat L By definition, Q = mL , [L] = [ML2T-2] / [M]

i.e., i.e.,

or

. [L] = [L2r2]

[ML2 r

2

As this equation is dimensionally correct, each term on either side will have same dimensions i.e., [a/V] = [PV] or [a] = [ML-1r2J [L3] [L3] = [MLsr2]

and or

/T]

[P x b]

[K]=--~2

11 _~

[L ][0/L] [Kl= [MLr30-11

Its SI unit is W/m-k while practical unit is caVs-cm-C'. 7, Mechanical equivalent of heat J According to Ist law of thermodynamics work and heat are related as

l~rt. is es_ timat·e·d· ;h.at p-;;;;;_.in.~te ~~c-h ~~--_i_~i~~rth.: -~_;_e_"iv;;:;j about 2 c°;lorie of heat energy from the sun. This constaizr ~

1

called solar constant S. Fxpress solar constant in SI units._;_!

Solution. Given that S = 2cal/ cm 2 -min

W=JH

But as 1 cal= 4.18J, 1 cm= 10-2 m'and 1 min= 60 s

w

or

J=-

S=

H

= [ML2 r 2 ] ' 2

or i.e.,

[J] - [J]

·

[ML

2

r

or

]

S=l.4k-W/m 2

~~-----[i21"--,-;---.

3

r----------·· -- ---------------· I

iflnd_@IJ)~nsions_of a.,. where P~= pressure, t

So

[R]

[ML-1r 2][L3]

=- - - ~ -

[mo!] [0] or [Rl = [ML2r 20-1mol-1J So its SI unit is J/mol-K while practical unit is cal/mol-K. It is a universal constant with value 8.31 J/mol-K or 2 cal-K. 10. van der Waals constants a and b According to gas equation, for one mole of a real gas

__J

[a.t 2 ] [a.]T2

= [M0 L0T 0 ] = [MOLOTO] [a.J = r2

= [ML2r 20-11

So its SI unit is J/K and its practical value 1.38 x 10-23 J / K 9. Gas constant R According to gas equation, for perfect gas, PV=µRT

= lime.~_

Solution. Exponential and trigonometric function are dinlensionless.

2

i.e.

~

Ip= P0 e-cot2J

E =-kt 2 2 [kl= [ML r ] [0]

2x4.18 -1. 4 x 10 3_J_ c10-2 mJ2(60s) m 2s

r . .d:E,xa•:"u.;.,i;e: 12. t.-·--··--~:\ii.~-:..=.:.~.-J

= [M 0 L0 T 0 ]

i.e., J has no dimensions. Its practical unit is J/cal and has value 4.18 J/cal. 8. Boltzmann constant k According to kinetic theory of gases, energy of a gas molecule is given by

-Le·

= [PV] = [L3 ]

[bl= [V]

.

i.e.,

= RT

a ab PV+--Pb--=RT V V2

or

So its SI unit will be J/kg while practical unit caVg. 6. Coefficient of thermal conductivity K According to law of thermal conductivity, Heat transferred per second dQ =KA dB dt dx

Va2 ] (V - b)

f-=Exam,,.,.il,e - -- --"-""' ___ ;! l j~ ~

_____ ___ ------,.

2

J

', a . a-ct The dimension of b in the equatwn P_= _b_x_ whe-re P = pr:_es_sur_e,_~ =_g~p_lacement aml L.=.. time _ _ __

Solution. [Pl

=[

b:]-[ c::]

By principle of Homogeneity, [P] =

[b:] = [ c::]

[i]i=[ML-1T-2] [i]=[Mr2]

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(uNIT AND _DIMENSIONS Method-II

r=-

.

.

----- -

= p•pb V = kP"pb:

V

---,

The position of a particle at time t, is given by the equation, I

x(t) = v; (1-e.-ac ), where v 0 is~ constant and _dimensions ofv 0 • & ct are respectively.

ct>

t

[1r1] = [ML"1T-2]"[M1-3

0. 'The\

1 a=-

_. __________ ;

Solution: [v 0 ] = [x][cx] = [M 0 11 r 1 J and [ct][t] = M01°T 0 , [ct]= [M0 1°r11

=>

kJ;_:>S.QnilRl&J 17

E,E~o:~f~,t~J.ii];;>' i

'When a solid sphere moves through a liquiq, the liquid opposes the motion with a force E The magnitude of F depends 10n the coefficient ofviscosityT] of the liquid, the radius r ofthe lsphere and the speed v of the sphere. Assuming that E is !proportional to different powers of these quantities, guess a iformulafor F using the method of dimensions. Solution: Suppose the formula is F = kri"rbv'

~-;;-ng,; mod~lus oi s;eel is 19 XJ010N/m ,clyne/cm

Equating the exponents of M, 1 and T from both sides,

•

Solving these, a = l, b = l and c = 1 Thus, the formula for Fis F = /crirv.

Thus,

[Y] =

Express i~-ij

Force (distance) 2

[Fl = [M1F2] = [ML1r2]

or,

JE

2

:r\~:r

1 2 : ; ~ =(\:g)(1

so,

or,

b~E?

Thus, 1 poiseuille = 10 poise

[P] = [ML-1r2]

... (1)

[p] ': [M1-3]

... (2)

]

Fmd how many poise (CGS _unit of viscosity) is equal to 1 lp_oiseuill~(SI unit of.viscosity)]__ · ___________ Solution: T\ = [M1 1-1 r 1 1 1 CGS units= g cm-1s-1 1 SI units_= kg m-1s-1 = 1000 g·(loo ~mi-1 s-1 . = 10g cm-1 s-1

Solution: Method-I

=>

•

Here dyne is the CGS unit of force.

This suggest that it has dimensions of

· -a-c = -2

[1 r1] = [P112p-112]

2

Solution: The unit of Young's modulus is N/m 2.

a=l -a+b+c = l

Cc!LP-112r>.-1/2

2

[12] 2 N/m is in SI units, , So, 1 N/m 2 = (1 kg) (1 m)-1 (ls)-2 and 1 dyne/cm 2 = (1 g) (1 cm)-1 (ls)-2

= [M"1-•+b+cT-a-c]

tc)_P-11~P.112

f~

[12]

Then, [M1r2] = [M1-1r 1]"[1]b[½]'

/If P is-~he pressure o_if a gas andp. is its d.!dimension ofvelocity. (a) pl/2p-112 (b) pl/2pl/2

1

b =-2.' 2 . [v] = [P1'.2P-112]

[v] = [P112p-1121

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\ji

s

'j

i.

V

-~~----h-:-·~ ~

~")

1""'

In two systems of tmits,"the relation betw~en velocity, . by acceleration ,': ,:, iµid ''. ,force : is ' given 2 ' 'v·e '· · , , F • , . v 2 =-1-·-,a:i=·a1E-r,F2 =_!.; where e and 't are 't_

I,.~

.

.E't

, .•,'

1

constants then.find in this new·,system: m2 . . . . ', (b.) L2, ( a) . . . . . ', '

ml

' '

'

.

'

'

, [Ans. (a). ·}

E -r

'

'. . ~I

~ ; (9)

;1 · 3

'.

't . •· . .

.

a

.

. nRT - - . · · 2. In the formula P =- - e ·Rn,;, find the dimensions of . ,: V-b . ' . · a and b where p ' = pressure, n = no: of moles, T "" temperature, -V = volume'. and R = universal gas. constant. . , . · · [Ans. (a),.;. [ML5 T-2 inol~1 ], (b) = [L3 Il 3. The loss of.pressure when a fluid_ flows through a pipe is'!liven by P-;= -kp"l Vbd'µ .where d and l are diameter arid"length\,ftlie pipe respectively, p, d and µ are the ma~s, ·density arid coefficient of viscosity of the fluid, V is the mean ·velocity of flow through the pipe and k is a huinerka! ·constant. Find the ~alues .of a, .b and c.

[Ans a·=ib'"h'= . M°gh·, ·~.;. i· . ' c:'='-21.

•

•

i__ ,

,.,

nxYT - - ·

.

.

. .

4. P =--. e .nxi;; where n is number of moles, P is Vo ·, . . . . ,, . . pr~ssui-e, Tis te~perature, Vo i; vol~~e, ,M is mass, g ~epresent~ ~ccel~ration iliie to g~avity and h is height. ····Finddiinehsioii'ofxandvalueofy., _·. · · · ,, [Ans. [Mi.2 r 2 K-1 mo1-rJ; y = 1l · 5. The uni~· CU) ~fvel~city,:;ic~eleratio~ ruid force in two systems are related as under : . 2

(a) U~

Cc)

'

= ~Uv p ', .

u~ = [_!_];F . ap.

. (b)

.

u: = ,.

'cct')

.

.

I-'·-;,'

.

' i:_. •., , '' '~

·.' ; ' ..

u~ = (p\ )uPJ

6. Specific heat of hydrogen at constant pressure, · C P = 29 joule _kelvin-1 mo1-1 •

(a) Find dimensions of C P. (b) Unit of length is changed 'to 50 cm, unit•of ti.me is changed to 2 sec, unit of temperature is changed to 2K. keeping units of mass· and amount of substance same. Find the value of specific heat of hydrogen in, . new system of units: [Ans. (a) [ML2 T-2 K-1 mol- 1],(b)928]

m_oniy~o;;';"Altefu';;tiC'ilifoI@~ 1. E, m, L, G denote energy, mass, angular momentum & ·gravitation constant _respectively. The dimensions of EL2 ""'"s'2 will be that of :

mG

(a) angle (b) l~ngth (c) mass (d)' time , 2. The dimensional formula for which of the following pair is not the same ? . '' (a) impulse and momentum ' (b) torque and work (c) stress and pressure (d) momentum and angular momentum 3. If the speed of light (c), acceleration due to gravity•(g) and pressure (p) are taken as fundamental units, the dimensions of gravitational constant (G) are : · (a) [c2g3p2] (b) [cog2p-'] (c) [c2g2p-2]

(d) [cog p-3]

never be a meaningful quantity? (a) PQ - R (b) PQIR

,

'All th~ pnme, by length r from origin and. angle e is generally measured from ,p6sitive x-axis (Fig. i.3). and x = rcos0'.'

y

] ]·

r=~X2+y2

and.

A'

(c)

(b)

(a)

Fig.1.5

in kinematics. 3. In nature, no fixed bodies exi.st and consequently there can be no fixed reference systems. A fixed reference system i.s usually assumed to be a system of co-ordinate axes attached to · the earth.

4"'

A

A

!

The displacement from a point A to a point B is a vector quantity. Its magnitude is the straight line distance from A to B; its direction is that of an arrow that points from A to B. Points B and C are equidistant from point A but the two displacements are different because they have different directions. Two displacements (vectors) are equal if they have same length and same direction (Fig. 1.5).

VECTOR NOTATION A vector quantity is represented by a bold later with an arrow above it or a bar above it.

e

~::::==::=;~x x=·rcose

...,

A or A :

...,

The magnitude of vector is represented as IA I and it is

...,

Flg.1.3

= rsin0 tan0 = ~ y

X

4. Trajectory of a particle y denotes the actual path followed· by it. Path length s(t) is defined as the distance travelled along a trajectory in time t. It is measured '--------•x, from the_ starting point Flg.1.4 of the motion at t = 0. Path length is the total laistance covered; it can only increase with time as a particle moves, hences is·a]ways a positive quantity (Fig. 1.4). 5. We always express results of our measurement in terms of a number, e.g., room temperature is _25°C. The value 25 is called the magnitude of the quantity. Some quantities do not have direction associated with them; such quantities are called scalar quantities. Motion is a quantity that involves direction as well as magnitude. We say a car is movin~ with velocity 10 km/h eastward. Such quantities which have direction as well as

referred as modulus of A. Geometrically a vector is shown _by an arrow dr& wn to an appropriate scale. The direction of arrow represents direction of vector and length of arrow represents magnitude of vector. Displacement Vector Displacement vector represents change in position of a moving object. A car starts from Kota and travels north-east to reach Delhi. Its displacement vector will be represented by an arrow joining starting point Kota to terminal point Delhi. --- ---- - l

. _ _ ___ Fig._ 1;6 _______ ..•

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, -DESCillPTION OF MOTION :__________ - ·-----

21,

_, _, _, 7

Displacement vector is straight line segment from initial point Kota to terminal point Delhi.

ii'

2:::·

Significance of Resultant Displacement A team of hikers begins from A travel to B, C and D the successiVe destinations. Net displacement, resultant

--;

displacement, total displacement mean same thing. Resultant displacement vector is straight line segment from initial point A to final point D.

A

--;

--;

Draw-A first and then B --;

~---~A

o'.i··:..~--~ . r.__, .,

.

A

--;

,._1 --;

--;

Draw B first and thenA

Fig.1.9

·A" ...

v,

a=B/2

aLJ~

~

-> v,

->

p

vectors to be added are placed tai~ to tip.

Fig.1.11 Fig.1.15

Significance of Vector Addition Illustration 1. Consider a Fa 100N block that is pulled by two boys simultaneously. Each boy exerts force of 100 N. We can easily ~Fa100N guess that block will move at 45° Fig. 1.12 angle. From rule of vector addition we can see that resultant force is

t ··"

FR

··~·.·········...-

V2

(2)

->

a

v,

:

~

->

v,

Correct representatio'n Fig.1.16

=..JlQ0 2 + 1002 = 10Q-J2 N

Vectors to be added are placed tail to tail such that they represent consecutive sides of a parallelogram. Diagonal of parallelogram represents resultant.

Net force is ·vector sum of all the forces acting on the object.

1/

....v,

Illustration 2. A boat moves with velocity Y8 in still water [Fig. 1.13 (a)]. Velocity measured by observer on

+

v;/"'(~·-··/; ~ ....

=

v,

ground will be y8 . But if water flows at a certain rate it will make the boat moye f~st':r ~s!

Draw B first then draw A Resultant obtained is same in both cases.

'

, --+

=V airplane~ V wind

i

Fig, 1.19

l

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DESCRIPTION OF MOTION

'

....

~-:

-

' . ""ii]

----·--·

·,,_

1

p (X, y).'

v,!L'• Qi

v, -->

A Fig. 1.20

I".. •

v,

'i

t t0.

• t ., ,_v, • \lL+·'v; ·c;i

-

A vector sum is independent of the order in which the vectors are added, Le., it obeys the associative law (Fig. 1.21).

it., ? .,.

,..'il"

.

,.// C,

I 1 '1

)( )'~

r"

•

~ rU -4i+3j V=-,,==== .J42 +32 4, 3 • =--1+-j 5 5 Thus velocity vector of. boat is

= Fz

- -i-j

,

1~~

.

) '' -~-t=-c_:~~-~f~-~1n_~j !

,*;

0.

'

" .

ir ·

1.

W.

'.

·sw

45°

1 ' , ~ ·

ti='sin81+cos8j 8 . .

E·

C

i.

Fig."1.38

s Fig. 1:35

a= - cos0i + sine]

b = sin8i+cosej www.puucho.com

!

!

'!

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-~ ·---·- ·--------·---27 .

DESCRIPTION OF MOTION

....

.1:r -

Concept: Vector A is indc;·endent of choice of cuordinatc axes, although components ofvecrordepend on the choice ofco· ordinate axes (Fig. 1.39).

l:2 _ 1!1

- final position vector - initial position vector At time t 1 , the particle is at point P1 , and 'its position

rl = X1 i + )' 1J. Similarly at time . . = x i + y j and displacement of

vector can be expressed as

....

t 2 position vector is r2

....

particle Li r

= (x 2 -

.

2

2

x 1 ) i + lv 2

-

.

y 1 )j.

Average Velocity Speed is a scalar quantity which describes rate of motion, but velocity is a vector that gives the direction of motion as well as rate. A body's average velocity is defined as its displacement divided by time M. Velocity vector is parallel to the

Fig. 1.39

....

displacement S. y Path of particle

t + t>t

'

Magnitude of a vector is independent of the choice of,' coordinate axes, hence it is a scalar; whereas vector 1component depends on the arbitrary choice of a coordinate! axis; it cannot be a scalar; we call it simply a vector: component. 1. If a vector is zero then all its components are individually zero

A,i+Ayj +A,k. = Q then

A X =A y =A Z =0

If two vectors are equal, then their components along

2.

---.--------. --. -.-,: ,Ix

y

1~

L . . . . .J"' v,

-1----------'--X (x + t>x) X

Fig.1.41

Velocity components vxor vyequal to the change in corresponding coordinate Lix or Liy divided by Lit Lix Liy V =- V =x At' y At

Position Vector of a Point with Given Coordinates Consider a point' P' with Cartesian coordinates (x,y, z) relative to the origin 'O' then the position vector of' P' is given by

the rectangular axes are also equal i.e., if

A,i + Ayj + Azk. "B,i + Byj + B,k. (Ax -Bx )i+ (Ay -By)j + (A. +B. )k = 0 Ax -Bx= 0, Ay -By= 0, Az -B,

:t>y Displace~ent

. .

_,

=0

OP = r = xi+ yj + zk • l

(x, y, z)

y

p

POSITION VECTOR The position vector of a particle is a vector drawn from origin to the position of particle (Fig. 1.40). For a particle at the point P(x, y) position vector is

....

.

Y

.

displacement

vector · Li 1 is the difference in position vector.

r

"'--------------x Flg.1.42 _,

If vector OP makes angle o:, pand y respectively with x, y

r=xi+yj

The

_,

Path of a particle

0

Fig. 1.40

and z coordinate axes then components of vector are

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rx ry rz

= r cos ex = rcosp = rcosy

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[38________ ---- -- _ ·.:.,---'------ . - - - - - - ' - - - - ' - - - - - - - - · : - '-·_,_MECH~IC~:U Angles a, p and y are referred _as direction cosines. thus cos 2 a+ cos 2 P+ cos 2 y = 1

r

The direction cosines of are, 'X y cosa =-,cos!}= and r r

Z

cosy=-, r

where,

r =-1-r!"" ~ x 2 + y~ + z 2 position vector of A w.r.t. B defined -->

as, rA/B

-->

-->

= rA- rB

Position

vector

of

point

A(x,y,z)

with

w.r.t.

B(x 2 ,y 2 ,z 2 ) is given by,

Where ·cosa, cos!}, cosy are known as direction cosines, these are the cosines of the angles chat the vector makes with the ·x, y and z-axes respectively. How to Obtain a Unit Vector in a Given Direction ? If we are given two points in space, we want to define a mrit vector along a· line which begins at the point 'A'(x1 ,y 1 ,z 1 ) and passes through the point'B'(i2,y 2,z 2). First we find position vector of point ':B' w.r.t. point' A'. Position vector of A is given by

, 1A/B = (X1 - X2)i(y1 + Y2)J+ (z1 -z2)lc

. Rectangular Resolution Dimensions · .

-->

of a Vector in

Three

Where rx, ry and rz are the magnitudes of components · !)long x, y and z-axes respectively. By the geometry of Fig. 1.43, r = l,2 + ,2 + 1'.2

,._

rB = X2i + Y 2J + Z 2lc ,--+

--+

--+

rB/A =

(X2

-X1)i+(y2 -y1)J+Cz2-z1)lc

rB/A = rB-rA

Now unit vector in the direction of tl:iis position vector is given by,

·.1

-,=======.====~~==~= i1)

• rBJA. (X2·-X1)i+(y2-Y1)J+Cz2-Z1)lc rB/A = - - = 2 2 2

ri!B/A I

z

y

X

•

Similarly, position vector .ofB "is

-->

Suppose the vector r is to be resolved into three inutually perpendicular component vectors along the directions of x-axis, y-axis and z-axis. In accordance with polygon law of addition of vectors, Le., r = rxi+_ryj +rzlc

.Y

•

rA =x1i+yd+Z1K'

~(X2 -

X1)

+ (y2 - ,Y1) + Cz2 .-

Sinlilarly unit vector in the opposite direction of this position vector is given by, -->

rB/A

--+

--+

r A/B = ----::;-- = - rB/ A lrB/A I

r

Note that vector A/B is opposite to vector 1B/A. Shortest Distance Between Two Points If the rectangular Cartesian coordinates of two poims' A'

' X

Flg.1.43 ·

.

C •• '

l Jand 1c- respectively then

rx = r cosa, ·ry = r cos!}, rz = r cosy So chat

r

r

cosa = ..l = x .r ~r2 + r2 + r.-2 X y z

.

r

.

· sinrilarly cosp = Y · J 2 2 2 ' -vrx +ry +rz and-

1'. COS"(=

.

---+

--+

__,

-->

AB= rB/A

if a, p and y are the angles which r makes the direction

of

and' B' with position vector rA/0 and rB/O relative to the origin 'O' be (x1 ,y 1 ,z1 ) and (x2 ,y 2 ,z 2) respectively, then --+

--+

= rB;o- rA/o

Where and Therefore AB= rB/A

= (X2 -

i + (y 2 -

Zt) le Therefore, the shortest distance between the points (x1 ,y1 ,z1 ) and (x 2 ,y 2 ,z 2 ) is IABI = ~r(_X_2---X-1-)2~+-(y_2___Y_1_)2~+-(z_2___Z1_)_2

z

+ r2 + r.2 "Jfr2 X y Z

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X1)

y 1) j + (z 2

-

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i DESCRIPTION OF MOTION -----!

.

29 or Also

-- - --- ---- - -- --•

j

'A bird ,moves with velociry 20 m/ s in a direction making an; :angle of 60° with the eastern line and 60° with verticalj lupward~present the velocity vector in rectan!(Ular form. __ _j

Solution: Let eastern line be taken as x-axis, northern as y-axis and vertical upward as z-axis. Let the velocity v makes angle ex, pand y with x, y and z-axes respectively, then ex= 60°, y 60°. We have cos 2 a+ cos 2 ~ + cos 2 y = 1. or cos 2 60°+ cos 2 ~ + cos 2 60° =1

=

or

cos~= ->

1

.fi. A

A

,.

· v = vcosexi+vcos~j +vcosyK

=l0i+1o.fi.J+10k

LEx:a!ltn'""'te -127......._ ~ --·---:S::I:!::. ---------- ----- ---·· -- -----

or given

--..:::i~

,--

;Two vectors, both equal in magnitude, hav.e their resultant!! !equal in magnitude of the either vector. Find the angle './!etw_g_en the vectors. ·--·-- _ _ __._!

Solution: Let 0 is the angle between the vectors A 2 =A 2 +A 2 + 2AAcos0 1 which gives cos0 =- -

r-

(Q +P)(Q-P) =144 P+Q =18

18(Q-P)=144 or Q-P= 8 Now from equation (ii) and (iii), we get P=5N Q=l3N ""'

],> · ---1

!The sum of the. magnitudes of two forces acting at a point isl i 18N and the magnitude of their resultant is 12N. If thel :resultant makes an angle of 90° with the force of smallerj bzm1itucle, what are the mwmit!!..d§..9f.l&.lWO ft,mc.,cc,,,escc?_ __,

Solution: Method-I: Let P < Q and 0 is the angle between them. tan 900 = Qsin0 P+Qcose

----------,

:;;;;-;:;r-;:,-_- -

-

,____J

,around and walks 5.0 m back towards the classroom. He stops 15.0 mfrom the door. Tota1time of motion is 25.0 sec. What is his average speed and average velociry?

-~- 5~-:~: ::?a:-2~:-~

'---- -----'

--- Fl~--1_E_.4____

I

l._~_~.:,., >· I

_,_I

_______ j I

Solution: According to definition of average velocity, /!,x

I

Vav

i't \\···-......Q

... (iii)

!origin, walks 20 m down the corridor, then stops, turns

!~

-.s- ----------------··

Q

... (i) ... (ii)

--·- ·---------·-···-----· --------·--·-1 IA student starts from his physics classroom considered to bel

0 =120°

•••••••••••••••••••••

->

r e-.·f:: ,=2$1!eJI$\~ g"'-4~i>.,_ i

2

1 LExa~""-'e = ~.:..,:.~-1:'!:~M~ 3

->

resµltant of P and R is equal to Q. p2.+R2 =Q2 or Q2-P2 =R2 =122 =144

=20[.! i +-2... j + 2 .fi. 2

or

P +Q =18 given or Q=(l8-P) From equation(i), Qcos0 =-P Substituting Q =(18 - P) and Q cos0 =P in equation (ii), we get p 2 + (18-P) 2 + 2P(-P) =144 P 2 + (324+ P 2 - 36P).:. 2P 2 =144 or P=5N Which gives Q= 18-P = 13N and Method-II: It is clear from· the figure that the ->

.!k]

... (i) ... (ii) ... (iii)

P+Qcos0=0 (·:tan90°=oo) 2 2 R =P +Q 2 +2PQcos0

=-=

Xf - X,

t.t t1 -t, . - 15 ·0 - O.O 0.600 m/s 25.0-0.0 total distance travelled Average spee d = ---.- - . - - - tune mterval = 20.0+5.0 l.00 m/s 25.0

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r-,30

-

------------

---- - -- - --~---

-+

-+

..., """7-+

-+-+-+

"""?-+

,subtractions: (a) A+B, (b) A+B+-

fCJ

tci, +

T..;

J

1\vo unit vectors i and are directed along x-axis and y-axis' respectively. What is the magnitude and direction of the (d)

vectors

Fig.1E.5

Remark:-------------------

i_+ J and i-) ?

d

y

............

h

In the figure shown, -C =A+ B

j

:

h

i

PR (a)

...,

The x-, y- and z-components of vector 2A+ 7 B+ 4C are respectively x-component = 2Ax + 7Bx + 4C x y-component = 2Ay + 7By + 4C Y z-component = 2A. + 7B. + 4C.

X

a ~-'-:--,-+--x

h

-j

f =(F)+Fyj +F,k)-(Jxf+ fyj + f,k) =(Fx

=(25.0) j

The displacement is given by

:-) , A

You can see ):hat the components of vector sum (F+ f) are the scalar sums of the respective components of the individual vectors. Similarly,

are --+

:,

->

xJ

'--------..... -~-

Solution:

/"~,(~

/~

~--------~------- ' -Azk :

2. Vectors in cartesian form may be added or subtracted provided they are of same type, representing same kind of physical quantities such as displacement add to displacements, forces add to forces. For example

= (Fx 1

.I

=aA)+aAyj +aA,k

F+

P

j

AxtAy],A,k.

The multiplication of a

--------·---------------- ----------~

ri

---- YA --- - -- --- ---- --, I

->

I 1I+ I j I cos C-90°) a= -45°

or

VECTOR MULTIPLICATION 1. The components of a vector are 1 Components scalar quantities. The multiplication of l-) - ~ - " the unit vectors, by the cartesian ,A=:A,_:i +:A,:j +:~:kl 1 components leads to a vector sum of 1 \ __ . I three mutually perpendicular vectors. ,_ _'29· 1.44 (a)___J In the figure shown the

i

i-J=i+(-j)

2

10.0

- fx)i+ (Fy - fy)j +(F, - f,)k ->

->

Hence the components of the vectors F- f are the scalar differences of the respective scalar component of the individual vectors. 3. In vector addition and subtraction the vectors involved. must be of the same type, While in vector multiplication there are two distinct ways; each of these ways has its own set of rules notations and applications in physics. ·

=(-2.50! + 2.50j) \11/s www.puucho.com

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MECHANICS:!'.] .

THE SCALAR PRODUCT OF TWO VECTORS Multiply a vector by a scalar produces another vector. The scalar product is different from this multiplication. The scalar product is a way to multiply two vectors to yield a ,..

-+

-+

scalar result. product of any two A and B vectors is ..., Scalar ...,

(iv) When the angle 0 between vectors is acute (0 < 90°' cos0 > 0) the scalar product is positive; if the angle is obtuse (0 > 90°, cos0 < O) the scalar product is negative. The scalar product of two vectors that are perpendicular .(orthogonal) is zero.

written as A · B and defin~d as

..., ...,

A·B=ABcos0

...,

...,

where A and B are magnitudes of the vectors A and B, and 0 is the angle between them when they are drawn tail to t.ail. (i) The angle between two vectors always is taken to be the smaller angle between the vectors when they are ·drawn from a common point. In Fig. 1.45 shown

...,

90s 8 > O;·so scalar prOdvqti~," Po~itiVe , })'\,,,

J.,f

L

...,

angle between vector A and B is 0. With this convention 0 is always less than or equal to 180°. _

f'lg, 1.45

';"s 90°

(ii) In order to geometrically interpret the scalar

...,

~

Az~~

O; So scalar product' is

...,

product we draw A and.B drawn with their tails together.We .

.

-+

-+

drop a perpendicular from the dp of B to line containing A.

...,

The quantity B cos0 is called projection of B or component of

...,

...,

...,

...,

perpendicular to A then the shadow of vector B on the line

...,

...,

So we have

component of B on line of A.

j-i= 0 k·i=O

,

...,

-.,.

perpendicular,"to'A , , ' - •

The li~e

-,

cont~~~1ngA. , .·.

f ....

· ' af'·-'... /2'. -Th~projec~n ~

~ ofBalongAis · 90~}

8

·v, ,"

A·A = (Axf+Ayj+A:k)·(~xf+A_,j+A,k) = AxAx (i-i) +AxAy(l · 1) +AxA, (i· k) +AyAx cj · i) + AyAy cj · 1) + AyA, cj · k)

: . : The hne

~ : containing A

- -i-A,Ax (k -iJ + A,Ay (k- j) + A,A, (k · k)

,_

or

A2

We can also take projection the other way around.

..., ...,

-+ -+

-+ -+

(iii) The scalar product is commutative. The scalar product is also distributive, Le., -+-+

---t

A, (B1 + B 2

-+-+ -+-+ r= A· B1 + A: B2

= A2 +A2y +A2 X

Z

A= '\J1A X2 +Ay2 +AZ2

A· B = A (B cos0) = (A cos0)B A·B=B-A

_,

taking the scalar product of A with itself

Flg.1.46

---~.~--

i-J=0. i-k=O J-1=1 J-k=O k-1=0 k•k=l

(vi) We can find magnitude of any vector A by _...,

Acr:Je ~-;;;, ,t

__J

i- j = (1) (1) cos 90° = O._

also

i-i= 1

Light

I

•

(v) The scalar product of a cartesian unit vector with itself is unity. For example i- i = (1) (l)cosO = 1

containing A has length equal to the projection of B or

...,

, - ,

Flg.1.47 ,.

B on i:he line containing A. Imagine light shining

'.

cos 0,< O;,SO scalar" ,. eroduct is negative

(vii) When two vectors are expressed in cartesian form the scalar product becomes -+ -)

A

• ',.

A

A

A-B = (Axi+Ayj +A,k):(Bxi+Byj +B,k)

= AxBx + AyBy + A,B, (viii) Angle between two vectors can be calculated with the help of dot product:

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r

DESCRIPTION OF r~OTION - ... ---+---+

k1!,~~~)~l~:19f~

A·B cos0=-AB ---+---+

Where

+ AyBy + A,B,

A· B = AxBx

~!:!~h~i;jr_onent of

A= ~A; +A; +A;, B = ~BX2 +By2 +B z2

and

---+

(ix) Work done :

Solution: Let b

---+---+

-:i': 2i + 3] along·~;~; ~irectio~:j •

•

= (i + j) ---+

W = F- s =Fscos0

---+

The component of a along b

(x) Angle between the vectors: ---+---+

We have,A- B = AB cos0

acos0b=[-;~b},

---+---+

A-B

cose = - -

(2i + 3J) · (i + Jl (i + Jl )1 2 + 12 . )1 2 + 12

AB

= -'-~====,,....=-'· ,=;;===;;=

A 1 B1 + A 2B2 + A3B3

=,=;;===§===§c-'i=;;====§===;;= ~Af +A~ +A~~Bf +B~ +B~

2x1+3xl (i+Jl =--~./2. -/2 5 • • = -(i+ j)

(xi) Component or projection of one vector . along other vector : i----· --------+··-------·-

!

B

/

2

L1;:¥~,~:t~~ I--+_

•

•

---+

•

•

:If A= 3i + 4j and B = 7i +24j, find a vector having the ~

i

~me

'L

Solution: The required vector is= BA

;:

B=)7 2 +24 2 =25

Fig.1.48

·-------· -

---+

--------+

---+

(a) component of vector A along vector B

and

A=A= A

A cos0B = AB case B

•

•

3i+4j ) 32 + 42

= .!.(3i+ 4J)

B

5

-[A·BJ.

-

~

"

mamftude as B and paralle!Jo A - - - - - - ~

• 1 • • BA= 25 x-(3i+ 4j)

-- B

5

B

....

= 15i+20J

---+

(b) Component of vector B along vector A

s

r----·-

IO LiJ_cos e

C

..

.-----

-;;·------·------

.

!Under a force (10i - 3 j + 6 k) newton a. body of mass 5 kg! /moves from position (6i +5]-3k) m to position! • • I 1 j(IO i - 2j + 7 k)m Dedu_c_e the work don!__ ___

~

....A

Flg.1.49_ _ __

Solution: As displacement

i.e.,

---+

---+

---+

s = r 2 - r1

s = (lOi- 2j +7k)- (6i + -j- 3k) = (4i-7J + 10k) m

Bcos0A = ABcose A A

=[A~iJA

So i.e.,

w= i.1 = c101-3j + 6kl-c41-7J + 101ci ---+

W = (40+21+ 60)

= 121joule

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g· _es "h-.. -,.;l'·~ --·-1.~ - -l··~ ~· f'F~am,,i!.ia 12fo-· .

~ parti~ie tli~vd.in th~ ·x-~~la~~'µha~i ~he action of &force ~ such th~qlie value of its lin~~r,n;t_~1~iitum p at any time tis ~- -~ . ·~-" . -~ p';,:",_ '2.cost and.Py = 2 ~int. Wha£is the angle 0 between F i··_

~

,!/'\;·,_.','.

,,.,·

.

.

'

7

l~~i; at:~

-._,,~·,::,:·_",·i_·

",_

.

~~ t?

given

.

,,

..

'

.

'/

··

..., ...,

~.

'

.

,

..., ...,

...,

Now as,.

.

..., --+ --+

..

• ( , 'I'

'

•

..

'

~

.. . . •

.

-

•

A

rY

A

=.!.= xi+_yj r

r

i, .= i~os0+ }sine.

•

•

-+

a=-Psin0 . cos0 Since 10 is a vector of unit magnitude : a2+P2=1 Therefore, p = + cos0 and a= -siri0· Thus, le = -lsin8 + j cos0 We have taken p = + cos0 and not the other solution, p = - cos0 because we define a system i,, i0 vectors in the directions of increasing r and 0. Or,

: ~~::~[t·;]c:::~-~2[co~t)](:s:::~1= 0 0 Fp2x2 . --+ .

"~~:,::;;;r:-:-1,--,,..,

~€!fjl~J\~J~

13 . ~

b~~:tijlfl'fl~~ 15 ~ [v1-·. two , ;~~-;e;.~ .vectors A' :;:;~-:~ -t

~olution: Here v =

31 + 2j + 3k.

Let 0 ·be the angle between the line and the velocity-of the particle. Then 3-2+3 4 , .. . cos0 = -T22 -./3 ./66. ~

-+

-t·

.-:.,, '., ="22-=-.

~ut

'.,.

., ' '

"i-}+k·

-t-t

. '

,

-t-t

(A+B) · (A+B)

-t

-t

-+

-+ -t '-+ -t

-t-t

-t-t

=(A-B)·(A-B)

-t -t

-t

-+

-t -t

-t ~

-t -t

orA-A+A·B+B·A+B·B=A·A-B-A-A-B +B·B ... (1) -t-t-t-)-t-t

2

-t-t•.

Substituting _these values in equation no. (1) .

.

,, /i

Vect~r comp~nent

-t·

(A+B) =; (~-B) Taking self product of both sides

We know that A·B= B-A; A·A =A and B·B= B 2

· · '

: .-0..,;=. -~

-t

Solution: Given

-t

-If'

·.. .-. ·c. ' . '·.·. -./3 . Vector component =c./3 d is the unit·vector along the

obey

(. .

[A+B = A:-B, the angle,betw~enthem_is :. 1(g}_l2Q."...:._.,_·_(b}_90° .· . (f).-60° • '(cJ)~0~0- - ~

. Compon~nt of velocity along this line=lvlcos0 .. •. .- ' · .. · 4 4

line, :

·.··· . • .

H,

i.e., 0 .= 90° i.e., F and p are orthogonal,

lli

.•

lie of unit !71Cig7Jitude normal to _the ,vectori;a.nd)yi~_(h'tli~ · ' " ;x,yp)ane.' __ · · __

Flg.1E.14 ', ' where a and. p are coefficients to be determined. ·· ·, Using the definition of scalar product; we have · i,.i 0 = (icos0 + j siri0)(ia + JP)

Now as ·· F'.p = FxPx +FyPy

0

'T,'~-"'l'.~"1,..--::t

x = rcos0, y = tsinB }low; Let_ le= la+ jp,.' ·

IFI= ~[(-2sint) 2 + (2cost) 2 = 2

.! .

I··

Or,

F = i(-2sillt) +j (2cost)

with

" .

\

dt ' ,..., d •. . .. • F·= dt[i(2cost)'+j(2sint)]

So,

, .---·

I~ (xf + y J),maFirig an an~l-~. 0 with;th,e_jcccms. F/~1:~,teff~r ,ir · of unit rr/agqitu!fe in the qirectl9n of} vector _r and·'! 'vector

ir

IPI= ~[(2cost) 2 + (2sint) 2 ,;, 2 --+ .-'> dp _. F=-·I,

'

,--+

= i (2cost) + j (2sint) '

,

Solution: By definition

Solution: As p=ipx+JPy .

r-

(A point P)ie-s i!l rl1e x-y plan~. Its po~itiq/l ca~:be ip~iifi,,ed cy litf -·~· y ~tgoi;,dinates. orby 1;·i11,HaJ1y direc1,i~/~~ptok

=. j3 (f - 'j + ~) .

-:·:'•-i

--~l · .

-t-t

-t-t

. A 2 +2A·B+B 2 ':'A 2 -2A·B+B 2 -t -t

or

-t

4A·B=0

...,

...,

or

-+

C·

.,

4IAIIBlcos0=0 '

Because A and B·are non-zero vectors hence.-, " ' ·~;;~' www.puucho.com 1:r

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.... IA!;,, 0

....

(iii) If~ is the angle between vector (2i + 3]) and (I+ j)

and !Bl¢ 0

cos0 = 0 = cos90° 0 = 90° Angle between two. vectors is 90° hence alternative (b) is correct.

,.

.....

then by the dot product

ry· ~

A

--·---------- .. - ---"·-·:7

Jand Jare unit. vectors along x,axi;; and y-axi;; respectfve(y._l IWhat is the magnitude and direction of the vector i + j and:

iY

IfI - ]? What are ·_th_e. magnitudes of c.omponents of a vectoj • I :; = 2f + 3] along the --directio~ of i + j and i :- ]? ---- --------. --------~J: r;·----

Solution: Use the relation;

IY

....

'h

....

A

(i + i)

~-~----

la+hi=.Ja 2 +b 2 +2abcos0 bsin0 and a= tan· 1 a+bcos0 (i) Angle between i and is 90°

I(2i + 3])/ cosp I(i + ]JI =(2i + 3J) · (i + j)

.i

li+]I 2+Sj.i+3 5

j Fig. 1 E.16

--=--- = -

(a)

2

../z

y'

_ 2i-i+ 3J-i-2i-J-3j. j

-

../z

acos(90°+1i)'= -~ .

v2

THE VECTOR PRODUCT OF. TWO VECTORS (i) The vector product or cross product of two vectors yields another vector. The vector product of two ~

~

~

~

vectors A and B is written symbolically as Ax B. The magnitude of the vector product is defined to be

.... .... .... ICl=IAx Bl= ABsin0

'

' .

where 0 is the smaller of the angles between the two·

....

vectors, The direction of vector C .is defined to be

....

....

perpendicular both A and B. Keep the two vectors until the tails of the two vectors coincide. The two vectors then define a plane as shown in Fig. 1.50. The direction of the vector product is perpendicular to this plane,

7 ,_-J

J) ·(i- j)

../z

1. sin 90° l+l.cos90°

, IJI sin(-90°) tana =-.~~.---Iii +Ul cos(-900) -l-sin90° -1 tan a.' - - - - = 'l+lcos90° l+O r = -1 = tan(-45°)

-J~~

J) is

.

=.Ji 2 + 12 + O =../z units

~-

Magnitude of component of a along the direction of

(i+

=-1-=ll+O tana = 1 =}a= 45° Ciil .-. Ii- j I= J~Ii-12-+~I J-12 -+2-1,-.Il~Jl-co-s-90-0

j

../z

....

.

umts

. / (2i + 3J) Icos(90°+~) = (Zi + 3

Iii sin0 •• lil+Ulcos0

A

2i-i+ 2i-]+3J-i+ 3j. j

I (2i + 3j) Icosp

=.J1 2 +1 2 +0 Ii+ Ji= ../z units . -1

D

a-b = abcos0. (2i + 3])- (i + ]J =I (2i + 3])/1 (i+ ]JI cos~

. _,_,

Ji+ Ji=~/ i./ +1]1 +2i.l. ~os90°

a=tan

A

,_ !:!~: _1E.16Jc)_. j

........

Since these.are unit vectors therefore /i/=IJ I= 1 2

A

(i -

....

(ii) The direction of ve~tor C can be determined by

: = 90° { -J 1

!

_ _ Ftg.1E;1~(b) ___ J

a vector product right hand rule. Curl fingers of your right hand and imagine swinging them from ·the directilln of first vector in the vector product to the direction of the second vector as shown in Fig. 1.50. The extended thumb of your right then indicates the appropriate direction of the vector product. · /

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r-'·

1·. 1i~·,._;:._:·-=-~---~-~--"'--'-,,-~-~-"--'-------'-'.h

r~----..

-

)·

A

A

A

A

'

Ax B = (Axi+Ayj +A,k) x (Bxi+Byj+B,k)

'! '

I Ii

--+

--+

~

= (AyB, -A.By)l+(A.Bx -AxBz)J +(AxBy -AyBx)k In determinant form we can write it as j k -i -, -,

"l

l

: -. ,· :· ____ ____ ,

Flg.1.~0 ___ _

AxB= Ax Bx

j

Ay

!1.

By

B,

-,

(iii) The angle between any vector A and itself is 0',

hence the magnitude of the vector product with itself is zero. -, _, , IAxAI= Msin0'= 0 rhe vector product of any vector with itself is zero. (iv) The magnitude of the vector product is the· area of the parallelogram formed by two vectors, also, the magnitude of the vector product is twice the area of the tri~ngle formed by connecting the tips of the vectors. (v) If the vector product of the two vectors is zero and neither vector is of zero magnitude then the sine of angle between the two vectors must be zero i.e., the two ·vecto;s_ ~re·either parallel or antiparallel. · ,' 1

--+

. --+

If

AxB=0 -,

:...

_,.,·'

B

~

0, then

-,

--+

'. . --+

--+

--+

--+

::--::7

.....

-,

_, _,

-,

Solution: Resultant vector= a+ b + c

= 1+2] + 3k:-l+2j +k+ 31+ j

= 31 + sj +4k

.

_,

Unit vector"tl. by the property is given by = ~ _, IAI Unit vector of resultant 31 + sj + 4k

--+

•

--+

=

--+

';'

_,

·

jixJI= Clf (l) sin 90'= 1 From right·hand rule the direction of vector product is perpendicular to both i and J, i.e., parallel to k. Similarly JxJ=O

_,

-,

.

.

,._

. .

,._

+3kx 2j +ix k+ 2jxk+ 3kxk , = 0+2k-3] + 2k+ 0-61+-j + 2i+ 0 = '-4l-41+4k = --4(i+ j-k)

itself is zer-o.

Jxi=-k

s./2

-,

= '-ix i-2j x i-3kx i.+ ix 2j + 2j x 2f

--+

(viii) ix i = 0, as vect_or product of any vector with

ix'J=k

(3i + sj + 4k)

r = axb = (i+ 2j + 3K) x (-i+2j +K)

Ax(B,+ B,) = Ax B,+AXB2

'ixi=O

~(3)2 + (5)2 + (4)2

=

Vector perpendicular _to vector a and bis cross product

,--+.

--+

3i + sj + 4k

_,

AxB=-BxA i.e~, · . .. ' The order of the terms in a vector product is important. (vii) The vector product obeys distributive law as long as the order of the terms is preserved. --+

--+

ft= 3l+sj+4k J3i+ sj + 4kl

Although Ax-Band Bx A have.the same magnitude in accordance with right- hand· rule their directions are opposite . .- ,- · .: •

,,.

-,

, implies sin9 = 0 Thus 9 is either 0' or iS0'. --+

.,,.,

un_it vector_ r w_hicli is nonnal to both a and I>. What is thej4 --+ ~ [nc!ingtio[l_oi_r and c? . ·· ' , ·

Ax B = ABsin9 =·o

--+

--+

'

_, and -

and

---+·

Given a =i+2j +3k, b;= -i +.2j tk and c = 3i + j. Find the argle of t.,_;ultant :with ~:aii,s: Also find a unitv_'iicto~ in the direction of the resultant of these vectors. Also.find a

ixk=-J

--+ --+

--+

--+

r-c=lrllclcos9 -4(i + j- k) · (3i + J) = l--4Ci + J-klll (3i + JJI cos9 -4(i- 3i+ j · 3i-k- 3i+ i- j + j ·J-k· j) = ~42[1 2 + 1 2 + (-1) 2] ~(3) 2 + 12 cos9 =}

]xk=i

kxi=j kx]=-i kxk=O (ix) The vector product of two vectors expressed in cartesian form ,as '

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../3o

1

e~cos-

(~)

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[~sc111PJl~N o(r.ijT101(__ -_-___________

~>i~mi:ili Gal> ,- -

X

.

,If~= 2i -

3] + 6k and b = 6i + 3] - 2k,

:

--+

find the angle,

Xo -••••••••••••••••••

--+

,between vector a and b. Also find unit vector perpendicular to ->

->

poth a _and b.

Solution: ->->

a- b

(i)

lo

=ab case =a, b, + a2b2 + a3b3

Fig.1.51

where

=~(2) 2 + (-3) 2 + (6) 2 and

b =~(6) 2 + (3) 2 + (-2) 2

Concept: 1. Note that in this manner we obtain the coordinate x of the point on the given moments but not the distance travelled.

and b =7 7.7 case= 2.6+ (-3)(3) + 6(-2) 49cose =-9 -1 -9 e=cos 49

a=7 or or

(ii) We know

->

2. The distance travelled can be found from the coordinate x only in the case when the particle moves in one direction.

->

ax b represents a vector which is ->

->

perpendicular to both a and b.

i j -> axb= 2 -3

->

Now

6

3

k 6 -2

=-12i+ 40] + 24k

-> r--------ax bl= ~(-12) 2 + (40) 2 + (24) 2 = 4v'145

-> 1

->

->

• axb . -Unit vector n : : : _, ->

-12i+ 40] + 24k

-3i+ 10] + 6k

4v'145

"'145

n=---=~--

In the graph of x-t show the coordinate of point cannot be greater than x 0 although after time t 0 the distance s(t) travelled by the point exceeds x 0 while the coordinate x becomes less than x 0 •

Concept: If a car moves constant(y in one direction the distance traveled ls equal to the coordinate describing the motion but when the direction of motion changes to the oppo.site direction the distance travelled still in creases while the coordinate decreases.

Velocity of a Particle in Rectilinear Motion

lax bl •

The distance x can be measured by taking snapshots of the moving particle at definite moments with a fixed cemera.

RECTILINEAR MOTION OF A PARTICLE When a particle is in a rectilinear motion it moves along a straight line, its distance from a fixed point on the line increases or decreases with time. In such cases we associate a reference frame with that _straight line and consider fixed reference point as origin. In order to completely determine the law of motion of particle the coordinate x of the point with respect to origin and as function of time must be known. To plot the graph of the dependence of the coordinate x on time t we choose a certain length scale and put values of the coordinate x on the axis of ordinates and time t on axis of abscissas.

The velocity of a point is a physical quantity determining rate of change of the coordinate with time. The magnitude of the average velocity is equal to the ratio of the distance travelled by the point to the time taken. If particle is at x 1 at time t 1 and at a point x 2 at time t 2 then its average velocity is

Concepts: 1. Average velocity depends on the time interval for which it ls computed. ' average velocity ls the same 2. If in a given motion, the for any time interval the motion has the constant velocity and ls said to be uniform. 3. In case of uniform motion which starts from the origin there ls no difference between the value of the coordinate and :that of the path travelled. 4. If a particle travels unequal distances in equal time intervals its motion ls seid to be non-uniform. In a non-uniform motion the average ve_locit.)r ls no longer a

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~[3::_:s=··==·=·=:0='>·::::·:~::;1~::L$::>:'.:;====:=:'·::7!7::;·,.1::::1_;::·-~;:::===::::::::::~::;:::;:;{li(\r,~:·=~--'~~r·;t""';-~~~~-·~-~·---"--·-::=~·-:,;~:._~}> 5

constantqii~nt.ityanddependso~ti~eintervalforwhit:li tt-;;J • d . ;~ . . l . ' co.mpu..te : F,·._P___ r,_a. non-um,_,orm. mo.ti.on av_e.r.age ve oc1ty·c-·annot1· describe. the:V9riations of motion of.the body. Such '.a IJ!(?tion can be described adequately by' instantaneous velocity. . ..

v(t) = Lim v· M--+ 0

av.

-/ :~~·.'

'ME($~1,CS-:I'

= Lim t,x = dx(t) . dt

M--+ 0 --6.t

i.e., the instantaneous velocity is the derivative w.r.t. time of the position function. +In the graph of xversus t the slope at each point (at each instant of time) is equal to the value of instantaneous velocity v at that point. Note that slope at t is zero, i.e., instantaneous velocity is .zero.

~--~-,-- -~= .... - ... - - ~ - - - - ··---~ 1. Average velocity: Fig. 1.52 (a) shows the positions of a car at times t 1 and t 2 •

Average velocity is defined as the ratio of displacement t,x to time interval .M.

2

X

.j

(a) F>O~itlQn_ yersu~_ume,:_g;:apti,.

:.: ...'........ ~:.' _, ... 7i:~>~2

+Fig.1.52 (b) depicts average velocity graphically. Join initial point P1 and final point P2 by a straight line. Slope of this li;,e is t,x. Hence, the average velocity is .

t,,~

.

the slope of the straight line connecting points (t 1 , x 1 ) and (t 2 , x 2 ). 2. Instantaneous velocity: If we decrease time interval M, for very small time interval, the line P1P2 will be tangent to the curve at point t 1 • The slope of this line is defined as the instantaneous velocity at time t 1 • Insta;,taneous velocity is velocity at a single instant of.time. Mathematically, it is defined as

I ... X

r ''·'· i

lt.

J''"

• •

-~~ ·S ,£1 ·...

:,c-,.-;~-. ..

.•

,,i~ . . '.

R2

A_tri

•

'

:

+The speed of an object is the magnitude of its velocity. _, d r · .. Speed= I_V I= I dt I Since sp~ed is the magnitude of a vector, it is .a scalar quantity that is never negative. 3. Acceleration·: Acceleration of an object signifies how rapidly the object's velocity is changing, both in magnitude and direction, whether the object is speeding up · · or slowing down.

..:.'·.:jJ

V2 - V1 Average acceleration =--"-~ t 2 - t1

·..

;', .. I• '. I[,_. ·~~·~·F~ig~1.5~.:_~~

I

. a Instantaneous acce1eration

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=

,iv

M

L;~ ,iv t..,-,q M llll

-

=-dv dt

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--as;;i

+ In velocity versus time graph (Fig. 1.55), the slope of line P1P2 gives average acceleration and the slope of tangent at point P gives instantaneous acceleration. + In accelerated motion v aJ?,,d a are in the same direction.

I

I

I

ds2

l ds

L,'____F1~g._1_.s_1_ _ __,

Similarly for distance

Jds =total distance

Jas= as,+ as 2 + dss ;t-, .. +as. => total path length or distance From Fig. 1.58. Displacement vector can be expressed as

·.,!1 Fig.1.55

_,

_,

4. Equations describing motion with constant acceleration:

vector sum of d x and d y vectors

=v 0 + at 2 x(t) = x 0 + v~t + (1/2) at

ds=dx+dy

· v(t) v

2

=v~ + 2a(x -

x

=x 0 + (l/2)(v 0 +v)t

_,

... (1)

_,

_,

... (2) ... (3)

x0)

... (4)

v(t)--, velocity at time t v O --, velocity at time t =0 (initial velocity) x(t) --, position at time t x 0 --; position at time t =0 a ~ acceleration +Thus displacement in time interval tis x(t) - x 0 • +For constant acceleration, the velocity varies linearly with time hence _the average velocity is the mean value of-the initial and final velocities Vav. = (1/2)(v 0 + v) This relation is valid only for constant acceleration motion.

Total Displacement and To~I Distance Consider a particle that moves along path represented by arrows. Entire path can be divided in very small displacement segments. --+-

Path of

particle. ""'. -

as =Id-; I= -Jcdx) 2 + Cdy) 2 J-Jcdx) 2 + (dy) 2 =total distance or total path~length Graphical Representation of Motion in One Direction If maximum power of xis 1 and maximum power of y is 1 graph is straight line. ·

• y=mx+cmx X

I

~:--i I

'

---->

vector AB represents net displacement --+

'~~~·1,; :

y2 =4ax 0

x2 = -4ay

-- ----'

Flg:1.60 :

y2 =-4ax -

'... . - . ·--~------~-'---'

If maximum power of x and y is 2 graph may be circle, ellipse, two straight line ~tc.

Flg:1.56

·--+

[,

-~~~

-

--+

.

V '

i ·;· "'-""'.._.~·l.:.n.• -~ -,1 . _...___ ___ .. .,,~~-~L----------'--~Flg.:1'.59·

x2 =4ay

--+

.

:iz····,;,v.

- i x ~ x " •'; -~, ,/'' -··_ '·_'''x

L----~-'-'- - - - ~ - ~ -

--+ sR

Ex, . ',

()/.. r) ...

,,..--,._-..'::::,.:_~~- si

.'

X

,,

_, i~S1

$~,

y-=:mx-c

If maximum power of xis 2 and 'maximum power of y is 1 cir vice-versa graph is parabola.

•.

Jd s =net displacement -,..; dS3

Fig.' 1.1!8

--+

= Jds=ds1 +ds 2 +ds 3 + ... +dsri

s-t

curve If we puts on y-axis and t on x-axis for every value oft we have a value of s.

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(b) Uniform acceleration:

We have a particle moving with uniform acceleration a and initial velocity u. Its displacement s at any time t can be represented as 1 2

s =ut+-at

1. The average velocity from time t 1 to t 2 will be

v av = s2 - s, = slope of ll~e joining p 1 and' P2 ' . t2 -t,

For a particle moving along a straight line when We plot a graph of s versm t, v av. is the slope of the straight line that connects two particular points on the s(t) curve: one is the point that corresponds to s2 and t 2 , and the other is the point that corresponds to s1 and t 1 • Like displacement, v av. has both magnitude and direction (it ,is another vector quantity). Its magnitude is the magnitude of the line's.slope. A positive Vav. (and slope) tells us that; the line slants upward to the. right; a negative v av. · . (and slope), that the line slants downward to, the right. 2. Instan~ani,ous velocity: According to definition . P2 s ------,·--·-.. t,i $ ,,2 "' ' v= 11m-

t;

M--)o .1t

In ·curve, if M ~ 0 the i s, ., ... point p 2 comes very close ,t,"b , :· to point p 1 .•

t-

Notg;)------~---~::;::==~:;:;:;

ts,

If velocity is· uniform slope of curve must remain unchanged. Curve with uniform slope is straight line. If velocity is 1 ms-1 => s = vt => s = t tane = 1

r-·· -,-.. -

-., ..--- -----··· .......... - ·. ·""'" ..•. -,

a!

1kcydiststa,;t;1ngfrom a point Atravels 200 ~ due' north'to !point B, at constant speed of 5 ms"'' He. rests at B for, 30i :sec?ndf q~d ~he,:, :travelsc300_ :m 90° Note: 0 is always with +ive x-axis.

[kl..i:=2&9me;'fec-.:....~~ . . . --r::-1----

L r:·----- -. -.. . ---- --------- ·---- . - --- -A particle moves .in a straight line with constant velocity of 5 '. rr acceleration l !ms · for 2 seconds. It then moves With a constant accele~ation'

(2) For uniform straight line curve: tan 0 = acceleration For increasing velocity:

tan 0 = acceleration For decreasing velocity:

1

Flg.1.68 (a)tj

3. Total distance will be sum of areas without sign. 4. To plot straight line using equation of motion.

it

lv '!

I,

8

t-+ii

Fig. 1.68 (b)

of-2 ms·2 for 8 seconds. Draw velocity-time graph for 10: !seconds of motion and find. ' t!l) _Ejrzql 11_ijgs:_ity_ ___(/J) Qisp_lgc_e_me_nt__ _(c)_ 1htal_dfstance '

1

Solution: Area (1) = 5 x 2 = 10 ,r·

!

decreasing

l

1:......:

~ __ Fig.1.68(c)

5

I'

a

·1

I

s-.! x 2 x s x s

V

--- 2--_-·p-:--7

I10

2

i

_

·A particle is travelling in a straight line. It has a. initial/ ;velocity of 10 ms-1 • When it is subjected to an acceleration of: -2 ms-2 for 8 seconds. Find displacement and distance] :tn:1Ver~ Differentiate y

= e'inx.

Solution: Here the inner function is g(x) = sinx and the outer function is the exponential function f(x) = ex. So, by the chain rule. dy d . . d . - =-(esmx) = esmx -(sinx) = esmx cosx dx dx dx Note:·-------------------

We can use the chain rule to differentiate an exponential function with any base a > o 8 x =(elnay =e(lna)x

the chain rule gives

r:.,E-~fl-~Rf~ _: ~1 )>

!!.... (a')=!!_ (el lna)x) =e(lna)x !!__(Ina )x dx

:Differentiate y =_ (x~ -1) 100 .

Solution: Taking u = g(x) = x 3 -1 we have dy dx

and

n = 100,

= ~ (x3 -1)100 =100(x3 -1)99 ~ (x3 dx

Find f (x) ,

if f(x) =

dx

In a =ax In a

because In a is a constant. So we have the formula !!....(a')=a' Ina dx

1)

dx

= 1OO(x 3 -1) 99 3x 2 = 300x 2 (x 3 -1) 99

lJ;:~~~-~gJ~:

dx

=e(fna)x_

RECTILINEAR MOTION

G'~.L> 1

~x 2 +x+l

(MOTION ALONG A LINE) We will assume that a point representing some object is allowed to move in either direction along a coordinate line. This is called rectilinear motion. The coordinate line might be an x-axis, a y-axis or an axis that is inclined at some

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45

LD~S~RIPTION OF MOTl~N__ angle. We will denote the coordinate line as the s-axis. We will assume that units are chosen for measuring distance and time and that we begin observing the particle at time t =O. As the particle moves along the s-axis, its coordinate is some function of the elapsed time t, says= s(t). We calls (t) the position function of the particle, and we call the graph of s versus t the position versus time curve.

For example, in figure the rabbit is moving in the positive direction between times t = 0 and t = 4 and is moving in the negative direction between times t = 4 and t = 7. 2. There is a distinction between the terms speed and velocity. Speed describes how fast an object is moving without regard to direction, whereas velocity describes how fast it is moving and in what direction. Mathematically, we define the instantaneous speed of a particle to be the absolute value of its instantaneous velocity; that is, instantaneous ] [ speed at time t

Particle is on the

=[v (t) [= Idsl dt

For example, if nvo particles on the same coordinate line have velocities v = 5 m/s and v = -5 m/s, respectively, then the particles are moving in opposite directions but they both have a speed of [v [= 5 m/s.

positive side of the ori in Particles is on th egative side o

the origin

L E;:x~_t:r\p}~

Flg.1.69

Fig. 1.69 shows a position versus time curve for a particle in rectilinear motion. We can tell from the graph that the coordinate of the particle at time t = - s0 and we can tell from the sign of s when the particle is on the negative or the positive side of the origin as it moves along the coordinate line.

INSTANTANEOUS VELOCITY

L.~5_i:-->

Let s(t) = t 3 - 6t 2 be the position function of a particle moving along an s-axis, where s is in meters and t is in seconds. Find the instantaneous acceleration a (t) and show the graph of acceleration versus time. v (t)

Solution: The instantaneous velocity of the particle is = 3t 2 -12t, so the instantaneous acceleration is dv a(t) = - = 6t -12 dt a

40

The instantaneous velocity of a particle at any time can be interpreted as the slope of the position versus time curve of the particle at that time. The slope of this curve is also given by the derivative of the position function for the

6

8

particle. , Concept: 1. The sign of the velocity tells us which way the particle is moving a positive velocity means thats position of particle w.r.t. origin is increasing with time, so the particle ·is moving in the positive direction ; a negative velocity means that s is increasing with time, so the particle is moving in the positive direction; a negative velocity means that s is decreasing with time, so the particle is moving in the negative direction (Fig. 1.70).

s(t) s(t) increasing v(t) = s' (t) "° 0

s(t) s(t) decreasing v(t) = s' (t) < O (b)

(a)

Fig.1,70

-40

Acceleration versus time Fig. 1E,35

and the acceleration versus time curve is the line shown in Fig.lE.35. Note that in this example the acceleration has units of m/ s 2 , since v is in meters per second (m/s) and time is in seconds (s). Concepts: 1. A particle in rectilinear motion is speeding up when its instantaneous speed is increasing and is slowing down when its instantaneous speed is decreasing. An object that is speeding up is said to be "accelerating" and an object that is slowing down is said to be "decelerating", thus, one might expect that a particle in rectilinear motion will be speeding up when its instantaneous acceleration is positive and slowing down when it is negative. 2. This is true for a particle moving in the positive direction and it is not true for a particle moving in the

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f45 ..· ..... ··- ··-· L--------·

~

·-····-· -- ----, -·- - - ...

;negative direction-a particle with negative velocity is speeding •up when its acceleration is negative and slowing down When: its acceleration is positive. i ' ' i " This is be.cause a positive acceleration implies an: 'increasing velocity and increasing a negative velocity: idecreases its .absolute value : similarly, a negativei ;acceleration implies a decreasing velocity and decreasing a_l ;negative velocity increases its absolute value. '

3. Interpreting the sign ·of acceleration

A:

'particle in rectilinear motion is speeding up when its velocity ;and acceleration have the same sign and slowing down when · , ,they have opposite ,signs. , 4. From the velocity versus time curve and the! !acceleration versus time curve for a particle with position functions s(t)=t 2 -6t 2• I

I

.

'

• Over the time interval O < t < 2 the velocity and' !acceleration are. negative, so the panicle is speeding up. This· iis consistent with the speed versus time. curve, since the· speed! is increasing over this time interval. Over the time interval! 1 i2 4 the! !velocity and acceleration are positiVe, so the particle isl :speeding up, which again is constant with the speed versus ltime curve. . . ·' --·--·---·---·-·-----·~-

-

--

-·----·--·-·-·-·-·----

-~---·-·-·

,-

.

.

: MECIIANics:,-1 - ~---------··~-,----- ··-········--· ·1 ·,,_)

- -· .. ·-·-. .

Concepts regarding position versus time: curve; The position versus time curve contains all of the j ·significant information about t/le position and 'velocity of a 1 'particle in rectilinear motion. I : !· If s(t) > 0, the particle is on the positive side qfthei

;s-axzs.

_

_

_

1

· 2. Ifs (t) < Q the particle is on the negative side of thei s-axis · ·· ; 3. The slope of the curve at any time is equal to the 1 :instantaneous velocity at that, time. · . /· I 4. Where the curve has positive slope, the velocity is 1 :positive and the particle is moving in the positive direction. 5. Where the curve has negative slope, the velocity is negative and the particle is moving in the negative direction. 11 1 I 6. Where the slope of the curve is zeta, the velocity is ;zero, I :and the particle is momentarily .stopped. · ' : 7. Information about the acceleration of a particle in 1 :rectilinear motion can also be d~duced from the position! 'versus time curve by examining its concavity. Observe that ilie :position versus time curve will be concave up on· intervctl.s ;where s' (t) > Q and it will be concave down on intervals 1where s" (t) < U But we know from ( 4) that s' Ct) is the 1instanta11eous acceleration, so that on intervals where "the ,position versus time curve is concave up the particle has a, ,positive acceleration, and on inte_rvals 'where it is. concave' !down the particle has_ a negative. a~celeration. -- , __ · _• '

j

l

J

_._,d _ _,

Summarizes Our Observations About the Position versus Time Curve

i Position versus time curve

kl lli!

_______ , M,!

I

I

'

.

lo . ,

-- ,,---

·~· '

.

•

_·

'

'

I

I

l I

I

tI

'

"

·- ··- .

-

!

!

, ... : -_to:

* Curve has positive slope * Curve is concave down *s(t 0 )>0

* Curve has negative slope * Curve is concave down *s(t 0 ) < 0

* Curve has negative slope * Curve is concave up

'

'

6;

,_

>0

,

,'

~

*s(t)

- ti.

. .. !o __ i --~ -

.

Characteristics of the curve at t

.. ,

=! 0

Behaviour ofthe particle at time t

= t,

* Particle is on the positive side of the origin. * Particle is moving in the positive direction. * Velocity is decreasing. * Particle is slowing down.

* Particle is on the positive side ol the origin. * Particle is moving in the negative direction. * Velocity is decreasing. * Particle is speeding up * Particle is on the negative side of the origin. * Particle is moving in the negative direction. * Velocity is increasing. * Particle is slowing down.

I

.

I

lI

*s(t 0 ) > 0 * Curve has zero slope * Curve is concave down

* Particle is on the positive side of the origin. * Particle is moving stopped. * Velocity is decreasing.

-~

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DESCRIPTIONOF tAOT!ON described schematically by the curved line in Fig.I. 71 (c). At time t = 0 the particle is at s (OJ = 3 moving right with velocity v (O) = 60, but slowing down with acceleration a (OJ = -42. The particle continues moving right until time t = 2, when it stops at s (2J = 55, reverses direction, and begins to speed up with an acceleration of a (2J = -18. At time t = 7 I 2 the particle begins to slow down, but continues moving left until time t = 5, when it stops at s (SJ = 28, reverses direction again, and begins to speed up with acceleration a (SJ = 18. The particle then continues moving right their after with increasing speed.

Conceptual Examples: Suppose that the position function of a particle moving on a coordinate line is given by s(t) = 2t 3 - 2lt 2 + 60t + 3. Analyze the motion of the particle for 1 ;;,, 0 . Solution: The velocity and acceleration at time t are v (t) = s' (t) = 6t 2 - 42t + 60 = 6(t - 2)(t - 5) a(t) = v' (t) = l2t - 42 = 12(t -7/2) At each instant we can determine the direction of motion from the sign if v (t) and whether the particle is speeding up or slowing down from the signs of v aad a(t J together Fig.1.71 (a) and (b)]. The motion of the particle is

V

5 0 2 O++++++O- ________ O+++++++++ s;gn of v(t) = 6(1 - 2) (t - 5) .Pos_itive

direction

Negative

direction

Positive

60

Direction of motio_n

40

direction

20

-Analysis of the partiCles direction (a)

a

7 0 2 2 5 t 0++++++0- ________ O+++++++++ sign ofv(t) = 6(t- 2) (t- 5) Positive direction

Positive direction

Negative direction

40

20

- - - - - - - - - -O++++t++++++++++++ sing of a(l) = 12 (t- 7/2) sloWlng speeding slowing speeding dOwn up ·down up Change in speed

1

-20

Analysis of the particle's

(b)

t =5 t=0

..

r___,,...1=....1,_2~ '--

:,' -------~'t=2

03

28

55

(c)

Fig.1.71

.

~.:,

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2

4

5

6

7

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f4a -"

is!

Concept: The curved line is above Fig. 1. 71 (c) 'descriptive only. The actual part of particle is back and forth· 'on the coordinate line. · ' -

·- "--~

·--.-' -1

'The position of a particle is given by the equation. s = f(t) = t 3 - 6t 2 + 9t :Where tis measured in seconds ands in meters. (a) Find the velocit,y at time t (b) What is the velocit,y after 2s ? after 4s ? '(c) When is the particle at rest ? ! :(d) When is the particle moving forward (that is, in thej positive direction) ? [ '(e) Find the total distance traveled by the particle during the: first five seco_nds _ __ __ __ __ _ __ _ _ i

- . -·- MECHANICS-I I -·-·-- ·-~·-,,~-- --·-· -'

lf(S)- f(3) l=l 20- DI= 20m The total distance is 4 + 4 + 20 = 28 m r--

1

INTEGRATION 1 Indefinite Integrals Fundamental theorem of calculus establish connections between antiderivatives and definite integrals, if f is continuous, then (f(t) dt is an antiderivative off and J:f(x)dx can be found by evaluating F (b)-J (a), where Fis

an antiderivative off J f(x) dx is traditionally used for an antiderivative off and is called an indefinite integral. Thus. f(x) dx = F(x) means F' (x) = f(x) For example, we can write

f

2

Solution: (a) The velocity function is the derivative of the position function, that is , s = f(t) = t 3 - 6t 2 + 9t

v(t) = ds = 3t 2 -12t+9 dt (b) The velocity after 2 s means the instantaneous velocity when t =2_. that is,

dsl = 3(2) 2 -12(2) + 9 = -3 m/s dt t=2 The velocity after 4s is 2 V (4) = 3(4) -12(4) + 9 = 9 m/S (c) The particle is at rest when v (t) = 0, that is, 3t 2 -12t + 9 = 3(t 2 -4t + 3) = 3(t -l)(t - 3) = 0 v(2) =

and this is true when t = l or t = 3. Thus, the particle is at rest after 1 s and after 3 s. (d) The particle moves in the positive direction when v (t) > 0, that is, 3t 2 - 12t + 9 = 3(t -1) (t - 3) > 0

J x dx = x; + C because

x; + C) = x 2

So we can treat an indefinite integral as representing an entire family of functions (one antiderivative for each value of the constant C ) . A definite integral J:f(x)dx is a number, whereas an indefinite integral Jf(x) dx is a function (or family of functions). Any integral formula can be verified by differentiating the function on the right side and obtaining the integrand. For instance d 2 J sec xdx = tanx+C because dx (tanx+C) = sec 2 x

!

[F(t)] = f(t)

f f(t)dt = F (t)_ + C

and

are equivalent statements, TABLE

[)eriyative ·-,F~nnula

This inequality is true when both factors are positive

Equivalent Integration formula

~[x 3 ] =3x 2

(t > 3) or when both factors are negative (t < 1). Thus, the

dx

particle moves in the positive direction in the time intervals t < l and t > 3. It moves backward (in the negative direction) when 1 'y~-=-~i(',.r'~"tF:::::::;.~;:{; -·-

-

-

-~

;·.

••

and [O, it]• •• ' •-'"

.

fc:osec 2 xdx = -cotx+C

2

Jsec xdx=·tanx+C

\f secx tan xdx = secx + C f cosec x cot xdx .= -cosecx + C l l · 1 .- " · I l dx . x+.C ,;I l J·-.-dx'= tan- x+C · '·.r:---z ·1

=S!Il

x +1 ., .·' •. I 2

I

'

.•

. • •

-

..

•.

•

"•

--~-

.

-1

.

-v.1-x· . .

__

•

I(a) Find th~ a,rea undfr the CU[".e y = cos ;[o, -rr/2]

·f co.s:xdx = sin.x+C

'

fsi~xdx'.=,-cosx +C

'I

The most general antiderivative on a given interval is obtained by adding a constant to a particular antiderivative. Thus we write

a

,-

X

f cixdx=_':__+c Ina

+C

I

2

f (x-l)dx= 0

JeX.dx = ex

I

signed area is -A 2 = _ _!_Thus, 2

'

X

>

~

··--~·.

.----

over th~ int~rval) ·

" - · · - - · .••

. ••

. ·I

f_!_dx=-.!+c 2 x

'

.

,

y

- - _-~:.

l.c:E•xa:1m1r:1fl e 1· -- - -- :_3. , . '

41__ j I. ~ -~

·J !b~·..='r--::;:.;;J-i,:r~--£:~;;\

J

iEvaluate Cic 3

' I

X

-:,--~--. ' .;---7 .. _

-

6x}dx

I_. __ -- -o_· .. , ...

Solution: We have I

J3(x

-,1

I

'!

3

3 4 2] 6x)dx = ~ - 6~

-

4

0

·}•

• l

Fig.1E.40 i - - - ,.,

----~

·-

.........

-·· .

(1 4

, a!

= 4.3 -3-3

.,,,,,-~.

= -81 -

Solution.Ca) Since cos x;:: 0 over the interval [O, Jt/2] , the area A under the curve is

2

0

2) - (14·0 4-3-02)

27 - 0 + 0 = -6. 75

4

n/2

A= Jcosxdx=[sinx]~/ 2 =sin2:-sin0=1 o 2 (b) The given integral can be interpreted as the signed area between the graph of y = cos x and the interval [O, it]. The graph in figure suggests that over the interval [O, it] the

portion of area above the x-axis is the same as the portion of area below the x-axis, so we conjecture that the signed area is zero ; this implies that the value of the integral is zero. This is confirmed by the computations.

i

iFind {

f:2(,zx,'

3 "-;-

0 .

L--- -~ ..... J.

X

+l

-~-···

. _.,:: ·

., "

· ·'

4

'J 1

ri+l, ._,

= .!x 4 2

= .!c2J 2

jU(x)'+ g(x)] ,ix•, : '

2

'

.

., ·.'.1 .. ' ..

·n;t-1

-·

-

1

x]

-3x 2 + 3tan-1 x]~

3(2) 2 + 3tan-1 2-0

=-4+3tan-1 2 This is the exact value of the integral.

\.aJJ(x)~'.,-Jg,(~)dx

···cc· . .n:t-1J . ' .. J:-cdx = lnlx!·f:C .. ••:, X,,. . ., ,

x .. · X "dx_ =_-·-_}-::.+

2

J(2x -6x+---i-)dx= 2x -6x +3tano X +l 4 2

fkdx=}:x+C !

. "" -- ,_,

-

3

Table of indefinite _integrals

CIE-x~mele.; .;;,-::--· -= ~~~..::::.,.:·

.

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"'1-:: .;_

'

'i

' --·-··-· .J

Solution: The Fundamental theorem gives

0

f c f(x) d. cff(x) dx

- ~--

~

I

•

---

·rcos xdx = sin x]" = sin 1t - sin O = 0 0

- -- -

• ). · 6x + + dx

l~

43 -~-:V

2

O

Anurag Mishra Mechanics 1 with www.puucho.com

-. -----·~~.'-.'-'-'"';,._----"'---,.._--- ~,~-' __ . _ _ ____ _;.·,_, ~--'·':'"'""'~- ______·_s1~.,I ,,

DESCRIPTION OF MOTION

Solution: First we need to write the integrand in a simpler form by carrying out the divi~ion 9 2 2r: 9 J2t +t i"t-1dt = Jc2+t112 _r-2)dr 1 t 1 9

r 312 c 1 · 3 312 =2t+----. =2t+-t ~ -1 ] 2 2

.

I

+-] 1 t

9

1

l] (

3 3/2 +9 = [ 2·9 +2(9)

4. If the rate of growth of a population is dn/dt, then 2 ' dn J-dt =n(t 2 )-n(t 1 ) ,, dt is the net change in population during the time period from t I to t 2 • (The population increases when births happen and decreases when deaths occur. The net change takes into account both births and deaths) 5. If C (x) is the cost of producing x units of a commodity, then the marginal cost is the derivative C' (x). So

3 3/2 +l 1) 2-1 +2·1

01 find the time when the velocity is zero. Find the displacement at this instan."'t·~-------

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Solution:

X

= 120-15t - 6t 2 +t 3

Solution: (a) Distance travelled is given by the area under velocity-time curve. Therefore distance = Iarea of A I + Iarea of BI + Iarea of Cl+ Iarea of DI (All areas are considered +ve irrespective of the nature . of the quadrant in which they lie).

dx

v=-=-15-12t+3t~ =0 dt Solving, we get · t = 5, - 1 sec. As t > 0, t = 5 sec. putting t = 5 in the expression of x, we get x=20m

~!il:m,i-ii,l,e~f7ol~ ~ ,_:: CE v & : ~ ~ -r- - - -

,

,___ ... - --- ~- - . - ~.----·

.•,,

(The figure shows the (V,t) graph for the train .accelerating' 1 ifrom resi up to a maximum speed of v . and then 1 decelerating to a speed of l Oms·- . The acceleration and deceleration. have , the same magnitude which is equal to z ., . . - . • , 0:5'·. . m s. • ./ -~--_ · .. ~=-.-"'--

ms-

I

v (ms·')

I

V

i

·' ,

~

1

:

i

Fig. 1E.60 (a)

·

..

IA birdfl.iesfor 4s with a velocily v = (t -

·-~J ~:;l!;!e::~e;e~~;i;:e~f::i~

§h6'0h,

t (s) t

0

1

=-x2x2+2x2+-xlx2--xlx2 2 2 2 =6m

10

o .,

=~x2x2+2x2+~xlx2+~xlx2 2 2 2 =Sm fb) Displacement= area of A+ area of B + area of C + area of D (Proper signs of areas are considered according to the nature of the quadrant in which they lie.)

2) m/s in a straigh~i

Ca.lc~late the displace~ent,arutj

v2 -u2

= 2as

s=--2a

Solution : The displacement is given by 4

-·----·--v,(rps·1)

4

s=J 0 vdt=J 0 (t-2)dt .

', •V

·v 2 -o 2 ) = -.- - = v

=1t:-2t["=o

2x0.5 ', For the motion between A . and B 102 -V2 ,Distance (s 2 ) = - - - 2x(-0.5)

The velocity of the bird become zero at; 0 = t - 2 .=> t = 2 s

10

JJ: vdt HJ: vdt I ' =Js:ct-2)dt J+Js: (t-2)dt I

Distance s =

= v 2 -100 Total distance = s1 + s2 = 2V 2

1.00

-

=1t:-2ti: + [t:-2.ti:

particle

2 2 =1 :-2x21+1(~~2x4)-( :-2x2)/

The velocily-time graph for a travelling along a straight line is.shown in the Fig. 1E:61. Find -·-

. ' =2+2=4m Graphical Method The velocity-time and speed-time graphs of motion of the bird are as follows:

. ' v (mis) 2 ···········~--~

A: 0

-1

I 1,

-2

1.

2

B C 3

4

6 5

o:'

time (sec)

'' ' ---------------------------- ' Fig. 1 E.61

·-------~~-------(a)·distance travel(edfrom zero to 6sec. _ __,__, '@Jlispjactimentinft

= (v,; i + Vy.j) + (ax

...v ·;:::: ...vr + ...at· 1

Ball 1 is released from the top of a smooth inclined plane, an~ at the same ins tan~ ball 4 is projected from the foot o};the' plane with such a velocity that they meet halfway up the incline. Determine: - - - - - - - - -1

Simllarly,

2

We may write the above equations for final po_sition vector.

li !: ~

-+

-+

2

__,

where

A

J

--·'

(a) Accelerations of both the balls are

2

Yd

A

vi

a t

A

rr =xii+ ...

!

1-+

---+

= r; + v, t + -

r1

(a) the velocity with which balls are projected and · (b)Jhe velocity of each ball when they mee,,,,t.~------

Solution:

2

1

hi

Fig. 1E.80

1 2 Xf =X·+V l XI-t+-aX t

-t+-a Yi =y-+v l yt 2 yt

i

4.LJ"'-------'

i + ay j )t

A

=v.ni+vyij

...

A

A

i + ay j

a= ax

For a particle moving in a plane we may write the following equations (note that we have assumed initial position to be origin, i.e., xi = 0, y, = 0).

a1 = g sin0 and a 2 = - g sin8 down the incline.

Ball 1:

.!. =(0)t + .!. g sin8t 2

... (1)

Ball 2:

.!. =v- t + .!. (-g sin8)t 2

... (2)

2

2

2

'

2

V,;f

Adding eqn. (1) and (2), we get 1 l=v,t or t=-

=V,; + a/

Vyt =V_y;+ Of.

vJ = vJ + 2n,(x1 - x,)

v.J.=;; + '2il,0/1-Y1l

v,

Substituting it in eqn. (1), we get

i2 =.!.2 g sine (_l__) v, or

or For ball 2:

[

vJ = vf + 2ax

-+

A'

Since

'1

.

•

I

-+

V

A

A

a= 6i+4j=constant

......... =U+at

Therefore using

V

,...,...

"'"

= 4i+3j +(6i+ 4j) X 2

...

A

A

=> v=16i+lljm/sec The displacement is -+-+ 1-+2 S =Ut+-at

2

A

A

1

A

•A

=(4i+3j) X 2 +-(6i+4j) X 4 2

A

A

,..

...

or v 2 =0 Two-Dimensional Motion with Constant Acceleration 1. The position vector for a particle moving in the ,y-plane is given by r =xi+yj Velocity of the particle is obtained by ...

-·-------.---------------+ ,.. ,A

Solution.

= g1sin8-g1sin8 = 0

A

-

_constant acceleration is a= 6i+4j m/sec 2 • Find the velocity: @!d disp)acement_ofihe JLarticle at t = 2 sec. ___ . _ _ j

vf = O+ 2g sine.!.2 v 1 = .Jglsin 8 = .Jii1 v~ = "f + 2(-g sin8) .!. 2

...

-- ---------- -

The velocity of a particle at t = 0 is"u. = 4i+3j m,/sec and aj

v, = .Jg1sin0 = .Jii1

(b) From equation,

For ball 1:

2

A

v=vxi+vyj Because "i is assumed to be constant, its components ax and ay are also constants. Hence we may apply equations of kinematics to the x and y-components, e.g., ...

A

A

Vt =(V,; +axt)i+(v_y;+ayt)j

'The accel_erqtio.n of a moving body at any_ ·me 't' is giv_en_ by: -+ " '-+ ' • ' ! 2 ,. 2 • If ti =·0thenfind the velocity, a= (4t)i+.(3t . )j · m,/sec . . I oftheparticleaJ4:sec. _ ________ .. . , u_

t

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Anurag Mishra Mechanics 1 with www.puucho.com

MECHANICS-I ...,

A

1

A

Solution: Since a= (4t) i+(3t 2 )j is time dependent

..., ..., ...,

· therefore we cannot apply v = u + at We can solve by applying calculus

...,

dv

...,

-+

f

.iuti= ..., 2xl = 1;

-=a dt

Thus ;::

fo

or

dv=

4

A

A

(4t i+3t 2 j)dt O

~2 ~t 21= ~2 X (2) X. (1)2 I

'

=1

=.{:J \dt }1; {3Jt dt }J 2

0

0

...,

A

1;1= ~c1) 2 + c1) 2 =..J2 m

A

v = (32)i+(64)j rn/sec

Equations of Motion in Vector Form The equation of motion in the case of a uniformly accelerated motion is written in vector form as-

..., ..., ...,

(i)

v.; u+ at

(iii)

-+ -+ V·V

=

(ii)

-+ U·U+

~

-+-+

½°BF ....................... :""'>

1-+2

.s=ut+-at

~~t

tan(j>=

2

. iut

J· $

1

lutl Fig. 1E.83 (b) . =45° Displacement is at an angle 45° with the direction of initial velocity. ~

co}

-+ -+ 2a· S

l2...,

2

.:

''

· ,--'(Application only when acceleration vector is constant, motion ' • may be along straight line or along curved path.) '

·..+iam~ii 03 ~

I

'fA_ pCl!tide 0J'.n1~s 1 kg has a~ velocity· of2171/sec. A cons~ant

'

if9rce, of 2N acts;on· the. particl~fer l sec in a direction 1p'erpe~dicul11",;.:'.t~ .1ts initial ve/qcirg.. Find. the velocity:r/nd fdisplac~mentoj't/t'e particle at th~ end ofl sec. ' ',, ,' ' F

2

-+ V

..., ..., ..., v=u+at ..., -+

'

...

where

ju!= 2rn/sec;

...,

~. at

. .

V"

r:··i;:····--,;··;··: i ' r

.

=2rn/sec

lvl= 2..J2 rn/sec or

... ...

8 = tan-1 Iat!

F:~. 1E.83 (a)

'.u

I

iul 1 = tan- (1) = 45° Hence the velocity of the pa1ticle after one second is 2..J2 rn/sec at an angle of 45° with its initial velocity. For the displacement equation of motion is ''

a

=10 , Va COS8 =10 10

10

V

5

< , •• "'

')
'I~~ 85

------~-~ '·.

~

i

'

•.

"

'I

••. -;,:

The figure shows the velocity and 'the acceleration t of' a point-like body at the· initial moment of its motion. _The 'direction. and. the absolute value of the acceleration rem1iin constant. Find,cthe time in secona.fwhen the velocity reach its . minimum value ? (Data : a= 6m/s 2 , v 0 = 24.m/s, · q, = 143°)

,, Flg.1E.85.(a) •'---..C..·- - - ' - - - - ' - - - - - - - ' - - - - - ' ~ ' - · - '

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Anurag Mishra Mechanics 1 with www.puucho.com '.DESCRIPTION OF MOTION

1--··----

Solution:

x-components ux = -v 0 cos37° ax= a

y-components uy = v 0 sin37°

Vx =- 4vo +at

ay

=0

V

0 =--

5

3v 5

y

2. Projectile motion: An object is flight after being launched or thrown is a projectile. We assume that the distance travelled by a projectile is much smaller than the radius of the earth so that the acceleration due _to gravity is constant; secondly air resistance is negligible. y

7°

/'i:'--"e!----Vo'

A

a=-gj

a

i when vx

..-· Fig. 1E.85 (b)

=0 4 24

t = - x - = 3.2 sec 5 6 lliustration 10: Consider a ball initially moving along the x-axis as shown in Fig. 1.82 (a). At time t O it gets a constant acceleration ay in the y-direction. At any time t > t O the x- and y-displacements are

1 2 y=.-ayC 2

The superposition of these displacements is a curved

_,

path. The total velocity vector v at any time is tangent to the curved path of the ball. Velocity vector is an angle e relative to the x-axis, given by 8 = tan- 1 (vy/vxl, · which

f-'---.----------,---.-lt-r_.,.,.... __ • a

~xi

a

1 ';' lfyJ=-Vyli

- -·· ----- - - - .. !~:!·~_(bl _______

_J

Fig. 1.82 (b) shows the path of a projectile with velocity vectors. Let the launch point be (x,, y ,); y is positive upward and x is positive to the right. Projectile is launched with initial velocity vi at an angle e. Since motion of a projecti)e takes place in a plane we will set up equations for x- artd y-components separately.

continuously changes with time.

~:~¼

,,.1,: 2 y

/'

Vx

"x = 0

a,=-g

vxf :::vxi+ Cl;.:t

V.>f =Vy;+ G/

=V;COS8;

1

Xt=X-+V ,t+-n• l X1 2 "".r"

V2

",, = ., t, Y2

=½ay t~

Uy1 = ay

1

•1

81

v

Q-2..+,,..... ..• Y = 0 ta Vx X1 = Vxt1 Strajght-line vy=O

' '

.t:''

X

:

,1

•

-~

- •.._:-- . .

1:,

v, sme,t--gt

v;, =v~ +

' '

... (4)

=.)'(+v_;.,'.-~'2af

= x, + v, cos8,t · ... (2) =y,+

v

X

2

2 ..

2a1 (y 1 -

I

... (5)

y 1)

: ; '

X

:

X2·=

Vxl2 X3 = Vxt3

Curvilinear motion

+av

motion

Yt

. IJ -~

V1

Y1 = 2aY t1 ------·.: '., ,•

2

ti(,,/ l

2

V

.... '.

---------------·.--

....'

;::; vi sin ei-gt

... (1)

L.:... "x

atto=O

3. Problem Solving Strategy: 1. Imagine the situation of problem; draw a picture which shows the object and its possible trajectory. 2. Choose a coordinate system, choice of origin is arbitrary. Generally point of projection is assigned the origin. If range is to be calculated along horizontal level, take y-axis parallel to acceleration due to gravity. For calculating range along incline take x-axis parallel to it and y axis normal to it. 0

Flg.1.82 (a) ____ _

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Anurag Mishra Mechanics 1 with www.puucho.com MECHANICS-Q 3. Identify the initial position, irntial velocity and acceleration. If irntial velocity and acceleration are not along assigned x- and y-axis, then resolve them into x- and y-components. 4: Identify the unlmown quantities and assign them letter symbols u, v 0, a, etc. 5. Make two set of equations, one for x-components and one for y-components. 6. Try to identify special condition of the problem that may define the object's position or velocity at the point of interest. 7. After writing appropriate kinematic equations in component form solve them. 8. Always remember that time of flight is same for both component~ of motion.

·ekam~'. a6 - ~

,·--------------------·----- --------IA bomber ii; moving horizontally at a speed v 1

= 72 m/s at a height oJ;h = 103 m. An enemy ta_nk is moving horizontally (x,axis) constant speed. At the instant .the. bomli is released a '.tank is at a distance_ X a~ 125 m from origin; Origin is directly below a bomber at the instant of release ofij bomb. Assuming the tank to be 3 m high, find the velocity v 2 , and. the time of fl.ight of bomb. ~----·-·------ -------7 I l V1 !

w;t6

I

I

\ I

i

l

'

I

L-C'---'

"'--~-"

-

Y• =Y; "': Vjif ~

1

j a_,t

=9 + Vo-.,,,,Sill ~J-f gt ;

t ·:

_ X - X; _

Vx

·--- >< (9.8)(1.85)2 ,cC'. (, -" .·2-~ ,; ,:,~,- .. -.. ·,~,i

t

rv~---"---,--,---'x

32.0m

17.32 _

-,

t

=4.52s

r

-,-:,

="1.sss' ~.

-'-,- ---- - -- - - - - - ' -

. , _ . , ; . l ... ..

•

•.. -----··-··-.._.:·... ··

'"S'.,..

· We have assumed y = 0 at 2 m from ground. So height of ball above ground is 3.73 m.

325 - 125 _ / i\ - 44 . 25 m S 4.52

1Aboj throws~ ball with v;loqty v0 =10,/2 mis.at an angle of 45° as ·shown in the figure.'After collision with the ball the vertical component of ball's velocity is unchanged and the horizontal.i:dmponent is reversed in d(rection. Where do~ the ball hit the ground? ··, · · · ,

'.

2 _ 100 -4.9

· 2

@exg~~~ i---~~--

------t=? y=?

',

V2----

,_:. '

=1.73m>

, 32.0 =--

,.__I

. For the tank:

'

,· ;, ''. · 1 ' .•. . -. -" f26:o}J@30°'d.ssJ;

t=~;.

cy=3m 3=103-(4:9)t

2··

2

•

When the bomb hits the .. tank,;

~325m

f;eom110nents

, X,=v~t

•

Solution: Initial velocity of the bomb will be same as that of bomber vxi = v 1 = 72 m/s; For the bomb:

Solution: We have to find timet when x = 32.0m. In the second part we have to findywhenx = 320m or we can say, find y at the time it crosses the goal -line. We take origin at the point of release, xi = 0, y i = 0.

• ".

\

L-~---..,.l.~ i g . 1 E : ~ - - - - - - _ J ___

lt:footballer throws d. ball from a height oJ 2.00 m 'aboye_ the Iground with an initial veloc_ity o/20. 6 m/s at an gngle pf 30° jabove the hof_fzont~L (a)How lonq,does the ball taki-wcro~ [the goal (ine l2:0mfrom the po~n~of_releas~?{b) Wh~tis the IJigll's heyghLabove·the grftundASJt cro§ses the.goal lute? -·

.

I

I

-

·~o-/2m/s -.

, _,,

Fig.1E.88

-~-'----'-----·--'-------~ ~

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Anurag Mishra Mechanics 1 with www.puucho.com .. . -· .• .. ... - ---

P>ESCRIPTIO~ ~-~~OTI_O_N______

---------- -

Solution. We divide the problem into two parts, i.e., motion from boy to wall and from wall back to ground. (i) Motion from boy to wall

x=v;t

.'Y"//'l,,Y"·

hl

Hi = tan 2 8 H2

and

A • :ti;;_··-..

-------

Iwo particles A & B are projected from the same point in, different directions in such a manner that vertical components of their initial velocities are same : (a) Find ratio of time of flight (b) Find ratio of range. ,Y

X

Fig. 1.84

If

the range of projection is n times the maximum height of the projectile, then angle of projection is given by Using equation (1)

8= tan-

1

(;)

Fig.1E.93

•

Solution:

. 4. Relation between angle of projection e and angle of elevation 0 = x(a-bx)

X

- - - - - X2------

Fig.1.89

Path of the projectile grazes the top of the walls, so coordinates of the top point of the wall must satisfy the equation of trajectory.

y = xtan0

g 2

x=0 and x = (a/b) = R (horizontal range) Maximum height is the maximum value of y-component of displacement, which is obtained at x = R/2 Putting the value of' x' in the equation of trajectory we get-

a2 a2 Ymax=2b- b 4

x2

a2

2

2u cos 0

h =xtan0-

x1

-

... (iii)

U=~i

y

If x1

1 cos0= r;--;, vl+a 2

and

g

2u 2 cos 2 0

x1_+ x 2 (s_UTI! of the roots) gives horizontal range.

95

[>

(a) Given equation of trajectory is y = ax - bx 2

... (i)

Let' u' is the velocity of projection and 0 is the angle of projection, then the equation of trajectory is given by y = xtan0

f x2 2u 2 cos 2 0

Comparing (i) and (ii), we get tan0 = a

A gun is mounted on a plateau 960 m away from its edge as

shown. Height of plateau is 960 m The gun can fire shells with a velocity of 100 m/s at any angle. Of the following choices, what is the minimum distance (OP)xfrom the edge of plateau where the shell of gun can reach?

I

A particle is projected from origin in xy-plane and its equation ,of trajectory is given by y = ax - bx 2. The only acceleration ;in the motion is' f' which is constant and in -ve direction of, 'y-axis. ( a) Find the velocity ofprojection and the angle ofprojection. (b) Point of projection is considered as origin and x-axis: along the horizontal' ground. Find the horizontal range: and maximum height of projectile. Projectile completes its flight in ho~onta_l p_~ane of projection.

Solution:

(Maximum height)

I ·----- --- - - - - - - - - r:::::7: ,~~~·~~J

I

sin0=~

and

2

Sm

sin2 S = 5 X (2 X 10) 12.5x 12.5

or

4

or

2g

or

+ tan 2 a;)

(1 + 9h2) 4a2

=~

(12.5 sin 0) 2

i:e.,

[,;1=xa-1mf,:;c~3 ~""k~ 105 i

----

"----

-·

J.~-

~

ball ~ projected· with velocity vO and at an angle ofi ~fojection a. After what time is the ball moving at right a'!gles! ltSLthe initial direction? _________ _____ ,, .. _________ · .• i

IA

·solution: Method 1: If initial velocity v 0 and velocity at time t are perpendicular, then the final velocity will be at an angle a; with the vertical. ,.- - - -- --- - - ---

&it1: ; ~:-

i

goo ··--••.

t ~ ·_-~---_.' :f"i:£'--..J

'(a) What must qe th~ verticd'l compQnent of the initia() ' 1 [_ .·., _, Fig; 1E.1-05 (a) ______ velocity oftf!e.ball?, _ - _., ·';::, . _. ·.; J' i(b) How many,ieconds after IJe teleases t_h~ ball Will it pass Horizontal component . of velocity is unchanged 1 through the.hoop?' .• · · · · - ,· I throughout the motion. 1 cc) At what horizoutal dista.nce infrqn~. o(t. he fo'op musthel v O cos a = v sin a Therefore I release thl!.11!#1? _ . · , · ___ , _. .: _·_ • · . · , or v = v 0 cote,;_ Solution: Two important aspects to be noticed in this problem are: (1) Velocity of projection of ball is relative to man in motion. (2) Ball clears the hoop when it is at the topmost point.

..,

..,

V ball, man

..,

V ball

= V ball

..,

..,

- V man

..,

Vertical component of velocity after time t = - v cosa From the equation v y = v O sin a - gt -vcos a= v 0 sine,; - gt v sin a+ v cosa t = -0" - - - - or g v 0 sina + v 0 cotacosa

·=-"---~---g

= V ball, man· + V man

~ v0

(a) Now we apply the above relation to x- as well as y-component of velocity. If ball is projected with velocityv 0 and angle e, then

g

..,

x-component of v ball = (v O cose + 10) m/s

..,

y-component ofvba!l = (v 0 sin0) m/s (b) Since vertical component of ball's velocity is unaffected by horizontal motion of car, we can use the formula for time of flight,

= Vo

[sin a_+ cos a] 2

2

sma coseca

g Method 2: We choose x-axis along the initial velocity. If after time t the velocity is perpendicular to initial direction, v x must be zero after time t,

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.·~..

Anurag Mishra Mechanics 1 with www.puucho.com ' .. . ..,.,_ 1a2 ..~.~···~·=-~_. ._:._J_-~'-":f__ ~-· ..~-~ -·.t·· ~ir~i:_.~-':-~---~-----~~

r--..-, _ .,;~:~~).·

The equation of trajectory of a projectile is

gx2

l~,i L_

Flg.1Ei105 (b)

i.e.;..._.----

0-v 0

or

t

j

-_

g smat--

=~ gsina

Method 3: Slope of trajectory at the point of projection, m1 = tan0

--~--_--7~-.;--,

.:7

'

I "

- - -.

. . ''

. ._, --~

Point (R cos p, R sin I}) must satisfy equation (1). gR 2 cos 2 1} Hence R sin P = R cos!} tan a 2u 2 cos 2 a gR 2 cos R or R(tan a - tan P) = ~c--"2u 2 cos 2 a gli cos p or cos a cos P 2u 2 cos 2 a 2

'

L,_~Fig.~5(c)·· _, ,

g cos 2 p

,

i-_---------- --·

Slope of trajectory after time t, ' dy dy/dt m2 =tana=-=-.dx dx/dt Vy

R = 2u sin (a - I}) cosa

or

Method 2: We take axes along incline and perpendicular to incline as shown in Fig. 1.9L '1n this coordinate system, components of velocity and acceleration along the incline and normal to incline are ux c:a u cos (a - !}), ax = - g sin,P uy = u sin (a - !}), aY = - g cos p

,'

. t~

___ (l)

y = x tan a - - ~ -22u 2 cos a

\;

\,

, '

u

y

'

I

'

' u, = u sin (a-~)

v sina-gt

0 =-=~---

X

Slopes are perpendicular, (

vosina-gt)ctana) = - l v 0 cosa _

or

When projectile lands at A, its y-coordinate is zero. 1 2 O=ut+-at y 2 y

t=~ gsina

PROJECTION ON AN INCLINED PLANE A particle is projected ry, ·: . ---;--:-;--- , ,. r from point O on the foot of an· inclined plane. The [ '. ~- · ,,u ~ath :fR~i~c\i;e : : ::-.< ,; /A'' . :_,'.. ,· velocity of projection is u, "'angle of projection a with - - : ,. ':· ( - .5' ,, R• ·,.. :i ,. ,,.Cl) x-axis, angle of incline P ,, .0::: ~: ,[see Fig. 1.90]. We wish to ,· B, ,, _x determine range along .,:;o.,, '_:,;· ~, Rcos 13----. I incline, . time of flight, 1 -·~' Fl 190 vertical height at which Lr_·_·_---~·--·-~---~ projectile strikes. (a) Range Along Inclined Plan·e Method 1: Point A where the projectile lands has coordinates (R cos p, R sin p}.

i

l

,

or

0 = u sin(a - J})t - .!_~cos J}t 2 2

2u sin (a - P) t=----~ g cos p This is the expression for time of flight from O to A. For motion along inclined plane (x-axis), 1 2 or

X

= Uxt + -

2

axt

, =UCOS(a-J})t,-.!_gsinl}t 2 2

Substituting expression for time of flight, we get 2 R = 2u sin (a-P) cos a g cos 2 p

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DESCRIPTION OF MOTION Method 3: We revert back to a new coordinate system with x-axis in horizontal and y-axis in vertical direction [see Fig. 1.92]. x=u coso:t 2u sin(o: - Pl = u coso: - - - ~ ~

Projection Down the Inclined Plane

y

y

-xFig. 1.92

gcosp

R = ~ = 2u coso: sin(o: - P) cos P g cos 2 p 2

and

x"' Fig. 1.93

From figure we have, ux =ucos(0+o:), ax =gsino: "Y =usin(0+o:l, ay =-gcoso:

(b) Vertical Height at Which Projectile Strikes Method 1:

From the equation y = u sino:t - ~ gt 2, 2

on substituting time of flight t, we get

Time of Flight

. 2u sin(o: - Pl - -g 1 (2u sin (o: - PlJ y =usmo:----~ g cos p 2 g cosp

2

As displacement become zero along y-direction in time

'T'.

1 2 O=uyT+-ayT 2

2u 2 coso: sinp sin(o: - Pl Method 2:

g cos 2 p y = :nanp_ u cos o: x 2u sin(o: - P) A =--------'---'--'-X tanp

gcosp

2

2u coso:sinP sin(o: - P) =----~=----'-g cos 2 P

or

0 =usin(0 +o:)T- ~(g cos·o:)T 2

or

T

2

g coso:

Range Along Inclined Plane (R):

1

R=uxT+-axT

(c) Angle of Projection for Maximum Horizontal Range Range R, is given by 2

R = 2u sin (o: - P) coso:

2

2

1 . [2usin(0+o:l] =ucos (0 +ex l[ 2usin(0+o:l] +-gsma ----g coso:

R=

2

u [sin(2o: - Pl - sinpi g cos 2 P

g coso:

2

2

u

g cos 2 a.

[sin (20 + al + sin a]

or

(20 + o:l

= 90°

or

0 = 45°-_ 45° l is the inclination to the horizontal of the initial direction of projection, for what value of tan $ will the particle strike the plane: (i) horizontal (ii) at right angle?

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Solution: Let the particle be projected from O with velocity u and strike the plane at a point P after time t. Let ON = PN = h; then OP = h-Jz. (i) lf the particle strikes the plane horizontally, then its vertical component of velocity at P is zero. Along horizontal direction, · h = (u cos )Ct) ... (1) Along vertical direction, 0=usin-gt ... (2) or usin=gt

·-----

or or

2tan •

~\J,

•

'

;'(4J-_45_o,}

,_

i

:"

___

De:-::--~:=·~~-·-, ,;:;E•""rU!l:Oi!Wf1 ei - 108 t,:,~

~--~---~~~-

j

0 = usin ( -

= -~_

,Jz or or

l

~_}~,

0

'

C

'

'

'

'

'.,

,_.-,7',--· -:--r ,;,

id'

N· -

f?/:f:;-. 1;0 ;_·_·.r.:~-~> :··.:, ,. b:fu,.;.L...::_ Fig: 1E:tOI! (b) , ' .

... (ii)

C

,'

,,

-:~ prOjeft(le /,t t/-irQWn, at ar( G71$/e ;~With an inf/i~ed plci.rtfOJ rlric(irtqtion ~- as .shoWrt. f:ig.: ).~:108.. Find ,the r~ldtio~ .bet:)l(een~q11i:!Bj(:. . _ ,,,, • ,,:',.' ,',, ';'/ C· . f(a) p"oJectf,le,stri~es the inclinedpCane,perpendicular1y1, {' ll;,),p:r:._ojectiie,~@X·8X1Sl '

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I _D~~~RIPJION O_F MO_TION

ssl -------- - - --------------------~

2u sin9 T = --g cosp u cos9 2u sine gcosp g sinP

as or

we also that

=>

=> 2tan9 = cotp

(bl If projectile strikes horizontally, then at the time of striking the projection will be at the maximum height from the ground. Therefore,

~:.

____________

0

u g sin 60°

10-./3 =2s x -/3 10 2

(bl Initial velocity along y-axis is zero. The velocity along y-axis after 2 s; Vy= Uy+ Uyt

= O-gcos60°x2 1 2

tf1(_______________________ . . . . _____ _

=-10x-x2=-10m/s (cl We have, v; = u; + 2axs Since and

2u sin9 tap=---

gcosp

=>

t=

=>

'

Fig.1E.108 (b)

=>

Vx=ux+axt 0 = u -g sin60°t

2u sin(9 + Pl tap=---~2xg 2usin0 2usin(0+Pl g cosp 2g

or

vx = 0

ax = g sin 60°, u = 10-./3 m/s O = (10-./3l 2 -2xg sin60°x (OQl OQ =

10

2

X

3

= l0-./3 m

-./3 2xl0x2

Distance

2sin0 = sin(0 + Pl cosp.

l -~-~~':Tl-r. "'7 I~cjJ.->

PO = O+ .!:. g sin 30° x (2) 2 2 1 1 = - x lOx- x 4 = 10 m

2

'lwo inclined plane.s OA and OB having inclination 30° and 60° with the horizontal re.spectively intersect each other at 0, as shown in Fig. lE.109. A particle is projected from point P with a velocity u = 10-./3 m/s along a direction perpendicular to plane OA. If the particle strike.s plane OB perpendicular at Q.

2

Therefore height h of point P, h = PQ sin 30° = 10 x .!:. = 5 m 2 PQ = ~P0 2 +OQ 2

(dl Distance

= ~(10l 2 + (10-./3l 2 = 20 m

l :!.~~G9!!11?J,f7: .G10 1;> u

30°

0

Fig.1E.109

Calculate (a) time of flight ,(b) velocity with which the purticle strikes the plane OB (c) height h of the point P from point 0 ( d) distanc~ PQ.

nvo guns situated on top of a hill of height 10 mfire one shot, each with the same speed 5-./3 m/s at some interval of time., One gun fire.s horizontally and the other fires upwards at an' angle of 60° with the horizontal. The shots collide in air at a, point P. Find (al the time interval between the firings and (bl the coordinates of point P. Take the origin of coordinate system at the foot of the hill right below the muzzle and trajectorie.s in the xy-plane,

Solution: Consider the .motion of particle along the axes shown in figure. We have ux = u, ax= -gsin60° Uy= 0, ay = -g cos60° (a) As the particle strikes the plane OB· perpendicularly,

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u,

---~

_,;..---ll,un 1 P(x,, Yr)

10 m (0,0)

P(x,y)

xFig.1E.110.(a)

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Solution: Let gun 1 and gun 2 be fired at an interval Llt, such that

t1 = t 2 + M ... (1) where t 1 and t 2 are the respective times taken by the two shots to reach point P.

For gun 1: · X' -~ X, -·.'Xt,

Method 2: We take point of firing as origin and xand y-axis as shown in Fig. lE.110 (b). Equation :of trajectory of '\ projectile is 2

y =

tan0-

X

gx 2v 2l cos 2 0

For gun 1 1 . 0 = 60°. y

anent

=11,t q>S 60° ~l

. .

./3 . . l i' · Y=Y·+-v-t 1 -~•t1

. . .

X

1 = )(.+,.;....·V·t1 \' 2 l

: •

2

I

I

z9•.

2gx2

(a) Now we can equate x- and y-coordinates of shots,

i.e.-,

... (2)

=x-./3-v?l

For gun 2, 0 = 0°. -gx2 y = 2v 2 · l

and ... (3)

or

Two shots collide at point P; therefore their coordinates must be same; i.e.,

On substituting t 1 from eqn. (2) into eqn. (3), we get -./3 1 2. . v; (2t2) + Z g(-3t2) = 0

2

or or

t 2 (-./3v,

t2

=0

-1

=0

gt 2 )

. and

t2

2

·Or

collide are

x=xi+vit 2 = 0 + (5-./3)(1) = 5-./3 m

and

3gx2 2v 2

= --

2v?!'

l

2v2

2(5/3)2

- -,J3g -

./3(10)

x--' --=~-

= 5-./3 m and

-gx2 _ (10) (SV3)2 y=--=--~~ . 2vr 2(5-./3) 2

=-Sm · If originis assigned at ground the coordinates of point P will be (5-./3 m, 5 m). Now We consider x-component of displacement for both the shots. Gun 1: x = 5-./3 m = v,t = (5-./3 m/s)t or t, = 1 s

2

y=y,-2gt2

= 10 - .!_ (10)(1) =Sm

x=O

V·

t 1 = 2t 2 = 2(1) = 2 s M = t 1 - t·2 = 2 - 1 = 1 s (b) Tbe coordinates of P at which the . two shots

2

v2l

-

-./3 g

Therefore,

1

x-./3 = - - -

=- -'-

=lx(7o3)=ls

and

or

2 2gx2 ~gx · = x-./3 - - 2v2l v?l 2gx2 gx2

Gun 2:o 2

X

= 5-./3 m= V;

COS

or t 2 =2s Time interval between two shots is Llt

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5,/3

60° t2 = - - t2 2

=t 2

-

t1

=1 s

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[ DESCRIPTION OF MOTION

. a~J

0~( .->

lJ~.j~}J~J~·~

-> V panicle, box

1

A large heavy box is sliding without friction down a smooth plane of inclination 8. From a point P on the bottom of thebox, a particle is projected inside the box. The initial speed of the particle w.r.t. box is u 1 and the direction of projection makes an angle a with the bottom as shown in the figure.

-> V particle, ground

->

= :V particle, ground ->

-> - V box, ground ->

= V particle, box + V box, ground

Applying above equation to x-components, o =u cos (a + 8) - v cos8 ucos (ex+ 8) v= or cos8 Method 2: The above y condition can be meet if the box covers exactly the same distance as the range of particles, i.e.,

Fig. 1E.111 (a) I

(a) Find the distance along the bottom of the box between the point of projection P and the point Q where the particle lands. (Assume that the particle does not hit any other: swface of the box. Neglect air resistance.) (b) If the horizontal displacement of the particle as seen by an observer on the ground is zero. Find the speed of the' box w.r.t. the ground at the instant when the particle was, projected.

Solution: (a) Motion of the particle will be reference frame of box.

0 2

Fig.1E.111 (c)

or or

Relative Motion Fig. 1. 94 shows an observer on ground, a balloon and an airplane, we denote them by G, B and A respectively. At any instant position. vector. of airplane for -an observer on ground, on balloon have been represented.

uy =u sin ex ay=gcose

a,=gsin0-gsin0=

y

1

=U/ --g/

2

2

Put y =0 for calculating time' ar flight. ' 2 1

x=u,! =ucosat

=U COSCX

J=v( 2gu:~~0cx)

1 . e(Zu sina) +-gsm 2 g cose u sin0sina ucosa= v + - - - cose cosa cos0 - sin a sine) v = u( - - - - - - - cose u cos(a+0) cose

in

x-component

or

( u; : : :

I

2u sin ex) ( g cos a

Q::;::usincxt _!gcos0t 2

u2 sin 2a

or

gcosa

t

2u sin a g case

(b) According to problem the horizontal displacement of the particle as seen by an observer on the ground is zero. If we analyse the situation in the reference frame of ground, resultant velocity of particle in x-direction must be zero. X

Fig. 1.94

.... rA/B

.--J

~

object observer

Q

.... rB/G

.--J

~

object observer

Fig.1E.111 (b)

....

rA/G

.--J

~

object observer

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....

r;y8 position vector of airplane for an observer on balloon

....

rB/G = position vector of balloon for an

observer on ground

....

rt\'G= position vector of airplane for an observer on ground

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_,

From figure

_,

(al or

Thus,

or,

VA/B

=

_,

_,

VA/G

V,rG

.j,

.j,

.j,

VP/E

_,

Rate of change of position vector is velocity.

_,

_,

_,

_,

rA/G = rA/B+ rB/G

Velocity of Airplane Velocity of Velocity of as observed by A for observer balloon for observer on balloon on ground observer on ground When we say velocity of airplane w.r.t. balloon or velocity of airplane in inference frame of balloon. it means

VP/G

_,

= VP/G-VE/G _,

_,

= VP/E+VE/G

which implies that absolute velocity of the passenger is the vector sum of his velocity relative to escalator and _, _, velocity of .escalator relative to ·ground. v P/E and v E/G both pciint towards right as ~hown in Fig. 1.95 (b)

~T~~,-Ir·~-~,

r- :

\, .. ! i

}

i

.I I

,JA/ 8 referred as relative velocity. Application of Advanced Concepts of Relative Motion

l

·I

River Condition Consider a swimmer in still water. The swimmer can generate a velocity due to its own.!'ffort. We call this velocity, velocity of swimmer in still water.

_,

Velocity of swimmer relative to water = v s/w Next consider a person with a life jacket in a river flowing with a velocity. Person makes no effort to swim, he just drifts due to river flow. Velocity imparted due to river flow is called velocity of water relative to ground, i.e., it denotes the rate at which water flows. Velocity of water flow relative to ground

_, = v w/o

----,_~,,.-----~------· ~,·- ~-~-·------,--~

'.(a) "Find but the motion of t;Je, ql~d (I~d ~Id IJlan(lS ~e~n by I . boy: ;'

(b) re)

..

·

· ,. ·. ·· •

·

··

= Vsjw+vw/G

Esc,ilator Condition Here is an analogous treatment. Just .. _, imagine an

·•; ...

.

Firi4 out,m~tioft oltree, bird, boy as seen.by,oid man, Find'qut11uitio[l.Pftre.§, DOYcartd old man as seenbjzoira. Solution: (a)With respectto boy: Vtree =4m/s (~) . vbinl =. 3 m/s (I) and O m/s c~)

Next consider a swimmer applying Iris effort in flowing water. In this case swimmer's net velocity resultant velocity will be decided by two factors Ci) his own effort (liY water flow. Thus resultant motion is obtained by vector sum of two velocities imparted to swimmer.. Resultant velocity of swimmer relative to ground = velocity of swimmer relative to water + velocity of water · flow relative to ground. _, _, _, VS/G

i'

;(b)

Fl~.1,.95

(b) With respect to old man:

=6m/s(~) = 2m/s (~) vbird =.6 m/s l and 3 m/s (I) (c) With respect to bird: vtree = 3m/s (.J,) and 4 m/s (~) and 3m/s(.J,) Void man= 6m/s (~) Vboy = 3 m/S (.J,)

escalator moving horizontally with velocity v E/G. A person _, . . begins the run with velocity v P/E fa the same dii;ection as . escalator. What do you think about resultant velocity of passenger? We assign a latter to each body P, passenger; E, escalator; G, ground. www.puucho.com

vboy

Viree

c~

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89,

, _DE_SCRl~TION OF MOTION

~-113 ",;>

f_-~?f9:'.T\P Ie

.

.

A helicopter is trying to land on a submarine deck which is moving south at 17 m/s. A balloon is moving at 12 m/s with wind into the west. If to the submarine crew the helicopter is descending vertically at-'5 m/s, what is its speed? (a) relative to the water and (b) relative to the balloon. See fig.

y-.

~E =iA/P

+4J__,,

Vp/E

Vp/E

8

1-----v='-------'=illF--,1+._-+•

EX Fig.1E.113

Solution:

Velocity of ant relative to paper

= Vsub/water + Vhel/sub = l7j + (-S)k = (17j-Sk) mis (b) V hel/ballodn = V hel - V balloon

(a)

Vhel/water

Fig. 1.96 (b)

=(17j- 5k)-12i =(-12i+17j-5k)m/s Ant Moving on an Ruler Fig. 1.96 (a) shows an ant scampering along a ruler. The

Girl Moving in a Train Illustration 11. Fig. 1.97 shows top view of a girl (G) walking in a moving train (T). Two observers one in the train and the other on the ground (E) determine the position vector of the girl._, _, _,

_,

ruler has been displaced w.r.t. Earth by

_,

SR/E,

the ant

undergoes a displacement SAfR w.r.t. the end of the ruler. The net displacement of the ant w.r.t. Earth (i.e., w.r.t. a fixed point P0 on the ground) is given by the vector sum _, SA/E

_, SAjR

=

-> + SR/E

rG/A

V A/E

_,

••. (1)

The position of the girl walking in the train relative to frame of reference of A is different from her position relative to frame of reference B (Fig. 1.97). Time derivative of eqn. (1) gives the ralation between various velocities.

... (1)

Taking time derivative of eqn. (1), we get the corresponding velocity expression _,

=rG/B + rB/A

-> VG/A

_,

->

=V G/B+ VB/A

... (2)

Ya

->

=V A/R+ VR/E

... (2)

G

Train

~t:;z=::==:f---Xa Reference B frame fixed to train

~------+XA QA Reference frame A fixed to Earth

Fig. 1.96 (a)

Eqns. (1) and (2) are valid irrespective of the direction of two vectors. Fig. 1.96 (b) shows the motion of an ant walking across a sheet of paper, that is itself being moved at a speed .f P/E. The ant is carried along with the paper so that it actually moves north-east w.r.t. Earth.

(a) - - - - - - - - v'G/r---tVelocity of girl relative to train

1Grr+~

-JTIE__,, Velocity of train relative to Earth VT/E VG,E---t Velocity of girl '--------'=--' relative to Earth ·1 ~

(b)

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Fig. 1.97

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,·

Velocity of particle G relative to reference frame A = velocity of particle G relative to reference frame B + velocity of reference frame B relative to reference frame A If the girJ, walks across the compartment, her resultant velocity will be as shown in Fig. 1.97 (b).

...,

V G/T

..., or

V G/E

i

. I ,_

. -··t:e ~

•••••

••••

\

H

•

->

VAJG

:

/

_;:f, ------(..

•••

Fig.1:98(b) ____ , ,

•

->

->

.=vs+vw Case (ii) Swimmer moves opposite to river flow

(upstream) when swimmer moves upstream. ->

->

...,

lvs;G l=lvs;wl-lvw/G I

A floating object like a wooden log move with the velocity of river flow.

' •

Step 1: Problem Solving strategy: Assign the initial point as origin of a coordinate system. r-:~---·y . ---,--,--·1 I, -+--.........,.~--,-

:

. '

.f

!

River flow.I

Iv s/q l=lv s;w l+lvw/G I

-••• •••• ••••• !

.

_-

~ ~ Vw/G ._vstG

Note:]------=-'-V£.s_-_v_,w"----------

Plane in which airplane moves

''

'''

~--1I

..., Iv s;w I=vs =velocity of swimmer relative to water ..., _I v W/G I= v w =velocity of river water flow

V AfW•

•

------·--:

Vsr,, ___,___

...,

i.e., velocity of girl w.r.t. earth (reference frame of ground) is vector sum of its velocity relaitve to train and velocity of train relative to earth. Airplane-wind Condition ..., Consider an airplane moving in still air, with ve!O(;ity

------- ____ / •• -,

~

1

..., ..., = V G/E-VT/E ..., ..., = V G/r+Vr/E

~AJW

MECHANICS,! ]

!_...... -· ...-_· .. '· -··----------.--. .x::,··· ........,

i

' · Ground plan~

->

.

W~E'

Vsr,,

swimmer begins here

~--,

#/ ·''"O

s ' Flg.1.98 (a)

~--- -·--- -- ··---· ...,------·- --- -

.

--·-

0

-

Fig.1.99

Wind· flows with velo~ity v W/G due east direction, Resultant velocity of air plane will obtained by equation -> V

->

A/G =

V

A/W +vw/G

~---- -------- ---V object/ground

.,

_____ ·-----

j

velocity of medium ·

• :Position P6sition where man where man heads actually reaches

I

A

B

..··

II

Initial direction of motion of

man

----,

j

= V object/medium + V medium/ground

Resultant velocity • velocity of object _ relative to medium

~---- ---

__ ----_---....·-=-===;~-:::-7

->

In all the previous real situations there is an object that moves on a moving medium Object Medium River condition Boat Water Swimmer Escalator condition Passenger escalator Ant-ruler condition ant ruler Girl-train condition girl train Concept

Step 2: Draw vector diagram

0

----

.

-··---~ ----- - - - - -

River Condition Revisited Case (i) Swimmer moves in direction of flow (down stream)

X

Flg.1.100

When swimmer reaches Bits x-component displacement is x whereas d represents of y-component of displacement. Step 3 : Apply component method of vector addition.

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DESCRIPTION OF -- -- MOTION ·- __ ,~,_.,

---· ·-------··-··----

-·

,

Concept Motion in x-direction is due to x-component of resultant velocity, similarly motion in y-direction due to y-component of resultant velocity. ...,

A

A

- - ·-

(vs;a\ = Vs sin0. Time taken to cross river d t=--VsSin0

Vs;w = Vs cos0i+vs sin0j ...,

------

Drift= (vs -vw cos0) x

A

Vw;a =vwi ...,

A

Vs;a

d_ Vs sm0

Concept: Note that drift can be zero if Vs= vw case In this case swimmer moves along shortest path. But above condition can be satisfied only ifvw > VsIfvs > vw, drift can be minimized but it cannot be zero. For minimum drift d -[vwcosec e+vs cot0] = 0 dw vwcosec0cot0-vscosec 20_= 0 Vs cos0=or Vw

A

=(vscos0+vw)i+vssin0j

x-component of resultant velocity (vs;alx =vscos0+vw x-component of displacement x = (vs cos0+vw)t Similarly y-component of displacement y = Vs sin0t Thus time taken to cross d width of river It= Vs ~n 8 Drift during crossing of river x= (vs;alx t (VsCOS0+vw)d x= VsSin0

A boat moves right across a river with velocity 10 km h-1

Concept: How to obtain time taken to cross river? y-component of displacement t=--"---'-----'------''---y-component of resultant velocity '

relative to water. The water has a uniform speed of 5.00 km h- 1 relative to the earth. Find the velocity of the boat' relative to an observer standing on either bank. If the width of river is 3.0 km, find the time it takes the boat to cross it.

...,

Solution:

What is drift ? Distance the swimmer is carried away along flow while crossing river. Xdrift

=[vSJalx

v B/R -, velocity of boat w.r.t. river

...,

v R/E -, velocity of river w.r.t. earth

...,

v B/E -, velocity of boat w.r.t. earth --t

---),

--t

VB/R =VB/E -VR/E

x time

--t

Position Position where man where man heads actually reaches A B •

...

·· ...

,,

·- ..

.

Actual direction of motion of man

..

Ay

······ ...

->

->

VRIE

VR/E River flow

lniUal direction of motion of man

Man begins atO

0

X

D

Fig.1.101

What happens if swimmer moves opposite to flow ? ---),

,,._

A

,.,_

Vs;a = -vs cos0i+vw i+vs sin0j (vS/G lx = (vs -vw cos0)

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---),

--t

or VB/E = VB;R+vR/E

Fig.1E.114

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Angle at which boat starts is given by

Hence, VB/E = ~V~jR

• ·

+ V¼E

·= ·/i0 2 + 52 = 11.Zkmh-1 The direction of v B/E is 8 = tan-:(VR/E)

....

....

.

Note that v B/E is resajtant of v B/R and v R/E • Effective

....

VB/R

cross

If x-component of resultant velocity vanishes the boat will move straight, along y-axis. Hence, VB/R sin8 = VR/E .

or

velocity of boat in y-direction is v B/R • to

river

is

Crossing River Along Shortest Possible Path ;_e., Moving Perpendicular to Flow In this case, x-component of resultant velocity is zero. ·---- -·,,. B. • -----1

cft=~-,J£1X0,mi.:i~5 ··' · ~r}\'.!,~~

[vs;alx

= O'

.

-v 5 cose·+vw = 0 .

V

cos8=_.!!',

i.e.,

Vs

From Pythagoras' theorem, VB/E

~-~lg.1;1~~----·-_j

-)

= VB/E

I

- x,

:_.·,.,A

Solution: Method 1: The boat must h~ad at certain angle upstream so that vector sum of ·velocity of boat relative to river and velocity of river relative to earth must be directed right across, -)

I

~7&G,.:

(relative to the riv.er and is to be iowards fight across, il(w/laq iiJir{ctiol!!.s/;lo.!!lsl.Jtb,ead? -':;,.:::.:,_ ' . . '/:_j

-)

v.·· :

, _.

Wthe b;;t-i>j'i,receding ,;:m,pl~ ,"ttd\>et_; with s~me 'jp~eij !

VB/R+VR/E

VR/E

sm8=--· VB/R

the

lQ

i.e,

2

vx =vB/Rsine-vR/E

-1(105) =tan-1(1)2

Hence time taken _!!__ = ~ = 18minute.

vB/R

= VB/E

Vy

VB/R

....

1

or 8 = 30° Method 2: , Consider point O to be. origin of a coordinate system x,y.

....

=tan

VR/E

sm8=--=-

Direction of swimmer's velocity relative to flow direction is (180° -8).

= ~rv-:~/-R___V_¼c--E= ~10 2 - 5 2 '= 8~66 km/h

->

·

~--

/2

2.

lv's;al=vssm8=Vsv1-l-;;';) =,iVs-Vw

I

Time taken to cross the river d d .d t = - - = - - - = ,==== Vs/G

t I.

v 5 sin0

~---~-0 .. -'"" E

y"

~v~-v~ '

,

.

~·

-~-·--1

nod

C!>nce~t: What happens if vs < v w .swimmer ca~ cross along sfiortest.path because v5 ,cos8 < vw•diift,will'.bej

always po1iff,ve .swimmer can . mqv~ ,right across qnly

if/

'-"-..'-'!-dre~----·-----~. ·- -·---·~----.'.....---·-··-~I v 5 >vw,

i'.

·

.•

,

Crossing River in Minimum Time If swimmer begins at angle 8 with µver bank, time taken to cross river will be given by d t=--Vs sin8 For tmin,

Sine must be maximum d

tmin = -

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Vs

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DESCRIPTION OF MOTION

93

Drift in this case N Concept: Swimming in a desired direction: Many times the person is not interested in minimizing the, time or drift. But he has to reach a particular place. This is' common in the cases of an airplane or motor boat. B

Flg.1E.116 ______ ....,. . - ----------------· .:.-:::Vmf ." _-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_ '

Solution: If bird is to move along AB, component of velocity of bird and wind perpendicular to AB cancel out . 3 (a) 4sina = 2sin37°:; a= sin- 1 ) 0 3 =} 37°+sin-1 with east.

(i

........... ·0················.

...,

v, Fig.1.103

The man desires to have this final velocity along AB in other' 'words he has to move from A to B. We wish to find the direction in which he should make an effort so that his actual' velocity is along line AB. In th~ method we assume AB to be' the reference line the resultant of v mr and v, is along line AB. Thus the components of vm, and v, in a direction: perpendicular to line AB should cancel each other. :

(io)

(c) .

...,

vm =vmr+vr

or

I,

-:J m = [vm, cosai + vmr sina]J +[v, cos8i- v, sin8j]' and, ~

Vmr

5

t = lOOx 5 8 + 2ffi .,

..

[·.- cos a= ffi] 10

5

=

250 sec 4 + ffi

··--,

.

A woman is running through rain at a speed of 5.00 m/s. :Rain is falling vertically at a speed of 20.0 m/s. (a) What is the velocity of the rain relative to the woman? (b) How far in front of her would an umbrella have to extend to keep the rain 'off if sh~ hold_s the umbrella 1.50 m above her feet?

Fig.1.104

...,

2foi = 8 + 2v'91

[,g-?::$

1B

i., -+

~- .

~ - - - - , - - - ----- - ............... --- -·------- ----.··-- I

:f~-4:··-·-

I

...,

VR/M = VR/G--.VM/G

4

(b) From Fig. lE.117,

-

...,

....... ....

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... (i)

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' DESCRIPTION OF MOTION

A river is flowing with a speed of 1 km/hr. A swimmer wants to go to point 'C starting from 'A'. He swims with a speed of 5 km/hr, at an angle ew.r.t. the river flow.' If AB =BC= 400 m. At what angle with river bank should swimmer swim ? Then the value of e is:

30° Fig, 1E.121 (a)

400 m -),

Solution:

"

=

VR/M

-),

Xj

=

--t

VR-VM

t

;M = 2v'3[cos30i+sin30j] = 3i+v'3J ...,

=>

,0

A

vR

rx

A

!

i-/3

Concept: Resultant path of swimmer is at 45° with bank therefore x-and y-components of swimmer's resultant' velocity must be equal.

(c)

Fig.1E.121

5=~3 2 +(x-v'3) 2 16= (x-v'3) 2 VR

A

Solution:

(b)

...,

C

Fig. 1 E.123 (a)

~ =>

400m

=-3i+(x-v3)j

'f

- Jo B

A

=>

Condition for reaching the point C 4+v'3 =

-->

X

VM

A

=-3i+4j

=>

tane = 3/4

0= 37°

·A pipe which can be swivelled in a vertical plane is mounted on a cart (see Fig. lE.122). The cart moves uniformly along a horizontal path with speed v 1 = 2 m/ s . At what angle a to the horizon should the pipe be placed so that drops of rain falling plumb with a velocity v 2 = 6 m/ s move parallel to the walls of the pipe without touching them? Consider the velocity of the drops as constant due to the resistance of air.

Fig, 1E.123 (b) Vy

tan45°= - , Vy=

Vx

Vx

(VR

+ VM COS0) = VM sin0 l+ Sease= Ssin0

On squaring, 1 + 25cos 2 e + lOcose = 25- 25cos 2 e socos 2 e + lOcose - 24 = o e = 53° On solving, We get

1__.§!',~~P'~ ·_124

Fig. 1E,122

Solution: Rain drops will move parallel to the walls of the pipe if their velocity relative to pipe is along the pipe.

...,

First we find v Rain. Pipe· --t

--t

V Rain, Pipe

= V Rain

--t

- V Pipe

-),

=V

-->

According to condition of problem velocity vector v must coincide with axis of pipe. This will occur if

v,

tana=-=3 V1

L>

The minimum speed with respect to air that a particular jet aircraft must have in order to keep aloft is 300 km/hr. Suppose that as its pilot prepares to take off, the wind blows eastward at a .ground speed that can vary between O and 30 km/hr. Ignoring any other fact, a safe procedure to follow, consistent with using up as little fuel as possible, is to: (a) take off eastward at a ground speed of 320 km/hr (b) take off west.ward at a ground speed of 320 km/hr (c) take off westward at a ground speed of 300 km/hr ((1.) take off westward at a ground speed of 280 km/hr

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----------·=·:__: MEC~~i~

veloci~:01

for

~Soluti~m"~!>nc;pt: Fin~i,,;;;:ae of l~irc~aft relanve to w~d. Iv'A,lwl> 30~~: .____...,:.. ____ ~J

Umin•

f(elmax = 5 ·

Umin=

'

16

'

_,

(3 sine - 4cose) = 16

5

VA =-VAi

.

3 sine - 4cose = 5 sin(e-a) = 1 e-a= 90" e = a+ 90° = 53°+90° e = 143° with river flow

_, A/W = (x+vA)t'

V

_,

I(v A/W) I= x + v A

~ 300

=>

=> VA~ 300- X x varies between 0 to 30 km/hr means v A in westward direction.

'

'

--

~

300 km/hr

----··- • -

' ._,,·J

IA man wants to !'.each poind3 on the opposite bank ofa :rivet/

in

l,zowing "at a.1peed. 4 mis as shown the Fig. lE.125 (a~.\ What minimrim:speed relative to water should the man ·have !.$othcit he can reafhl?oint B'.directly6yswiniming? ln which ~30 •...m..:_~---! ·.·. ::· ldirection_shouldhe,sWim? .~ • i. -., ! .I ,r---~==:"')--ll. I i : I I

i''. . I \

,

I

.

I

-

! • .1

.J .,

l

4m/s

40m

I I

,

JC--_ __,__ _ __

lL-,,====""""'-----""'""'""-~,.,.L--_-:_--::::-_-_-.:_:::: I _.-~~ Fig. 1E.125 (a)

:

~

-.. :so.both luti. ~---!l. .•,.-~--n.~e.pt:we .N.o.have te th·_·.a_ to ·.t.. s•_.·peed ~-,j swimme: _ ati._d. angle are: I(nknown determine function of ~.peed and maximize· ·;t .time ta~en 'far x-component qf,j

(.i.~~in1W~.fml>

oc~C:.Qllll!onent of::di:splace1JJ.l!.~-~l!.'ll!2,. ·

I

___ :. ,.,..

30';;;-·"·s'

.

fcrossing time: 'During the second. ,;tossi1!-&, his goal ~-Ni lminimize the distance that the boat is carryed downstream;}rt1 [the first case, the crossing time is-1'0 • -In the second cdse,

and

For B

urel Seel

-48g

= 0;

arel

1

= Ure1t +-are1t 2

->

=- 2

A hori2ontal platform is moving vertically upward with constant acceleration' a'. When velocity of the platform is 'v ', a particle is projected from the platform with velocity' u' relative to the ground at an angle 0 with the horizontal. Find the hori2ontal range and the time.of flight of particle on the platform. Concept:, In case of projection from a moving platform entire motion takes plane on platform, always use reference !frame of platform.

----.----

Velocity of projectile relative to the ground

g+a

t =0 T= 2Cusine-v) (time of flight) g-1:a 2ucos0(usin0-v) Range CR) = (u cos e)T = g+a

Projection of a Ball in a Horizontally Moving Trolley A trolley is moving horizontally with a constant acceleration' a'. When velocity of trolley is' v', a particle is projected with velocity' u' at an angle 0 above the horizontal from the position which is at distance 11 from the •y I front wall and 12 from the ; rear wall. This velocity and . angle of projection are hf v . relative to the ground. a 1·

_,

->

+'---,.,r-""'--.,,.,.-'----+! X'

->

A

I acceleration vector a are in the same vertical plane. Fig.1.106 _ --·--Find the time of flight and the horizontal range of particle on the trolley. Also discuss · the condition for whicli particle will fall (i) in front of point of projection (ii) at the point of projection (iii) behind point of projection

I__ _

la upward v j moving

~~,,~:~J

_,

Acceleration of projectile relative to the ground aP/gr

2

2(u sin0-v)

Velocity vector u, v and

= (u cos0) i + (u sin 0) j

->

2

or and

Projection of a Particle in an Accelerated Elevator

l

1 2 + 2 ayre/

=(usin0-v)t-.!.(g+a)t 2

t

= ISA/balloon 1-1 SB/balloon I = 480-270 = 210m

y

t

'

.

A

UYrel

0 = (u sin0-v)t - .!.(g + a)t 2

2

1 3g 2 · ' 2 2 SB/balloon= -27 g = -270m Separation distance between ' A' and 'B'

u P/gr

A

Yrel

3g

s,el =SB'balloon = 0---(6)

->

A

= -(g + a)j

a,.1

Y rel =

-480m

=

A

At the end of flight y-component of displacement of projectile relative to platform becomes zero.

2

(A falls off 8 sec) Srel =SA/balloon=

.

u,.1 = (ucos0)i+(usin0-v)j

1 2 = Urelt +zarelt (8)

A

Velocity and acceleration of the particle with respect to the platform

3g

s,., =0-21(3g) 2

_,

= (a) j

ap~gr

a,.,= -2 Sret

I

uP/gr

A

= (-g)j .'

->

and·

aP/gr

A

->

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A

= (ucos0)i+(usin0)j; A

= -(g)j

Velocity of platform relative to the ground at the time of projection and

I

A

a,1gr = (a) i

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.. ~· 102 .. ,..,-:o< ->

A

l----Concept:

A

u,., = (u cose-v) i+u sinej ->

and

A

·'A1

We assume that the flight completes on the floor of trolley. It does not strike the roof or the front wall or the back wall.

1

= UYrel t + 2 aYrei t

Y rel

2

If the di;ection of relative velocity is

,

.

!through .(ii) theWhenposition .direction of relative velocity does_ not pass of' A' then perpendicular' AN' f,wn the 1

':

.

· ·i

_4

Iposition of'.,!:' on the line of action of relative velocit_y(vB/A) !gives the m _ ·. _i_'nimum possible .,d_ista. nee between 'A_.·, a. nd 'JB' iduring their'_motion · , . ·. ·· , -···--•.. - _ _ AN= dsino: · · .. ,

0= (usine)t-~gt 2 2

2u sine T = ---

Thus time of flight is

(i)

:directed towards , th~ position of;A' then the body \Bl meets

A

= -(g)j-(a)i

a,el

MECl-l~NICSi!j

->

a is the angle which v B/A forms with y-axis.

g

For range ->

First we will determine velocity and acceleration of particle in reference frame of trolley. Horizontal range as observed from the trolley 1 2 Rre, =(ucose-v)T--aT

A

vB/A =-(vBsin0 2 +vAsine 1 )i

+(vB cose 2 -vA cose,)j (vB cose 2 -v A case,) tan ex = --'-.;...---"'---"--"-- (v B sine 2 -vA sine,)

2

As observed from the ground u 2 sin28 Rrel g

From tan a determine sin a and· cos a Time required to come closest is given by BN dcoso: t=--=--

Closest Distance of Approach Between Two Moving Bodies

->

lvB/~I

->

lvB/Ai

->

, Two bodies are moving with constant velocities v A and ->

v:i,

as shown in Fig. 1.107. y ~

Ll:E:~p~l~~ r1wo roads •_z_~.;erse~~ at righ/a_ngle-.s-._C_a_r-.A-.-is-s-itu_a_t_e_-a.a:·-;i which is 500 m from the intersection O on one of the roads.I Car B is situated at Q which is 400 m from the intersect/.6n on the other road. They start out ,at· the same time and, ti:aVel towarcls the intersection at 20 mis and 15 m/s respectively. What is the}ninimuni distance between them? How'long' do they tdke to reach it? _ _ _ _ _ · ·· ·

:' A_,; ::

. E. vi;~~-,~·

' '

:

..... i

:. : :-' .

...

.' .

..

.

:

.

~

••••• ')

~

:fd

.

. . . .S t .

J,

Vs :

..._.... _... --· i \

-'vl\.

a ,

'a

i

2

· vA

'i)J),JIV

500m

~ 20 mis ~ L _ - - - ! 0 P Car A

.

.. . -·········----····>x

400m CarB

B

Flg.1·.f07 ->

A

A

v A = (v A sinei) i+ (v A cose,)j

and

->

A

Fig.1E.138 (a) A

sine2H+ (VB cose2)j Motion of' A' relative to' B', is along a straight line in the VB= -(VB

->

direction of relative velocity (v B/A).

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DESCRIPTION OF MOTION Solution: First we find out A 20 the velocity of car B relative to A. e As can be seen from Fig. lE.138 (a), the magnitude of 15 velocity of B with respective : vA = 20 m/s, vn = 15 m/s, OP = 500 m; OQ = 400 m Fig.1E.138 (b) 15 3 4 tan0=-=-· cos0=-· 20 4' 5' . 0 = 3sm

0

B

5

3

= ADtan0 = 500x- = 375 m 4 .p 500 m BC =OB-OC e =400-375=25m BD = BC (cos0) 625m

OC

4

o c 375m

=25x-=20m 5

a e

Shortest distance = 20 m

..,

Fig.1E.138 (c)

_j

PD= PC +CD = 625+15= 640

Therefore, relative acceleration between them is zero i.e., the relative motion between them will be straight A line. Now assuming A to be at rest, the condition of collision .., .., .., will b e that V CA = V c-V A = relative velocity of C w.r. t. A should be along 0\.

..,

VA=

C ,n'

-

Vn=-5i-5v3j

VBA ..,

:. VBA

.. .

r.;:;;:;l • l}=~g,tp:12;1~ 11391.>

-

C

Am

Di

--- 30;60°·

A

~

1

r

10m

'T

r;;'

= -15i-5v3j

--'ss~=d=~o

10 d=l0-./3 m

Two towers AB and CD are situated a distance d apart as: shown in Fig. 1E.139 (a). AB is 20 m high and CD is 30 m high from the ground. An object of mass mis thrown from the top ofAB horizontally with a velocity of 10 m/s towards CD. ·

VaA

=-5i-5..J3j-10i

:. tan60°=

Bi

Fig.1E.139 (c)

-

5.Jam/s

10mis

lOi

..,

l~ABl=25m/S 640 t == 25.6 sec 25

'

D

Fig. 1E.139 (l>)

Fig.1E.139 (d)

l~c:;;;~m21~8> ,On a ftictionless horizontal surface, assumed to be the X·Y .plane, a small trolley A is moving along a straight line. :parallel to the y-axis [see Fig. IE.140 (a)] with a constant: ,velocity of ( ,Jj-1) m/s. Ata particular instant when the line'. :oA makes an angle of 45° with the x-axis, a ball is thrown ialong the surface from the oriiµn 0. Its velocily makes an' ·angle with the x-axis and it hits the trolley. ·y

Fig: _11:.139_ (a)

_..nA

Simultaneously another object of mass 2m is thrown from the ·top of CD at an angle of60° to the horizontal towards AB with the same magnitude of initial velocity as that of the_ first' .?bjec~. Tl'.e two ob!ects move in the same vertical plane, collidej m mtd-mr and suck to each other. : Calculate the distance d between the towers.

Solution: Acceleration of A and C both is 9.8 m/s downwards.

2

.· •'45° X

0

Fig.1E.140 (a)

:Ca) The motion of the ball is observed from the frame of the trolley. Calculate the angle 0 made by the velocily vector · of the ball with the x-axis in this frame. (b) Find the speed of the ball with respect to the surface, if • _ =_40/3_. --- __ _

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F ·' ,

. 1)1~(1!A~ld-il,S,I

- - - ~ · - - - - :::'::::::====::==· ·~';:;::;::==:'.:::::::::,~-_:'.:::.::..:J:::; Sol""•".' [,) ""A ~ml, fu, ®lley rnd B fud•lL ~· 1 I .

Relative velocity ~f B with respect to A(~_;!A) should be

alongOAfortheballtohitthetrolley.HencevBA will make 450 th an angle of wi positive x-axis. (b) tan0 = vBAy = tan45° or Further or

v BAy

v BAx = v BAx

... (1)

vBAy

=vBy -

v Ay

v BAx

= v Bx

O

VBAy

= VBy -(-J3-l)

tan.-(,

1

_:Column-2-.,:~:\',,bzt1§',

1. A dart gun is fired towards a Squirrel hanging from a

tree. Dart gun was initially directed towards Squirrel. P is maximum height attained by dart in its flight.

Three different events can occur. (Assume Squirrel to be a particle and there is no air resistance) .

.·~ -~·

Two projectiles are projected from a height such that they strike ground at the same time.

..· :

.... :

'

(B) :u 1 > u2;81 > 82 (Q)

v/:_f--::t.~"".'Tra)ectory of dart

0

J::-_..

rs"··· ...

d

Colurnn-1

'-'-''------- --------·

'

•

Two projectiles under standard' ground to ground projection such that horizontal range is

----------------~-

(A) Event-1 : Squirrel drops

itselfbefore the gun is fired.

(P)

same.

When dart is at P Squirrel may be at A

(B) Event-2: Squirrel drops '(Q) When dart is at P

itself at same time when the -gun is fired.

Squirrel may be ·at ,B Two swimmer starting from 'same point on a river bank such that time of crossing is same. u1 and u 2 are velocities relative to

(CJ :Event-3: A strong wind imp- (R) In gravity free arts same constant horizonspace dart will hit tal acceleration to Squirrel Squirrel.

and dart in addition to gravitational acceleration. Squirrel drops itself at the same instant as the gun is fired.

river.

(S)

(SJ :Dart cannot hit Squirrel in presence of gravity. 2. Column-1

shows certain situations with certain conditions and column-2 shows the parameters in which situations of column-1 match. Which can be possible combination.

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Person moving downward along slope in rain such that he ·observes rain vertically.

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--120 - - -- - 3. Figure shows a graph of position versus time graph for

-

y

a particle moving along x-axis. Parabola

X

Straight Line

t,

A

---! 0, Vy > 0, Clx > 0,_ ay < 0

,(Q)

:Vx

> 0, Vy= 0, a, > 0, ay < 0

'

'

' I l

...,1--,,~~-~-x· ;, _ Speed constant ,. y

' (P) greatest for Aonly

p

i

(B) ,uy/ux

: (Q) greatest for C only

(C) ux

: (R) ,equal for A and B

'

, (S)

'equal for Band C

'

...,S_p_e-ed-is~in-c-,e-a~si-ng•x,

-

(C)

-

-.

y

A . _ w a vector 'i:. at angle 0 as shown in the figure column-2. Show its unit vector representation.

p

--'---'-----•x

(A)

1

\Vx

I

(B)

(D) ux !l.y

',·(P) I

p

.o. (A) Time of flight

.

e

a

(S) t4 ->ts '(t)

.

- - - - J . - -..x,

(C) Moving away from origin (R) t3-->t4 (D) Speeding up

_,

,a·= acos0i-asin0j

,

Speed is decreasing'

- ------ - - I

ia

'., (P) _, = asin0i. + acos0j•

' > O,vy >0,Cix > 0,ay>0 (S) .lvx 7. A particle is moving along a straight line. Its v-t graph is as shown in figure. Point l, 2 and 3 marked on graph

are three different instants. Column-1 has fill in the blanks, which are to be filled by the entries in column-2.

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-

-

--

! DESCRIPTION OF MOTION ~---·----~--- -------~

i (S) :Magnitude of velocity

->

V

(D)

di rl

I

I '

. dt

i (T) :None --l-------r

10. For the velocity-time graph shown in figure, in a time interval from t = 0 to t = 6s, match the following :

Column-1

.. -v(mls) .

(A) a1 is .......... a 2

(P) 'Parallel to

(B) v1 is .......... V2

(Q) Anti-parallel to

i0

0

(C)

I

: (R) ,Greater

•

V3lS .......... V1

than

(in

magnitude)

'.

(S) ,Less than (in magnitude)

(D) a1 is .......... v 1

(A) Change in velocity

8. Figure shows a cube of edge length a. ·y Ht,1------,,G

!' (Q) .:_ 20 SI unit

(C) Total displacement

: (R) :- 10 SI unit

'

X

_,,;/ ..-.,

'

. ,' .:·9;

. ...,,,r; ·.:r.olumn-1 . -..i · .. . · Columr1,2)': !-----·~---~-------~-------·~;....;.,;..,:.;:JL (A) The angle between AF and x:CP) -axis

·_ 5 SI unit

i1:~,z,:.,_ .£'."- ..'*t~}J.l

.· -~~9.lu"!n-2

60° (A) ,M

'

'

(B) Angle between AF and DG : (Q) 'cos-1 _!. 3

(C) ,Angle between AE and AG , (R). 'cos-1

'

= 3 s I' (S)

s· N velocity is positive increasing. ' A -I when velocity is negative and increasing. R when velocity is positive and decreasing and R- 1 when velocity is negative ··-· and decreasing. Now match the following two tables for the given s - t graph :

C

D

'

11. Let us call a motion, A when

"-'A'-_ _,__ _...J..CB'-

z

i- 5/3 SI unit

'

(B) !Average acceleration

(D) Acceleration at t E ,----'f---1',F

; (P)

'

J_

f

i

(Q) IR-1

(B) ~N

I

(C) 'p

'(R) 1A

'.Q

I (S) iR

(D)

-.J3

! (P) iA-1

'

'

(S)

9. Match the following : -:::-:-

_, (A)

i (P)

Column-2 .

. '¼~;~

~-~~

Acceleration

I~

dt ->

(B)

:di vi

: (Q) 'Magnitude of acceleration

' dt ->

(C) dr

, (R) 'velocity

(C) Maximum height

, (R) '45°

(D) 1Horizontal range

I

,

I

I

'dt

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;

I

' (S) tan-I

( 1)

2

Anurag Mishra Mechanics 1 with www.puucho.com

122

----~-~--..;....________.;.;.___ M_ECHANl(S'\J

= lp + 20 t

13. In the s-t equation (s following:

·_~tf:c~_-r:,· ~:x_·'.'.?-ii:_ m) are allowed to fall from the same height if the air resistance for each be the same then both the bodies will reach the earth simultaneously. Reason : For same air resistance, acceleration of both the bodies will be same 4. Assertion : A body is momentarily at rest when it reverses the direction. Reason : A body cannot have acceleration if its velocity is zero at a given instant of time. 5. Assertion : A particle in motion may not have variable speed but constant velocity. Reason : A particle in motion may not have non-zero acceleration but constant velocity. 6. Assertion : A particle in .zy-plane is governed by x = a sin rot and y =a-a cos rot, where a as well as ro are constants then the particle will have parabolic motion. Reason : A particle under the influence of mutually perpendicular velocities has parabolic motion.

(P) /Zero

(C) Displacement of particle

**h.

: 'l":r1,s. ~~'%

~_,: o umn""'

(A) ;Distance traveled in 3si (P) - 20 unit

.

6-;:::;-·

rn

~e. .

~~-·~~~·i~-i-;-:~·

.I

,._,.«:.;:;:,µ;;q,..w,::;>iim:s;.a:c.

/10 SI units

(R) 140 SI units·

I (S)

'' J20 SI units

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[ DESCRIPTION OF MOTION

--

,-·-·--·

i

·123 ···--.-:----·

--- ---- ·-··-----~~--------

----~- -

AN8WER9 - ----- ---- - -- - -- -----------------

-

-------- -··

=

7

··_J'

------·

i;;vel-1: §niy One Alternativ£is Correct.~

1.

(a)

I

9. : (c) 17, (b) 25. ' (b)

10. 18. 26, 42.

49,

50.

57. ', (d) 65.

1

(b)

73.' (b) 81.' Cal

=-

3.

(b)

11.

(b)

I

(a)

I' 27. i 19.

(a)

' ''

58.

I

(b)

35, 43.

(b)

51.

''

(c)

4,

(c)

12. 20.

(b)

I (a)

5.

(!;,)

6.

(d)

7,

(b)

8.

(c)

I

(b)

13.

(d)

14.

(b)

15.

(d)

16.

(a)

(c)

21.

(b)

22. : (a)

23.

(d)

24

Cal

30. i (b)

31.

(c)

' I 32.

(d) I

39.

(d)

46. : (d) 54. I' (a)

47.

(b)

''

(a)

63.

(a)

71.

I (a)

(d)

79,

(d)

I

I' 28. !' (c) : '

(c)

I·' (a)

(a,b,c,d)

7,

I

13. 19. 25.

:

I

2.

(d)

45.

(b)

(d)

53.

(d)

II (a) (c)

61.

(b)

62.

69,

(c)

70.

i 76. I (d) 84. I Cal

, 77,

(c)

78.

Ca)

86. I Ca)

44,

I

(c)

52.

59. ; (b}

60.

,

' (b, c)

.I

I

! I

I

68.

3.

i I

(a, d)

14. I (a)

I

(a)

20. j (a, d)

: 21.' (a, b, c} : t' - -

I

I

Cb, al

I

I

' 85. :

I'

'

I' '' i

55.

I

40,

I

I

I

I

(b)

I

(d)

I

(b)

56.

(b) ' (d)

(c)

64.

(c)

' 48.

72.

I

'

(~)

' 80. I, (c)

! '

9. I {b) 15. I' (c)

I

'

4,

(b)

10.

(a, c)

16.

Cc)

22.

I'

(a, b, d)

I

i 'I

I

5. 11.

' 'I· 1 I l

(b) (a, b,'c)

I

17,

'

23. \' (a; b, c).

'i I

I

(c)

'

3. (b)

4. (a)

5, (b)

6. (c)

7, (c)

Passage-2: 1. (a)

2. (c)

3. (a)

4, (d)

5. (a)

6. (a)

7, (a)

Passage-3: 1. (a)

2. (c)

3. (b)

4, (d)

5. (b)

=sMatchhl!!~!Ype P_rob~-~~~ 2. A-P, Q, R, S; B-Q, R, S; C-Q, R, S 4.A-R;B-P;C-Q;D-S 6. A-P; B-Q; C-R 8. A-R; B-Q; C-P

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: ,6.

I

(a)

'

i ; '' 12. i,I (c, d),

i

.I I

I

I

I

2. (a)

A-P, Q, R, S; B-R; C-R A-R; B-P, S, T; C-Q, R; D-P, S A-S; B-P; C-Q; D-R A-P, R; B-P, S; C-P, R; D-P

I

I

'

Passage-1: 1. (a)

1. 3. 5. 7.

I

I

-~

(a, b, c)

' 8. !' (a, c)

(b, c, d)

' 38. '' (b)

(b)

(c)

83 . . Cd)

'

37.

i

67. ' (a) ' 75. j. (b)

I

(c)

36.

L~vel-~: Mcir~!han o~~-Aii

8=45°

~ - - - ---+--+ ---+--+ 2+1B1 2+2A IAl .B - = ,~~~~----+--+ --+--+ --+--+ --+--+

27. (c) We know if a vector makes (al+ bj + ck) an angle a, J3 and 'Y with x; Y and Z respectively then a b cosa. = cosJ3 = --;===== ~a2 +b2 +c2 ~a2 +b2 +c2

--;=====~=, C

cosy= ,===ea~== ~a2 +62 +c2 2 So, sin a.+ sin 2 J3 + sin 2y = 1- cos 2a. + 1- cos 2J3 + 1- cos 2y = 3 - (cos 2a. +cos 2J3 +cos 2y)

A+B

=..J5 33. (d)

Let angle between the two vectors be 8 3 2 =7 2 +4 2 +2x7x4cos8 cos8 = -1 => 8 = 180° Cross product will be zero. 34. (a) --t

-->

•

--+

A

A

A

--t

A

P+Q =lli+9j ~

A

A

-->

--+

IQl= 5 . --+

--+

--+

--+

AxB=12 A+B=7 A=4, B=·3 A=3,B=4

--+

Given A .l B (Le., component _of B along A is 0) --> -->

A.B=8+24-4x=O X=B

-->

... (i)

... (ii)

-->

:. Minimum resultant· is A- B (when· they are antiparallel)

30. (b) -->

BIIA

-->

--+

IA X Blmax= IAIIBI= 12

29. (a)

-->

--t ' .

Let vector be A and B givenA+B=7 (when they are parallel resultant is maximum)

Q=4i+3j

--+

-->

Area of parallelogram = IA ,x BI = 5 units 35. (a)

•

P =7i+6j

--+.--+

-->

-->

IA-Bl min= 4 - 3 = 1 i.e., unit vector

-->

=> B= kA => 61+ 16j:+-xfc = kC3i+ sj-2kJ => =>

A

-->

28. (c)

Given

A

A= 2i+3j, B= i+4j

=·3-1=2

=>

IAl 2-t1B1 2-2A.B

A-B

36. (c) Consider a hexagon with all sides equal

k=2 X=-4

E

F1

A

F,: B

0

31. (c)

The displacement will be maximum if he walks in the way as shown after walking 20 steps displacement is sJz :. He will walk 40 steps for displacement 16../2 m

,I'

/4~_;.-,/ '..

,.,,/e_....-·/ ('"

:.-···

IA.. ··-,··.. (

Bm

8m

8 ni . .

/

Sm/ •

, ,10s!eps /

,

•

//

--+

--+

--+

--+

--+

-:7

(By polygon law) Resultant of the five vectors F1 will in opposite sense of F2. . Therefore resultant of all the given vectors i.e. (SF1 and F2) will be F2 -F1 or F1 - F2 AB +BC+ CD+ DE+EF=AF

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- - - - - - - - - - - - - _________ fu]

D.ESCRIPTION OF MOTION

=KIN§Atj~

=>

·21 dl = 2x dx + 2b db dt dt dt

39. (d) Let magnitude of acceleration be a. Also let west to east be positive direction. Hence

=>

2l(u)

xA = Ju(l0)-½(a)(10)

2

= 2x(:) + 0

=(:)

I= JlOu - SOaJ

l

u cos0

=-U=--

Xn = Jcu -10a)(l0)-½ (a)(10)

2

X

I= JlOu- lSOaJ

(lOu - 50a) = 11

Let

xA = l11I, Xn = 111 -lOOaJ Hence x A may be less than equal to or greater than x B depending on 11 and a. [For example: If O < 100a < 211 then xA > Xn, if 100a = 211 then xA = Xn, if 100a > 211 then xA < xB ; under the condition 11 is a positive quantity. However if 11 is a negative quantity then xA.< xB]

41. (c)

Let both the balls be thrown with speed v O and let 2gh and height of the building be h. Hence vi =v

Alternative : Let the veloc/ty of the block be v upwards. Hence velocity of the block along the string is v cos 0 and perpendicular to string v sin 0. Hence VCOS0 =U

46. (d)

_, _,

d

\

T

of

aP-

a, 2

= 4ft 2

... (ii)

_,

as,- a,1

= 4ft 1 -

I;,,-;,, I=

4ft 2

~4 2 + 4 2 + 2(4)(4) case

= 8 cos~

.

d

2

=> Here 0 is the angle between ft 2 and-ft 1. 47. (b)

-+-

6

43. (c) Since both have same initial vertical velocity (zero in this case) and displacement along vertical axis is also same for both when they strike the ground therefore time of flight is same for both. 44. (d) 2 . 300 2 . 900 R - UA Sill d R - UB Sill A 2g an B 2g

Hence RA and RB depends upon initial velocity of projection which is not given Le. , information is insufficient.

z2 = x2 + b2

... (i)

= (4-Jz).Jl + case

So average velocity=~ = ,4 m/s

45. (b)

= 4ft 1

_,

-x2+-x6=-=>T=2 2 3 6

12

a,1

where ft 1 and ft 2 are unit vectors. Subtracting (ii) form (i)

d

2d

aP-

_, _,

m=--=-sec 3x4 12 Let body travels for next T sec then T

_, _, _,

v~ =v5-2gh =>vA =Vn,

law

u

= -COS0

Let a p, a,1 , a,2 be accelerations of the particle, fr3.!11e S1 and frame S 2 with respect to ground. Hence

5-

[Note that v A = v B also follows from conservation of mechanical energy] 42. (b) Suppose the total distance be d. Time taken for first d/3

=>V

;_u_ --------:_i

12 = u(l) +~(a)(1) 2 = u +~

2

2

... (i)

12 = (u+a)(¾) +½(a)(¾r

3u 21 =-+-a 2 8 a= - 3.2 m/s2

... (ii)

Solving 48. (b) __, __, _, r = r 0 (t -at 2 ). At t = 0, r = 0. Hence the particle returns back to initial position if velocity of the particle = dr = r0 (1 - 2at) dt So, particle will come to rest when v = 0, i.e., after time

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1 t=2a · Positio11 of particle at this m9ment

= r0( \,so distance travel

=>

¾[(u +~u 2 - 2u J-(u-~u = 3.,Ju(u -

!-

=:

-2u )]

2)

a

a x (~)2 )

2

52. (d) -+

= double of the above distance = _:9_

·-:

'-+

'::

/2

2

V =Vxl+VyJ =>lvl=\/Vx +Vy

dl;I = ( 2vx ~+2vy~)

2a

dt

.,fv2+v2 ~

y

X

"7

div I= Vxax +vyay dt fv2+v2 \I X y

=3x2+4x1= 2 m/s 2 .JJ2 +42 53. (d)

50., (b)

,_,' (dxJ·'i (dyJ· ' v = dt

"7

s=4t+.!.(l)t 2 =2t+.!.(2)t 2 2 · 2 4t + 0.St 2 = 2t +t 2

''

+ dt j = ai + a(l- 2bt)j

'

Solving we get, t = 0 and t

'

A,= 0 i + (-2ab)j ...

,_

'

-

s=4x4+.!.(1)4 2 =24m

So,

-+

·· · Hence acceleration A is along . ·, ·, .- ·, ··:' . ,-;, 'negiitive y-axis. Hence when A

= 4s.

.

2

-54. (a) "7

'

'

= (ay)i + (V0 )j Vx = ay and VY =.V0 v

"7

,,and· v enclose it 14 between them the velocity vector makes ·.angle ·It/ 4 with negative y-axis. Hence

dx -=ayand dt dy V0 -=-=>· dx ay , 1 . .2 -ay =V0 x+c

· tan 2: = a => [1- 2bt[ = 1 . · 4 1ac1-2bt)I . t = - or 0 b But when t = 0 the y-component of velocity is along positive y-axis, hence t = 0 rejected. . 51. (c) . Let at .~y time t the displacement of first particle b~ S; and that of second particle be S 2 • 2 ·. S1 =½at and S 2 =u(t-~)

For required condition S2 > S1 1 2 =>t 2 --t+- u ( t--1 >-at a 2 a a2

J

dy. dt

-=V0

f aydy= s·V dx .

·

0

.

2

1 ay 2, = "v x, 0

(·: (0, 0) satisfies)

2

' . 21'

y=± __o_x ---!!,_,, negative

-~2V 0x · y.-, a · Also for y to be real x must be negative. 55. (b)

0= 30t+.!.(-10)t 2 =>t = 6

2

=> ¾(u-~u -2u) < t < ¾(u+~u 2 ~2u)

2

Hence the duration for which particle 2 remains ahead of particle 1

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_.-F:'o'ES(RiP.Tfo":q~LM9YI~'.,.,,...___.....,. . · ~~-- .. , · => =>

../2

t · ~.

85. (a)

../2

2:f

,

J.

y 2·2t 2

80. (c) Let v be the velocity of the particle when it makes 30° with the horizontal. Then v cos 30° = u cos 60° => v = 20/ ..J3m/s

So,

j..

'

=>V

"'dy = 4t

· .•

-~dt ...._~· ::._.Y ,-, \

- -·

'V

·,

4t

+-='--2 = 2t ·v; '

Differentiating with respect to time we get, d0 •· .(sec2 0)-/= 2 ,__ dt ,, .. . ,. 2 ' d0 . d0 =>. (l+tan 0)-=2=>(1+4t 2)-=2 dt -· dt , ._ ,, . dB-;;· ... z' .' => · ,r,·" ;. ·dt · • 1;+:4t 2

R

15.4m

81. (a) Components of the velocities of both the particles m vertical directions are equal. Therefore, their time of flights are equal and their relative motion is in horizontal direction only. Thus the maximum distance between them is the difference between their horizontal ranges.

,,

· - , · tan0 =

v2 gcos30°=v2 R=--g cos30°

:;

,,,_,., 1·:;r dx . ·x = 2t;=> V = - = 2 X ' dt

=>

Now

decreasing

continuously.

or s ={h-(v-vsm0)t} 2,+(vcos0t)2 s is minimum when 2

=>

is

Hence

2

-rad/s p ,[::1:2 ~r+ ~2) 2 17

=>

f!6.

11

,

C~? ,:.. ~-

.,~ ·;.~ ·-::: 11

·,1·L

From V·S gti;'PP. ' "'

1

...:·nn '.

1 ;__.

'

1 ... ,

.. .

'·duds ·• V =·S l

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.... l'. l11;_'

•

=> ....,.....-= - =>a= V "dt dt

···~t''

l

Anurag Mishra Mechanics 1 with www.puucho.com

; MECHANIC5:~ . --+ r1-r2

--+

'.!.· ca, b,

c, dJ Graph (a) ~dicates two c\lsplacements at a given time, '. · ' which is imposs)b!e. . Gr~ph ·(b) 'iridj~tes. two velocities at a given time, which is impossible. (ri-r;)= (v-;-v;}

--t

(v 2 -v 1 )

/v 2 -vd

/r2 -'r1 /

R = l!2 sin20 => dR =·(u2 sin20)(-l)dg ' g g2

s.

-,

Fpf the particles to collide with each other, the particle 2 must be moving towards particle 1. Hence

4;. [b]

=>

-,

-,

r2 w.r.t.1 = r2 - rl

!

.Jv2 -v2 ~. V·I-~2 .... v2

.

1

I 8"· ..... .

l

Similarly . · .

ii'2- ii', 1

11, - 121

Tii~:lift is /ICC~lerating downwards with acceleration g. . · !ieic~. a~~~ler'!,tion of sto~e in lift frame is g - g = 0. 3. [a,'~, ·er . . . · (i) is blowing alongfIB. Hence total time T for the f X = -Jb 2 + a.2 + 2ab COS0 => la-bl:,x:,Ja+bJ . Hence 8 is the angle betwe~n ft 2 &·-ft,. 10. [a, c] If the velocity of 1 · · paclEastj _ parabolic with respect to ground as shown in figure. . 2 With . respect to 4m~s . ···-·>4m/sZ-7 l 2 . aeroplane the initial 1',· .. ••• O~s \ • I velocity of the packet is I' 1West c-'-----·-----······>East ,- . gro~risJ._ _ ___, zero and acceleration is as shown in figure. -1 5 -1 1 . a]'. west..of vernc 8 = tan - = tan 10 2 11. [a, b, c] 2 h _ (v 0 sina) h2max => (a) is correct lmax 2g cos8

rs

1

... (i)

p-B= bn 2

... (ii)

2v 0 sin a (b . · · T1 = - ~ - - = T2 => ) 1s c.orrect g case . R, =(VoCOSa)T, _.!gsiilBT,~ 2 '

R2

=. (v 0 cosa)T2 + .!2 g sinBT,,2 (R 2 -R1 }= g sinBT,2

=> (c) is correct v ,, & v ,2 are the velocities of the particles at their maximum heights. Let the particles reach. their maximum heights at time t 1 and t 2 respectively. Hence 0 = (v 0 sin a)- (g cos8)t 1 v 0 sincx t, = ~ -gcos8 v sincx · · t2 = 0 . Hencet 2 =t1 Similarly gcos8 Hence

v,, =v 0 cosa+(gsin8)t1 v,2 = v 0 cosa+ (g sin8)t 2 .

·

vt1 -:t=vt2·

12. [c, d] •

·Hence ft 1 and ft 2 ate unit vectors depending upon direction of acceleration of the particle with respect to respective observers. Subtracting (ii) and (i)

I

rs, . ,

=>

...,p-A ..., = an

A-B=bn 2 -aft 1

JA-BJ= Ibft~ +ac-11 1 )[°

=>

smrot

=>

x2+(y-a)2

=-Xa and cosrot =1..-· ·a-y ·=a2,

which is a circle. Hence (c) is-correct.

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, '7._,>;;:,~ (n .•

1 ,•,·. c1.1 1

"

-,ci

L..,.,,.

~fJ' \'

·c1x ., ' dy . vx = - .. = arocosrot and vy = -.-·= arosmrot 1 dt. ~0and-·· - > 0 and - 2 < 0 dt dt => s-t curve is increasing _and lies below its tangent. ·· ,

'

P' = p_p t . ~rv~f_+_V~~-+-2-'-V-1_V_2_C_O_S_CY.

sinu

·

dv

ds < 0 and d2s > 0 dt dt 2 => s-t curve is decreasing and lies above its tangent. ' · · ·, . dv t:9~. 10~ V > 0 and-> 0 . dt ds d 2s : ->0and->0 dt dt 2 => s-t curve is increasing and lies above its tangent. 17. [c] The ball will stop after a long time. The '.final displacement of the ball will be equal to, .the height. The motion is first accelerated, then retarded, then accelerated and so o~. =>

The velocity of the particle first increases linearly and then at the point of collision it suddenly changes its direction and then starts decreasing·in magnitude and the pro~ess is repeated again and again. Also every collision decreases the speed to half its value before collision. Hence graph given in option 'a' is v-t curve and th.at given in 'b' is speed,time curve. 19. [a] v 1 =a2 (t:t-t 1 ) (forcarB)

V +v

a t Vi a2 t + t 1 a1 .. ·(V+v 1)(t+t1 ) -= > 1 ==>a1 >a2

1 =1- -1---

vfinal

=s •

U2

For case B =V1,T=t1 +t, distance=s (V+v 1 ) = a1t 1 · (forcaseA)

. ,vfinal

dv

t:5~ 9~v < 0and- < 0 . dt

16. [c] For case A we C'l,Il write . , ,

=v.+v1 T = t 1 , distance

.

. t:2 ~ 5 ~ V < 0 and - < 0 ' dt ds · · d 2 s - 0and-= 0 ' ' dt ds · d 2s · ->0and-=0 dt dt 2 => s-t curve increasing and a straight line.

.

v1

20. [a, d] So, velocity of first particle

' ..:.::' '

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I DESCRIPTION OF MOTION

,,. ., ,~ . '

23. [a, b, c]

zero. . 1 . el oCity=---=-=. . 16 9 7 14 m/s Therrreanvev

5

5

21. [a, b, c]

dv a=-=A-Bv dt """7 max. possible velocity is terminal velocity (i.e., when a= 0) => A-Bv=O """7 initial acc. is when t = 0, u = 0 a=A-O=Am/s 2

f

f,

dv dv- = dt -=A-Bv => v- dt OA-Bv o ..!1nA-Bv =-t =>1-.!l.v=e-"' B A A A(l -Bt)-e -v B 22, [a, b, d] ·

v 2 =2 2 +2xax~ 2

=> v2

= 4+ 142 -

v 2 =4+

v = 10 m/sec"""7 if AP=.! AP=~ · PB S 6 Let velocity at P is v 1 2 2 ' d . 142 -2 2 v 1 =2 +2xax-=4+---

6

Let time taken to reach mid-point from A is t 1 , and t 2 be time taken to reach B from mid-point. 6= 2+at1 ••• (i) 14 = 6 + at 2 .:. (ii) t 4 1 · ...!. = - = - => t 2 = 2t, t, _8 2

24. [a, b, c, d] Since the graph is like a· parabola :. let x(t) =At+ Bt 2 + C

(dx) dt Put in (i), we get

_,

dj vi = tangential acceleration dt

_,

dv ' dt

*0

6

=> v 1 = 6m/sec

nme

=0

192 =100 2

x(4) = 0=> 16B+4A = 0

distance > displacement :. Average speed > Average velocity

dt """7 In uniform circular motion

22

2

From graph x(O) = O => C = O x(t) = Bt 2 + At

Total distance Total time . displacement Average velocity = - ~ . - - -

Average speed =

d!vl dt

I

j

14 in/secl

142 =2 2 +2xaxd at mid0 point let velocity is v

So, relative horizontal velocity is zero. So their relative velocity is vertical only. Since both particles are moving under gravity, so their relative acceleration is

_,

.1

~ - " - - ~ -B IJ. 'A- - -'p

=>

_,

13sl

~sec

= s 1 +sJ

dj vi = net acceleration

r

- •.·' d '

= 3cos30°1+3sin30°j 12: 9: =-1+-J 5 5 velocity of second particle = 4cos 53° i + 4sin 53° j 12: 16:

5

t-,.

[Q]

"""7 In circular motion from pt. A to Pt. A again Average velocity = 0 (at any time) lnstaneously velocity ;e 0

o

... (i)

=1 =>(A+ 2Bt),=0 =1 '

1 4

B =-- -

t2 X=t-4 max. x coordinate·= 1 (from max. and min._) """7 Since motion is a straight line motion """7 total distance traveled = 2 x 1 = 2m(

. 2 . Average speed= - = O.Sm/sec 4 25. [b, d] Separation between. them will be maximum when both particles have same velocity. This situation come at t = 2 sec, but just after it, first particle comes to rest and second 1 m/s. So first particle will again gainthis velocity in next one second. So, maximum separation will ocCIIr after 3 seconds.

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j'---1 c.:3~6-·-------~---~_;_________ ~._ __,___ __o..',.,_.-...;·.__..;___._··...:.M_EC'--H-'-A.,__Nl~Cs_:1..:J! . Maximum separation = Displacement of seconds particle - Displacement of first particle in first 3 seconds = (2x 2+ lx 1)-(.!. x 2x 2+.!.x1 x 1)

·

=5 -

2.5

2

2

= 2.5 m

• . ·&;e1.'i;:c;;;;"'eh=e=n~""i;-n""'"ii=--se"'"d""''i,-;-;;-b""j;=;"'~,;------. lll,::e::--z:r-,rc::::::e:r::!::'t·-n·

•

4

c;-=ur

~ j@&s,

Passage-1

·

dv

= 96-128 =-32 Relative velocity = 4 - (-32) = 36 mjsec 7. [cl 2 V = 12t-2t

a= 4(3-t) =>- = 4(3-t) dt

=>

f:dv= J;4(3-t)dt

=>

v=l2t-2t 2 2t 3 2 x= 6t - -

=>

5. [bl JA ~ 4 ,m/sec .\ At t=2sec (from Q. 2) [!~16m/~cj They meet for 1st time v(2)=24-8=16 Relative velocity = 12 m/sec 6. [cl At t = 8 they meet for 2nd f A.,_-.,:4m(sec; time !32.m/sec I B 1 1 v(8) = 12 X 8 - 2 X 8 2

3

= 18 -

1. [al

For particle B to stop v = 0 => 12t - 2t 2 = 0 => t = 0,6

2(t 2 - 6t + 9)

Passage-2 1. [al

2x 63 x(6) = 6x 6 2 - - -

500 t=-vcos8

3

5

=63(1-¾)=6: =2~6=72m

t mm · = 00 for 8 = 0° V

2. [al -

For particleBx(t) = 6t 2

2t 3 3

For particle A x 1 (t) = 4(t +

2

For u = 3 km/hr, v = 5 km/hr 500 t . = - - ' - - 360 sec = 6 min. mm 5x1000

J)

For particles to meet x(t) = x 1 (t) 6t

2. [cl

--

3

3. [alu = 3 km/hr, v = 5km/h to reach exactly Pt. B vsin8 = u => 5sin8 = 3 . 8 3 Slll = -

8)

2t = 4 ( t +3 -3

t=8,t=2 time interval = 8 - 2 = 6 sec

5

I .

~¥-·- --- - -··· --

5

= 0.5 hr = 60 X 5

4

4

. t =30 - =75 . mm 4 V

= 12t- 2t 2

4. [dl

x(6)=72m x( 8) = 128

U

Total distance;;,'d + d1 = 72 + ( 72 304 =-m 3

= 5 km/hr, V = 3 km/hr

again to reach Pt. B vsin8 = u => 3sine = 5

3

1 8 ~ )

sin 8 =

~ not possible 3

:. The swimmer can never reach to Pt. B

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··.··11

~ ·, .· :,

500 0.5 t=--=-vcos8 5 xj

3. [bl Required position x(t) at t = 8 or x 1 (t) at t = 8 1 8 =4(8+¾)= ~ m

4. [al For particle B

GZJ, r-~

-

....,.

,·

. .

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[_•._DE_SC_R~IP.J~l~~-O~F_'~~O~T!~O_N_._ _

~--·-·-----------------,____-·_··~~--,3IJ

5. [a] From question 3 required angle '0' is . 0 =3 sm

4. [d] At the top of trajectory speed = u cos0 = 5../s m/sec

5

i.e.

5. [b]

u 2 sin20 Range=--g

0 = 37°

6. [a] U

= 4 km/hr,

V

= 2 km/hr

=

d vsin0

t=--

drift

vsin0

l

=> Speed time curve will be ,- . -----7

~

!·~ -

----·

v = -J250 = 5-Jio m/sec usin0

--

r··--·1

t t / ! i~~~. '---~~-· -----~_:'.'.~_~j

!

_,s-[fo "

u 2 sin 2 0- 2x !Ox 12.5 = (5-J5) 2

u sin0 = .J125 + 250 = .J375 and u cos0 = 5../s = -/125 tan0 = ~375 = F3 . 125

-t 4 ) '

(B) (x) ( : ) < 0 to return (P,S, T). (C)

x(:)

> 0, i.e., in (Q, R)

(C)

Slope of v 1 is+ve } Slopeofv 3 is-ve

anti-parallel

(D) Slope at sis +ve

a1 > 0 :. parallel V1 > 0 a1 and v 1 cannot be compared.

-+AA-+

16. AF=ai+aj+ak

AA

DG=ai+aj-ak

b= i (A)AF · b = AFcosa 1 cos a= F3

=> a=..ffacosa

0 = 60°

->

. ->

(B) DG ·AF= (AF) (DG) cosp

3. [b]

· and

3

(D)Speed in increasing in (t 1 -t 2 )(t 4 -t 5 ) 15. (A) Slope of a1 and a 2 is +ve parallel Slope of a1 > slope of a 2 R (B) Both v 1 and v 2 are +ve parallel v 1 < v 2 (obviously)

PQ = .J15 2 + 20 2 = 25

15'[5-·---··-- ·'"

4

· - - _:,______:__i

'-··

v 2 sin 2 900 v 2 25=----=g 10 2. [c]

l~Ol · I 1,i,(1 1,; I,, i speed decreasing in·(t

------ u=2I

Let speed at P = v Range

25.)3 m

n/C',u,.l

From this we get sin 0 = .!. => 0 = 30°

.

lOOx sxF3 2xl0

11. (A) Corresponding w graph will

x=(4-2cos0)x~ vsin0 . . . For x to b e maximum or m1n1mum -dx = 0 d0

Passage-3 1. [a]

10

Matching Type ·Pr~em?i;;--:--_,

= (u - v cos0) x _d_

7. [a] Here v>u Minimum drift = 0 => 4sin0-2= 0 0 = 30°

=

(10v'5)2 sin 120°

ucos0 = 5../s 0= 600 u=

5../s = lOv'S m/sec 1

2

a 2 = F3a F3a cosp ->

•

,.

(C)AE =aj+aK ->

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->

. ->

•

•

AG=ai+aj

AE ·AG= (AE)(AG)cosy

cosp

= -1

3

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...,_ Tl

\,~

\\;@ r

,. -·

"

~

,

\:

\

\ }

\\.,(''V.:'_y ::~--,>':'~=r-· : .-.:":;: ~:b"'"."'t:' · ..;.;..._. :· 'iI{ff,!j{{fliil:;;, >: ' "G:l~~-... """ ~, Fwp the weight can be eitherl in a downward accelerated motion or in an upward: decelerated motion. More precisely, when the acceleration is· directed downwards the velocity can have an arbitrary, 'direction; it can go upwards or downwards or even form an angle with the vertical. The direction of the velocity at a given; moment has no direct connection with that of the acceleration: ,while the acceleration itself is completely and uniquely :determined by the acting forces.

F,p

. ' ~

If the acceleration of the weight is equal to zero, the sum of the forces acting on it must be zero; in other words, in this special case the force Fwp of the action of the palm on the weight is equal in its magnitude and opposite in its direction to the force of gravity Fwe. In these circumstances the weight can be in a state of rest or in a uniform rectilinear motion with any constant velocity. CONCEPTUAL EXAMPLE-3 : A weight is suspended

from a spring attached to the post placed on the table, we consider the interaction of three bodies: the weight, the spring, and the Earth (as has been said, the Earth together with the table and the post form one body). The forces taking part in this interaction are shown. The earth acts on the weight with the force Fwe (the force of gravity of the weight) and on the spring with the force F,e (the force of gravity acting on the spring). The weight acts on the spring with the force F,w and the post (considered as one body together with the Earth and the table) acts on the spring with the force F,p. According to the third law, we always have the equalities Fwe +Few= 0, F,w +Fw, = 0 and Fsp +Fp, = 0 Assuming that the magnitude of the mass of the spring is negligibly small (and only under this assumption) we can write, on the basis of the second law, F,p +F,w = 0 Condition shows that the force of tension of the ("massless", i.e., 11inertia-free11 ) spring is in all the circumstances the same at both ends of the spring. In this approximation the magnitudes of the forces acting on the ends of the magnitudes are equal to those of the forces acting on the ends of the spring are regarded as being precisely equal. Further, by the third law, these magnitudes are equal to those of the fores Fp, and Fws with which the spring acts upon the bodies stretching it.

"

' '"

nmnmm mmmm

Fig, 2.7

Concept: Thus, an "inertiafree 11 spring 11 transmits 11 a force without changing the later irrespective of whether that 'spring rests or moves. Any body whose mass is negligibly small possesses this property; for instance, in our discussion we tacitly imply that the threads connecting the bodies in 1 ,question are 11 massless 11, 11 inertia-free" and possess the indicated property. That is why when speaking or a tension of a spring or of a thread we mean the magnitude of the ,stretching force which is considered the same for both ends of the spring or of the thread. 1

The force Fwe with which the Earth acts on the weight (the force of gravity of the weight) is no longer equal to the force Fw, with which the spring acts on the weight. The difference between these forces determines the acceleration of the weight. It should be noted that if Fws > Fwe at a certain time this does not necessarily mean that the weight moves upwards; this only implies that the acceleration of the weight is directed upwards. The force of the spring Fws and the force of gravity Fwe are not equal to each other (according to the second law). It is the difference between these forces that produces the acceleration of the weight. When the weight and the spring are at rest their accelerations are equal to zero; aw = a, = 0. Then the force Fw, with which the spring acts on the weight is equal in its magnitude to the force of gravity Fwe and, by the third law, to the force Fsw with which the weight stretches the spring; in the state of rest the force Fsw coincides with the force of gravity of the weight. Thus, in the state of rest the absolute values of the three different forces Fwe (the force of gravity of the weight), Fw, (the force of tension of the spring) and

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Fl

FORCE ANALYSIS

------

--

-- ---

F,w (the force with which the weight stretches the spring) are the same. The force of gravity of the weight and the force Fsw are simply equal to each other: Fwe

Nonna! component N ---·-

= Fsw Force of the string on girl

Force of the string on the wall

: Horizontal : component

. - . ----Ffric.

Fig. 2.10

The system is the wall

push is the normal force of the table upon your hand. The component of force parallel to surface is friction, discussed later in this chapter. + Normal force in perpendicular to the contact surface as shown in Fig. 2.11

The system~ is girl

'' (a) Force of the wall

The system

Force of your friend

:n thi strtn~- -----.--_J_ _------~~-t_h_·-~rg

8

.'!\: _- - - - - - - - - - - . - - - - . - - - .• ' (b)

Fig. 2.8

A

Ideal String An ideal string is considered to be massless (negligible mass), inextensible (does not stretch when pulled), pulls at any point in a direction along the line of the string, can pull but not push. The force with which one element of the string pulls on its neighbouring element is called tension in the string. A girl pulls a string tied to a wall. The string will exert a force on the girl in a direction opposite to the force the girl exerts on the string. The string exerts a force on the wall in a direction opposite to the force exerted by the wall on the string (Fig. 2.8).

8

B l!---+---+-+Ns ----"t---t--+-+Ns

A

+

Whenever two surfaces are in contact they exert forces on each other. Such forces are called contact forces. We resolve these contact forces into components, one parallel to the contact surface, the other perpendicular to that surface Fig. 2.10 shows contact force on finger by a tabletop as it slides on it. The component of force perpendicular to the surface is called normal reaction. The force resisting your

A

~~ --~ Fig. 2.12

+ + +

Contact Force

Fig. 2.11

If direction of contact force cannot be determined, it should be shown as two components (Fig. 2.12).

Ideal Pulley An ideal pulley is assumed to be massless, frictionless. Action of the pulley is to change the 7B;•,Tlcieal pulley direction of force. The ideal pulley . does not change the magnitude of tension in the rope. Tension is same in the string on both sides of Fig. 2.9 the pulley. If there is no stretch in the string, the speed at which rope comes onto the pulley is equal to the speed at which it leaves the pulley (Fig. 2.9).

A

When contact between two bodies breaks, the normal reaction vanishes. The weighing equipments measure the normal reaction. Normal force is a variable force; it can very in magnitude as well as direction. In Fig. 2.13, normal reaction passes through centre of gravity of body in the absence of any external force. Line of action of normal reaction shifts to the right when an external force is applied, as shown in the Fig. 2.14. At the instant the body is about to overturn, it passes through the edge of the body about which

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'

L..f1~44_,~'--'_·,_·'_~'------------~"-~'--'~~---~----ME_C_H~ overturning talces place. For a block kept on an incline, N = mg cos0. If angle of incline is gradually . · increased, the normal !eaction decreases. ;:c

'~.~---,~

. :·1 .N

earth to be separate systems, the weight -is, external force on both the .bodies. + Internal forces always act in pairs. + Vector sum of all the internal forces on a system is ,,' zero . 7

L, Fiilternal

System

=0

Problem Solving Tactics By Applying Newton's Second Law mg:

mg

(c)

, (a) Fig, 2.13,, r-.., '

. ·_ _ L 0

_·

m!!

Fig. 2.14

- ,I

L Identify the object you are considering; make a simple sketch ofthe object. 2. Draw arrows on your sketch ·to show the direction of each force acting on the object. Arrows are drawn to represent direction of forces acting on' the body. This diagram is called direction of forces acting on the body. This diagram is called a free body diagram. Only external forces (forces exerted by the other bodies) acting on a body are shown in the free·body diagram.. .·,··:,-·

Concept of External and Internal Force Consider a boy pulling two toy cars A and B connected through a string. In Fig. 2.15 (a) our system includes A and B; the· pull of the boy comes from outside the system,

:.~~\::!.~¼~.' ,. Pull_ "'~-_-:•.'!;.:.--:,

'c;E'NA

·, '

,,

. msg

''.Ns} .. _-

-1,

,·

..

[

/

.. , "•'

Groun!l,.-"..,.,- . - - •,,,

_F_i9~·_'2_.1_s~--------~

-,.. •

,,

. r -· :_:- '

'

'

.

(b)

(c)

.fig. 2.11 · :,

internal force. Note that this tension is paired and acts on both the toy car as well as B. In Fig. 2.15 (b), the pull of the string on the toy car B is external force, because string is not part of the system. Now .. consider. a ball projected upon the .surface of earth. If we include the ball and in our system, then weight is internal force. If we consider ball and·

earth

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In the Fig. 2.17 (a), force Pacts on block A; it.must be shown only on A. Block A presses the body B with certain force, which is represented by a normal reaction NA , which acts on both the bodies. Neither weight of A nor force P should be shoWn on B. Whatever force A exerts on B is communicated through normal reaction. Similarly body B presses the ground with normal reaction NB downwards. ·

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C

FORCE ANALYSIS

---·- -

3. Assign a coordinate system to your free body

diagram. Coordinate axis is assigned according to convenience in resolving forces and accelerations into components. For exampk in Fig. 2.17 (a), x and y axes point in horizontal and vertical direction respectively. In Fig. 2.17 (b) it is along incline and normal to incline. For a particle moving along curved path tangential and normal axes are assigned as shown in Fig. 2.17 (c): 4. Resolve all the forces acting on a body into its x aod y- components. 5. Apply Newton's law in component form as LFx == max, .EFz == maz

LF'y == may,

:_dL-= o·-·;:r;-x Pulley System A pulley system allows you to lifr an object while exerting a much smaller force in a more convenient direction and with greatly improved control over the object's motion. In a single pulley system (Fig. 2.19), the rope exerts equal tension force at its two __,ends. __,At one end, tension

1

T(x)

L

rJ 1 j_

X

j

w

= Mg x

Fig. 2.20

When a pulley is used to change the direction of a rope under tension, there is a reaction force on the pulley. The force on the pulley depends on the tension and the angle through which the rope is deflected. A string with constant tension T .is deflected through angle 28 0 by a smooth fixed pulley. What is the force on the pulley?

At the other end,

......

Block

Element of rape at hand . (b) Fre"i9:-body diagrams,'

(a)

r

_Reaction Force on a Pulley

Fig. 2.18

Object being lifted with

vL

__/~

At the bottom of the rope the tension is zero, while at the top the tension equals the total weight of the rope Mg.

v...

= - F.

,, :~.,,

L

'y

balances the force you exert: T

The force diagram for the lower section of the rope is shown in the figure. The section is pulled up by a force of magnitude T(x), where ' T(x) is the tension of x. The downward force on the rope is its weight W = Mg(x/L). The total force on the section is zero since it is at rest. Hence T(x)

6. Solve the set of equations for any unknowns. l:F

145·

--··. -- ··---

T /18/2

the aid of a single pulley.

Fig. 2.19 -+

-+

---+

-+

tension balaoces the objects, weight, T = - W. Thus, F = W. The single pulley is useful because it allows you to pull downward rather than upward, but it doesn't reduce the necessary force.

Tension in a Hanging Rope

Fig. 2.21

Consider the section of string between 0 and 0 + t.0. The force diagram is drawn below, center. t.F is the outward force due to the pulley. The tension in the string is constant, but the force T at either end of the element are not parallel. Since we shall

A uniform rope of mass M and length L hangs from the limb of a tree. Find the tension at a distance x from the bottom. www.puucho.com

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. MECHANICS-I 2F - (M + m)g = (M + m)a (i) Minimum force is required when box moves with constant velocity, i.e., a= 6, thus

shortly take the limit t.e -, 0, we can treat the element like a particle .. Fo.r equilibrium, the total force is zero. We have Af/ -

2T sin t.e = 0

For small t.e, sin(t.e/2)

= t.e/2

2 Af/ =

and

2T t.e = Tt.e

Fmin

2 Thus the element exerts an inward radial force of magnitude T t.e on the pulley. The element at angle e exerts a force in the x direction of (T t.e) case. The Tt.0 total force in the x direction is LT case t.e, where the sum is .... ~- .... over all elements of the string which are touching the pulley. In the limit t.e -, 0, the sum ---~ig.2.2_2_ __

=

(M+m)g 2

(ii) If F > FmJn, then acceleration of the system is 2F

a=---g M+m

(iii) For calculation of normal reaction will have to consider FBD of man. Considering the free body diagram of the man, we have from Newton's Second Law,

F+N-Mg =Ma F +N - mg= m[_l!__-g]

or

M+m

N=[M-m]F

or

becomes an integral. The ' -- -

· total force in the x direction is therefore Tcosede = 2Tsine 0 •

J-•o•o

M+m

b~§~~P{!?:

[~~2ci'iR!~~.J> ;A

ma~ of mass.M stands on a.box of mass mas ; 1shown in the Fig. 2E. l (a). A rope attached to 'the box and passing over an overhead pulley : ,allows the man to raise himself and the box by , !pulling the rope downward. ' °(i) With what minimum for¢e should the ' man pull the rope so as to prevent himself , fromfalling down. · (ii) If the man pulls the rope with a force F : '---'"'--""' greater than the minimum force, then : Fig. 2E.1 (a) determine the acceleration of the ·- -· --- -· · . (man + box) system. .. \ '.(iii) Determine. the normal reaction between the man and the .. trqlley. . . . .. ·- .... • .. .

,12-le>

1· ..... ... .. . . ... . .. .. ·- .. ,A heavy block of mass M hangs in equilibrium at the end of a :rope of mass m and length l connected to a ceiling. Determine 1tlie.. temiq_n in_ the rope qt aAisJance xfrom the _ceiljng. ....

Solution : Procedure: When a rope has mass, due to force of gravitation it tension in it will vary, separate the part of string and block on which tension is required : . --

.

-

-

-

-

--

)

T

l~

(e-x)g

lMg

'j, ..

Free body diagram of the block anii rope of lengtt, (f- x)!

,j \ ·

Fig. 2E.2 (a)

'

Using the condition of equilibrium,

:EFy

=0

m

/

T--(l-x)g -Mg= 0 l or

(M + rn)g

(b)

~,

m

e

Solution: Procedure: Draw free body diagram of box and man apply Newton's second law separately to them. Let the whole system moves upward with an acceleration a. Applying Newton's Second law, · f'

~--

X

I

.

'

(c)

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(/- X)

T=Mg+mg - -. 1

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r FoRce ANALvs,s I-·----- - - -

- - -- --

•-----

--- - - - - - · - - • - - - - -

T

T

Mg 1 - - - - - - ~

x•l

0

X 0

X I Tension is constant along the length of a massless string. (c)

Variation of tension T as a function of x. (b)

Therefore ITx = - Ta cos 8 + Tb cos = 0 ... (1) ITY = Ta sin 8 + Tb sin - mg = 0 ... (2) _Tacos8 From eqn. () 1 , Tb - ~ - cos On substituting Tb in eqn. (2), we get Ta cos 8 sin Ta sin 8 + - - - - - - - mg = 0 cos mg or Tb = - - - - - - sin 8 + cos8 tan8 ,----

4

Fig. 2E.2

-- -

Tension in the rope is minimum at the bottom, at I = x i.e.,

T

= Mg,

and the tension is maximum at the ceiling, at x = 0 i.e., T = (M + m)g Let us consider an idealized case of massless suing When the weight of the string is already small compared with the other force involved, we consider the suing to be light. For a light suing, tension is constant throughout its length. T = constant O :S: x :S: 1 If the block would have been suspended from a light string, then the tension would be T = Mg, constant everywhere.

---

Fig. 2E.4 (a) _shows a block of mass m1 sliding on a block of mass m 2 , with m 1 > m 2 • Find (a) the acceleration of eadz block; (b) tension in the string; (c) force exerted by m1 011 m2 ; (d) force exerted by m 2 on the incline.

\;:\

Fig. 2E.4 (a)

Solution : Fig. 2E.4 (b) shows free body diagram of each block. We will apply Newton's second law along x- and y-axis shown in free body diagram. Block m1 is heavy, hence it slides down whereas m2 slides up.

A bucket is suspended by two light ropes a and b as shown in Fig. 2E.3 (a} Determine the tensions in the ropes a and b.

-

-

y

Fig. 2E.4 (b) mg

(a)

(b)

Fig. 2E.3

Solution: Light rope implies that weight of rope is negligible as compared to the force it exerts. Since the bucket is at rest, its acceleration is zero. Thus Newton's second law gives ITx = 0 and ITY = 0

Block 1: ITx = m1g sin 8 - T = m1 a ITY = N 1 - m1g cos 8 = 0 Block 2 : ITx = T - m 2 g sin 8 = m 2 a ITY =N 2 -N 1 -m 2 g cos8= 0 From eqns. (1) and (3), . m1g sin 8 - m 2 g sin 8 a= --'-"-------"'CC--m1 + mz And T = m 2 a + m 2 g sin 8

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... (1)

... (2)

... (3) . .. (4)

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,"~

,+,. \.

i

I

. 8 = m2(m1g sin. e - m2g sin 8) + m 2g sm

I

' I,'

=

From eqns. (1) and (3), m 3 a= m 2 g

m1 + m2 2m1 m 2g sin 8

a= (m 2 /m 3 )g or From eqn. {4), F=T+M 1 a+N2

m1 + m2

From eqns. (2) and (4),

'

= mzg + m1 x (m2) g + m2 (. m2) g .

m3

m3

,

m = (m1 + m2 + m3) -2 g m,

Constrained Motion In unconstrained motion the moving body follows a path determined by its initial motion and by the forces which are applied to it from external sources .

.'

'' . '

Equations of block m3 : T=m 3 a iv~= m 3 g Equations of block m2 : T=m 2g N 2 = m2 a Equations of block m1 : . F-T=m 1 a-N 2 N 1 =N 3 +m 2g+T

... (1) ... (2) ... (3) ... (4)

lllustration-1 ... (5) ... (6)

Remark:------------------If m 3 has to be at rest relative to m1, they must have same acceleration. ->

->

->

am:,m, = am, - am, = O _,

an73

->

= elm,

In constrained motion, the moving body is restricted to a, specific path i.e. the path of the •body is governed by the· restraining guides e.g. a train moving ,tlong its track; a ball tied to end of string and whirled in a circle a lead gliding on a fixed wire frame. Kinematic Constraints: Kinematic constraints an equations that relate the motion of two or more· bodies. B) differentiating the kinematic constraints for the position 01 the particle in a system, the corresponding kinematic constraints among the velocities and accelerations of th, particles may be obtained. · In the figure shown the masses are attached to the inextensible string. At any instant, let the positions of m 1 and m 2 be x 1 and x 2 respectively as showri in the Fig. 2.24. then, x 1 + x 2 + 1tR = l (length of the string) = constant Differentiating with respect to tirile, we 'get-

.

m,!i ~

"\;

... (i)

... (ii) I

.-,~I

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[!:ORCE ANALYSIS -

---- - ----- --

The equation (i) and (ii) are constraint relations for velocity and acceleration. Negative sign denotes that their directions are opposite to each other.

llfustration-2 In the Fig. 2.25 the blocks 'A' and 'B' are connected with an inextensible string. The block 'A' can slide on a smooth horizontal surface.

VB X9

{

J_7

.

-

r-1-.

1491 ___,.1___ _,_

L ~~~!'.DP}":::!~_>A block of mass m1 on a smooth, horizontal· swface is: connected to a second mass m 2 by a light cord ovbr a light,: frictionless pulley as shown. (Neglect the mass of the cord and' of the pulley). A force of magnitude F0 is applied to mass m1 · as shown. Neglect any friction. '

'

IB

{

h

Fig. 2E.6 (a) a

(a) Find the value of force F0 for which the sy!tem will be in

equilibrium.

'

(b) Find the acceleration of masses andtensio~ in string if F0 :

Fig. 2.25

has a value which is double of thatfoun,d in p_art (a).

Since the thread is inextensible, its length remains constant i.e. )xi+ h 2 + xB = constant Differentiation w.r.t. time, we get,

XA dxA + dxB )xi +h2 dt dt

/

=0

As the ball moves, xA increases and xB decrease with time. dxA dxB --=VA--=-Vs Therefore dt dt XA and --;===== = cosa )xi+h2 hence v 8 = v A cosa.

Concept: If blocks are connected by an extensible string, ,component of velocity along the length of the thread of the any two point of the thread must be same, otherwise either length of the thread will increase or thread will get slack.

Solution : (a)

F0 =

zr = 2m 2g

i.e.,

(b) Concept: Movable pulley is massless therefore forces on either side of it must be equal.

r--r-:,.,___ T'~T

T-Zf'= Oxa

although pulley is accelerated

_

For 2T F'o T=m2g

m,g

Fig. 2E.6 (b)

For m 2 : T - m2g = m2 (2a) Component of For m 1 : F0 - Zf = m1a v, velocity perpendicular 4m e, 2 g - 2T = m1 a to the length of the Solving eqns. (1) and (2), gives thread changes the T = m 2 g[m 1 + 8m 2 ] angle of the thread. m1 +4m 2 If the thread is attached to a sliding 2T constrained body then at the point of T attachment of the Fig._2.26 thread, component of velocity of the body along the length of the thread is equal to the component of velocity of every point of the thread along its length. Fig. 2E.6 (c) v 1 cos8 1 = vb sin8 2 1· 1 sin 01 changes the angle of the part 'AC '?f the thread www.puucho.com and vb -:.,0 2 changes angle of the part 'BC' of the thread.

... (1) ... (2)

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\ \

ITTo- --\ ---- -- .. .

~

am, = 2a =

4m24g

+

m1

m2

2

=

m 2g m1 +4m 2 4 mzg acceleration of m 2 = \ m1 +4m 2 ace ~ration of m1

i

Fig. 2E.B (a)

'. . . _m___,"'g.:.[m-'1--_+_B_m~2""] Tens10n m strmg = 2 · m 1 +4m 2

1

. ;.app ly equatwn

-....., . . -~·-1 I. Exci.m::JP. l.e I 7 · __.. I

L:T- .- .•..

Solution: ·-,------------~--'-----! Concept: If a.body slides on another accelerated .mrjacej S rel = u rel .+ -I arel t i

I

·

2

. ~--

-~reJ_::= Oi are1_=:__~_- °:. 7 S9S 3_7_0 =-: N__ __ ma

A sm~;l ~:b~ca\~l~ck is p~ac:n a triangular block M.so thati ,they touch each other along a smooth inclined contact plane: 'as shown. ThJ, inclined surface makes an angle 0 With the I horizontal. A hqrizontalforce Fis to be applied o.n the block mi so that the two\ bodies move without slipping against each other. Assuming the floor to be. smooth al.so, determine the

I

·--'

N

,,.l· m(g- a) cos 37° mg 37°

ill~~//

.... Fig. 2E.. 8 (b) __

Fig. 2E.7 (a)

or

.(a) normal force with which m and M press against each' other and (b) the magnitude of external force F. Express your answers; in terms of m, M, a and g. _i

or

4 N=7x- =5.6N; 5

lxarel =7sin37°

3

arel = 7 X - = 4.2 5 1 2 2.l=-X2.1Xt 2

Solution: Concept: When' there is no sliding at any contact lsu,jace we may take c~mplete system as a single body. Considering motion of the system

A particle of mass 10 kg is acted upon by a force F along the I 1line of motion which varies as shown in the figure. The initial' ;velocity of the particle is 1oms·1. Find the maximµm velocity' ;attained.by the particle before it comes to instantaneous rest.

~F---,a £mg FBDofm Fig. 2E.7 (b)

F

t=lsec.

or

.,. =-·-------- ---,

: F(N) •

= (M +m)a

... (1)

,, 20---

From FBDofm N cos0 = mg F -N sin0 ~ ma

:' (0, 0)1---l--- - - -... , t (sec) I 10

... (2) ... (3)

and N = mg/ cose From eqn. (2) Solving eqns. (1), (2) and (3), we get

:

'15N--1---'----Fig. 2E.9

F = mg (m+M)tane

. F = 20 (0 ~ t ~ 10) a=F/m=2m/s Max. velocity will be attained at t = 10sec. because after that force stan acting in opposite direction

Solution :

M

:---1

~~~gmplg ! a : ._.-__,. =======->\._____ ~ f

.A block of mass l kg is kept on the tilted floor ofa lift moving; jdown wfrh 3 m/s 2• If the block is released from rest as shown,

.what will be the time taken by block to reach the bottom. '¼'hat is the normal reaction on the block during the motion? ; ·' - -. - - . - . .. . '

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dv dt

10

V

or

=2

fdv=f2dt 10

0

v

= 30m/s

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r-Foiice ANALvs1s L

- . -

-

-

- - 151' ___lj

!J~~.Gtnf:>!·~ r: 10~'-> A homogeneo!'S and flexible chain rests on a wedge whose side ·edges make-angle a and p with the horizontal [refer Fig. 2E., ,lO(a}]. The cehtrarparcuf the chain lies on the upper tip the wedge. With what acceleration should the wedge be pulled ,to the left along the horizontal plane in order to prevent the, displacement of the chain with respect to the wedge? [Consider all surfaces to be smooth]

o/

T Nsina

Tsin a

·~ P,

'="·A P1 cosa

.....B..+

..o

J',s,0 'I-

Fig. 2E.11

Solution: , Concept: Draw neat and clean FBD of fixed wedge and •blocks. Let reaction at comer on wedge is R.

Fig. 2E.10 (a)

. I

Equation of wedge:

Solution: Concept: Consider the parts of chain on either size of. incline as two different element, draw FBD. Apply Newton's law or these parts separately.

R+Tcosa=Nsina R =Nsina-Tcosa Equations of blocks : N

=P1 cosa

. .. (3)

P2

T-P2 =-a

... (4)

g . P1 - T + P1 s1na=----:-a g

(J,

mg/2

Fig. 2E.10 (b)

... (5)

-1)+(~ -sina)=o T

Taking comp. along incline

... (1) ... (2)

=

p1p2 (l + sin a) P1 +P2

R =Psinacosa- PiP2 (l+sina)cosa P1 +P2

mg sina-T = m acosa

2 2 T- mg sinp = m acosP 2 2 g[sin a - sin Pl on, solving we get a=~---~ cosp + cosa

= Pi cosa[(P1 +P2)sina-P2 -P2 sina]

P1 +Pz R = P1 cosa(P1 sin a -P2 ) P1 +P2

L~~~a~~~!~".f12 [> A body A weighing P1 descends down inclined plane D fixed of ·a wedge which makes an angle a with the horizontal, and, 'pulls a load B that weights P2 by means of a weightless and' inextensible thread passing over a fixed smooth pulley C, as: ,shown in Fig. 2E.ll. Determine the horizontal _component of, :the force (in Newton) which the wedge acts on thef/.oor comer E. ,

The pull P is just sufficient to keep the 14 N block in, equilibrium as shown. Pulleys are ideal. Find the tension (in .N) in the cable connected with ceiling. Upper cable

p

'

Fig. 2E.12 (a)

Solution:

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T1 =P T2 = 2T1 = 2P T3 = 2T2 = 4P

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MECHANICS-I

j

Upper

cable

a~ei

Tw~bl~1k.A ~;d·B h~ving ·;;:.;;;s-;~-~1-kg, ~: =4kg arr.anged as shown in thefigµte. The pulleys P imdQare light, ~and frictionless. All the blocks are resting on. a horizontal\ 'floor and the pulleys ate held such that strings remains just! taut. . At.momentt = OaforceF = 30t (N)starts acting on thei ipulley p along .vertically upward direction as sho,.;m ii'). thei [figure. Calculate. · · · j '(i). the time when the blocks A and B loose contact with! I ground. , ' (ii) the. velocity of A when B looses contact with ground. 1 (iii) the height raises]. by A upto.this instant. · (iv) the work done by the force F upto this instan(.

T,

'

p Fig. 2E.12 (b)

..

For equilibrium of block T1 +T2 +T3 =14 7P = 14 P~2K

= = --:

.. - . . ~7-.

v·

~"-~~~~R!c~ ·J 13

'

},

.

t,

r"

1

--------------

'

,For the equilibrium situation shown, the cords. are strong, enough to withstand a maximum tension 100 N. What•i$ the 1 largest value ofW (in NJ that J/)!lY can support as slwwn. ?.

.

•

=---53°

·,

'

·---·-

F -30t(N)

i

t

! I j

I

..

Fig. 2E.13 (a) __.,...:. • _,

Solution: -

- " ' .. \ V C = l m/s 1' where Ve is velocity of pulley C _

2 m

[downward]

Now acceleration of string, a, a, = g 1' [upward] [where bead is placed] Thus equation of bead

--

=

L,S-~fil-t,TIP J,!=!-_ I 17

dt

= -va Ve =2

In Fig. 2E.17 (a) shown, both blocks are released from rest. Length of 4 kg block is 2 m and of 1 kg is 4 m. Find the time they take to cross each other? Assume pulley to be light and' string to be light and inelastic.

= O.Sm/sl :. aa

1.;:>

= lm/s 2 .1. www.puucho.com

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154 ~---------- ---- ··· 1 -=c..=-=

=-=--1-- ·-- -- -- --- -·1

,11 ! mjQ0!4m

!2

L

Solution :

_,

_, v p/g

or,

1kg

ig. 2E.17 (a) •-,.,----_.,..- --=-~----....,__ ____ --···- --

--- --··- ---·-

C

¼bead c~n ;;,ove fre~ly ~n a h_o_nz_·-o-n-tal ro~Th;bead ~ I.connected by blocks B and D by a string as shown in l:he/ ifigure. If the velocity of B is v. Find the velocity of block D. /

r

T I

!•l I

T

2m!

-·

i. :9 I

t

A

1g

L_ - - - - - · - - - - - - - - - - - - - - - - · - · -

'

!

Solution: . [---~o.:~e_p_t_:_a_lo_n_g_thelength of s;ing ~~ocity component/

I

Fig. 2E.17 (b) -----

.)

-

From FBD of blocks A and B solve acceleration of each block ... (1) 4g-T=4a T-lg =lxa ... (2) 3g After solving eqns. (1) and (2), a=acceleration of A w.r.t. B

_is sa'!!:.f!'!..!11!..!1!:.J!Oi!_1-ts_onstrl11g. ~--

.I ~. l .I

5 6g aA/B = - = 12m s2

,

I

l

!

·i

'---

~ Vp

(b)

~ .,. Ve COS

(c)

--- -------- -- --·----VB

PULLEY CONSTRAINT · lllustration-3 In the Fig. 2.27 shown pulley moves 1 -- with acceleration P. Let acceleration of

lI

i!0'······.....'•,,

tD

Fig. 2E.18

A

__

37° I

I

____ _J

= Ve COS 53° Ve cos37°= v 0 from eqns. (1) and (2) we get · vB cos37° VB(4/5) Vv = cos53° (3/5) 4

2 , t = 1sec 6= o+I.xl2xt 2 .

blocks m 1 and m 2 w.r.t. ground are v 1 and

I. t""

Ve

I

6m

_,

...-o

,. ,.

I

5

If A will cross B then distance travelled by A w.r. t. B is

v

I

4g

4m

I

.a,

a

4~g

I'

_, v1/,+v~, = 2

Lltxamr.:.l·e . - -r,;_=~1--~_':---.. its I ~

r--- -- - ----- --I

j

MECHANICS-I

... (1)

... (2)

VD =-VB

3

i _,

! v,f

i;:g~me1~ai::>

i m,

¼-ii{; goes ~~-with lOmj~.A-pulley P ~-fa

A

,

1

2

Thus,

B

l

2

a=g/4 Thus, a= g/4i For system block A: Mg-Mg =Ma

or,

'.~~Lsysterii ---~=Mg~,]

-

37'

- -- - -

-----

--- .

-·--

•Three blocks shown in figure more_ vertically with constant! !velocities. The relative velocity ofA w.r.t. C is 100 rri/s upward! '.and the relative velocity of B w.r. t. A is 50 m/s downward. 1 :Find the velocity. of C w.r.t. ground. - All.l the. string are ideal. I .• ·- • ~

I

I

• I,

I'I

I

!-

Fig. 2E.22 (a)

Solution : Constraint equation a2 sin37°= a 1 cos37° or, a 2 = a1 cot37° = (9 x 4/3)m/s 2 = 12m/s 2

~

· ~. A~a

2

I I~ !a 1 sin37° a

i

Fig. 2E.21 (a)

Solution: Let velocity of blocks, A, B and C are -->

-->

-->

-->

•

-->

-->

•

vA,vBandvc

VA-Ve= 10Qj Vn-VA=-50j

cos 370

~

·:ct a sil137"

2

_ _ !ig, 2E.22_ (b)

--- - - .=-,e_-,.-1_

-->

a1

cos3r-

... (1)

From FBD of wedge we can see that N sin37°= Ma 2 Thus force enerted by rod on the wedge is N= Ma 2 10x12 sin 37° (3/5)

...(2)

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=200N

-

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15_!i

; FORCE ANALYSIS

L.-.

On solving equations we get T 4mg 1

Find the tension T needed to hold the cart equaibrium, if there .is no friction.

T

= 3,,J3 = 2mg

3Jj

2

a=__!_

3Jj

T

Concept: What is cause of a acceleration of bob? ,Resultant force on ball in x direction is (T1 - T2 ) cos 60° it cause acceleration in bob.

30°

•

•

Fig. 2E.23 (a)

Solution: Nsine

Rl*N T

-t'-i'(:-

w case

_•• J;,:.·

··)B.....

T

w

Fig. 2E.23 (b)

·A block of mass 10 kg is kept on ground. A vertically upward force F = (20 t )N, where tis the time in seconds starts on it at t = o. (a) Find the time at which the normal reaction acting on the block is zero. (b) The height of the block fr~m ground at t = 10 sec.

Solution:

N =Wease Nsin0=T

(a) When

or, T=W[.}372x~]

Wcos0sin0=T

mdv 20t-mg = - dt

r- - - ...

.

2~J>

v(t)

B

Fig. 2E.25

lO)dt

0

v(t) =lt 2 -10tl~ v(t)=t 2 -10t+25 h

10

0

5

f dh = f (t

60'

A

mg

C

Jdv = J(2t -

A steel ball is suspended from the ceiling of an accelerating, carriage by means of two cords A and B. Determine the 'acceleration a of the carriage which will cause the tension in-A ,to be twice that in B. 60°

N

t = 5sec to 10sec

(b) from

9

[J~~fl~J?..I e

F =wt

t == 5 sec

or,

2

T=Jjw -

N =0 20t = lOxlO

2

-

lOt + 2S)dt

.il.,.

10 t3 h= - - 10t2 - + 2 5 t1 3 2 l 5

125 =--m 3

Fig. 2E.24_ (a)

Solution:

Concept: When force is variable always apply calculus . T1 cos 60° - T2 cos 60° = ma T1 sin 60° + T2 sin 60° = mg T1 = 21'2 T1

y

Lx·

-

.. . (1) ... (2)

.

... (3)

'lwo mass A and B, lie on a frictionless table. They are :attached to either end of a light rope which passes around a ,horizontal movable pulley of negligible mass. Find the ,acceleration of each mass MA= lkg,M 8 = 2kg,Mc = 4kg. :The pull_ey P2 _is vertical._

mg Fig. 2E.24 (b)

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ffsa ,---

MECHANICS:::i-J its equilibrium position. If we hold the block in position x, form Newton's second law,

B

JL •

__Flg'._2E~6

'-...··· ··,.... : •

-+

4

-+

->

->

->

'

(a) ap = aA + a 8 ->

(c) ap = aA- a 8 (BJ Acceleration of A is : (aJ 3g (bJ 4g

->

': Fexternaf

Equilibrium position (a)

(cJ 2g 5

5

--¼

\.. / Frictionless •·· -· surface

->

(d) a, =2(aA+a 8 )

· (CJ Acceleration of B is : (a) 3g (h) 4g 5 ' 5 (DJ Acceleration of C is : (a) 3g (bJ 4g

I

1

i

System

!•] __

(AJ Constraint equation for pulley A is :

5

. -· . -- ....... - - ·r

1·-- _ .............. --- -··· ···-···

A

...

(c) 2g

System

j x>o,

Fspring

5

:

-· -· .. 5- . . ---~ ---

2 (cJ g (dJ ![ __ 5 ________ _5

, ,

. ,

,. .1

........... EqJ_ilibrium position

Equilibrium position

Solution : ... (1) ... (2) ... (3)

mcg - T = mcac

,

I , ,'t.xtemal J a .£me· I

,--f·1

"t.pnng /.. ··-\

(b)

(c)

Fig. 2.32 -> F external

->

+ Fspring

=0 ->

Robert Hooke experimentally found that F external is proportional to x.

...

Fe?(ternal

x>O

x

5

Fexternal

Where k is called spring constant and has unit N/m

Elastic Force of Spring

->

Spring shown in Fig. 2.32 (a) is stretched or compressed· by applying a horizontal external force on spring. We choose origin of coordinate system at equilibrium position where the spring has its normal length. In horizontal direction there are two forces acting on the system: ->

(1) Fextemal

= kxi,

->

(2) F,pring •

When we pull the block to stretch the spring, force of the spring is opposite to out pull [Fig. 2.32 (b)]. Ifwe push the block to compress the spring, force of the spri_ng is again directed opposite to our push [Fig. 2.32(c)]. Force of the spring is restoring force since it acts to restore the block to

Fstring

= -kx i

Therefore force of spring on block is proportional to the amount of stretch or compression of the spring. It is always directed towards mean position. It is independent of mass m attached to spring. An ideal spring has negligible mass as compared with mass m attached to it.

Series Combination Elongation or compression in different spring may be same or different but tension in each and every spring is ' . same.

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---159

\FORCE ANALYSIS

~- ---

Fig. 2._34

x represents is the total extension produced in all springs. x 1 , x 2 , x 3 , ... Xn are extensions produced in individual springs. If this spring is replaced by a single spring and same elongation is produced and tension developed is then this single spring is equivalent to combination and its force constant is equivalent force constant of combination. Even total energy stored in combination will be equal to energy stored in this single spring for same deformation. X1 + Xz+..... ... +Xn = X k1 X1 = k2X2 = .... : .. . = knxn . . . _ 1 . 1 . 1 X1,X2, ......... Xn - - . - .....•... .kl k2 kn 1 X

k1

X

l

Fig. 2E.27 (a)

Solution: Concept: Force of spring does not change instantaneously so find spring force at initial instant,

Initially m1g =kx When support is removed, spring force does not change.

k1

1

1

1·

k1

k2

kn

kx

-+-+........+-

T = k1X1 = k,qX 1 k1 l l X = k,qX

kx

M2g FBD litially

FBD when support is removed

- + - ........+k1

k2 kn 1 1 1 1 - = - + - ........+k,q k1 k2 kn

(b)

NewFBD

For m 1

or For m 2

k

Equivalent force constant is smaller than smallest individual force constant.

or

Parallel Combination Tension in different springs may be same or different but direction of · tension in each spring is same. Even· elongation or compression produced in each spring is same. Total tension in this combination k,.x and that produced in single equivalent spring must be same. k,q x = k1x + k 2 x+........ knx Fig. 2.35 k,q = k 1 + k2 +..... , .. kn k,q =:Ek Equivalent force constant is greater than greatest individual force constant.

(c)

Fig. 2E.27

1 -:E(l)

k,q

1

_,

The system of two weights with masses m1 and m 2 are connected with weightless spring as shown. The system is resting on the support S. The support S is quickly removed. Find the accelerations of each of the weights right after the support S is removed.

----c--~---,---X

I -

- -

:

m1g -kx = m1a1

=0 m2 g + kx =m 2 a 2 a1

(m1 + m2lg az = - - - - m2

An object of mass mis suspended in equilibrium using a string of length l and a spring of constant K(< 2mg/!) and unstretched length !/2. Find the tension in the string. What happens if K > 2mg /! ?.

Fig. 2E.28 (a) ·

Solution : The string is under tension and the system is in equilibrium, if Kx < mg

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l [herex = -]

K(½) < mg

i.e., for,

2

K< 2mg l T=mg-Kx l =mg-K-

i.e., if,

2

2

acceleration of 3 m will be zero.

If K > ~g, the spring force is more than mg, for x

=.!. Thus the system will ai:c~lerate 2

! -

#- _';-~ -;_::-,-~~ --- . --;.",-f0:~1

·· •

[The mass in the Fig. 2E.30. can slide on a fr.ictio11/~sJ lsurface.TIIe mass is pulled out by,a,distance x. The ..sp,;ifzgj )constants are k1 and k2 · respectively. l:i".d the force. pulluj; . . · .l •back on the mass and force on the wall. I

-~\·

;

. -·· ___ ..•. ...

Fig. 2E.3~ ·"'

.J ._

·)

''

Solution : Springs are in series Hence k = k,k 2 eq

k1

+ k2

and

Solution:

/. •.:

·· - · · · - · · ..___,,,_.The -;,;;e-m-sh~-:Vn. Fig. 2E.35 ( a) is given an ;acceleration 'a' towards left. Assuming all the surfaces to be jrictii°':l_ess_,_ji~d_t~e_J!)_rc~ _on ~~1!'_5_P.~"!_e._____ __ ·--.

i~ --;h;

i

g

I

':_N\

I

i i

....,,,_.,..(a_)_._.i.,

I:

i

I

·

aBA

a

Jj

i

;BG = ;BA + ;AG

Takethe~~~~:t;~in_Fig._~:~~-~)~. N,

(b)

N,

=

mg cos 30°

N 2 =ma+ N 1 sin 30° =ma+ (1.15 mg) x (1/2) = m(a + 0.58g)

l_ .. _. -----~

TI1e block B st-;;,;.fr~~.~~; and slides ;n-th~ ·,:;~e A~hich1

Block B:

!

move on a horizontal surface. Neglecting friction, 1determine (a) t/ie acceleration of wedge, (b) the acceleration ;af! t!Je _bJoctrelati)'Ll'Q tli

A

9 __ Fl~g.

mgcos9- MA =mAsin0 sin 0

-+--+aA

! ...,,.,...,...,.__....,

A·mgc~·se

sin 9 Substitute N I into eqn. (4) to get A.

7

'

-+--+•A=A.

1 8 mgsn

Wedge A:

lcoJ:E~p:~,w,~·J36~ ican

AN,

-+ x-component of acceleration aB, aBX = a - A cos 9 • -+ y-component of acceleration aB, aBy = A sin 9

= '1.15 mg

From eqn. (1),

N,cos

!

: -- . I :--···----~~--~:61~!--~-----·- . .J ~

Solution : This problem involves two branches of mechanics: kinematics (which deals with motion) and dynamics (which deals with cause of motion). >,\;--. First we shall analyze the. accelerations of wedge and "·'.',·block.::·.·.. · · Wedge A : It moves on horizontal surface; we assume its acceleration towards right.

mgcos0=mAsin0+ MA sin 0 A= mg cos0sin0 or msin 2 0+M . a I __ Fig. 2E.36Jc) From eqn. (3), ma= mgsin0+mAcos0

~:

vn

l

.

a=g sin0+A cos0 . mg cos 2 0sin0 = g sm 0 + -''-----m, sin2 9 + M (M + m)g sin 0 = M+msin 2 0 Note that axes of x and y can be assigned in another manner, as shown in Fig. 2E.36 (c).

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1

AG·

Fig. 2E.35

Thus

..

-

For sake of simplicity we drop the subscript G. Therefore resultant acceleration of wedge is vector sum of its acceleration relative to A and acceleration of A on ground.

-- ~olu~:;~~-F~rces actin;:n -;~er~ are ~ho_w_n i~-~e Fig 2E.35 (b). :r.Fx =N 2 -N 1 sin 30° = ma ... (1) :r.Fy = N 1 cos 30° - mg= 0 ... (2)

,

= aBG

where B stands for block B, A for wedge and G for gro~~s

~

N,·..

I +J!.

Bloc:ic B : ~o acceledrationsd~re sudpealrpose~ clino':1 it: itsd acce1eranon re1anve to we ge aBA rrecte oni:jm e an acceleration of :edge.-+ -+ .

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:_164___ --- . --- --- ·x-componenc of acceleration of B = a cos 0 - A and y-component of acceleration of B = a sin 0. Now force equations for block B are D'x = N 1 sin 0 = m(a cos0-A) ... (5) D'y =mg-N 1 cos0=masin0 ... (6) We can arrive at tbe same result by considering eqns. (5) and (6) instead of (3) and (4).

-

On substituting expression for Nin eqn. (4), we obtain mg cos0- MA = mA sine sin 0 . 0+MA mg cos 0 = mA sm sin 0

A= mg sin0cos0 m sin 2 0 + M

or

-- - --·--- - -- -- r - r

LJ=,~t;pHJ~J

37

l__>

'' In the F~. 2E.37 (a) shown, mass ;m, is being pulled on the incline of a wedge of mass M. All the surfaces are smooth. Find the acceleratiqn of the wedge._ .

A rod 'A' constrained to move in vertical direction rests on a wedge B, as shown in the Fig. 2E.38 (a) Find the accelerations, of rod A and wedge B instantaneously after system is released, from rest, neglecting ftiction at all th_e contact surfaces. '

. _JmL

m M

F

B

A

Fig. 2E.37 (a) B

I Solution : Fig. 2E.37(a) shows force diagram of tbe wedge and the block. Let acceleration of block relative to

wedge be

a'.mM

=

or

-+

Solution: In the Fig. 2E.38 (b) dotted line shows initial position of rod and wedge. If the rod is displaced vertically through y, then tbe wedge moves a distance x. y=xtan0 Therefore tbe relation between accelerations of rod and wedge is ... (1) a=A tan0

-+

= am - aM -+ -+ = amM + aM

(amlx = a - A cos 0 , Cam\ = A sin 0

·_r X

B

Fig. 2E.38 (a)

a'. and acceleration of wedge on ground is -+ amM -+ am

or and

M

N,

+-·f

if~~:~e Pis greater than IF Fmax., the block will have I a resultant force "· P - Fmax. on it. The block -----··;1.·,~,----Dynamic will accelerate in the iL direction of resultant ~· ta· force when · sliding 45° "-'-'--~--..,.P mdtion ensues.

a

are:

+

where µ 1c is d¢fined to be coefficient of kinetic friction. · · Fig. 2.38 shows variation of friction force versus external force graph. When condition of impending motion or sliding is not known. To determine friction force we assume static equilibrium and solve for the friction force F,. The possible results are: (a) F, < µ 8 N (maximum value of friction): Body is in static equilibrium. The value of F, can be determined from the equations of equilibrium. Jb) F, = µ, N: Body is in impending state or about.to move assumption of static equilibrium is still valid. (c) F,,> µ, N : This condition is impossible. Friction force cannot be greater than F,max. (µ, N). · + Normal contact force N and friction force F are two components of the resultant contact force R of the surface on the block. Angle between resultant contact force R fill.d contact force N is called angle of friction(¢,).

+

r.Fx = mg sin¢,, -µ,N = 0 Friction force opposes relative motion between two surfaces. In order to decide the direction of static friction, try to imagine the likely direction in which the· body will tend to move; friction force is opposite to it. In the figure, force P pulls block B towards left and A is pulled towards right. Friction force on B is towards right and on A is towards left. Important point to notice is that for two contact surfaces friction force is in opposite direction. It is intern.al force for two contact surfaces, so it must be an

.. ,

...._+-_,_- fe

i

.

ma9'

_Fig. 2,41_;_____ · - - - - - - "

~ f

,___ _F_;;ig. 2.39

'

'

.-,.~-"--IA T

tan¢,='~; R=~f 2 +N 2 When block is in impending state maximum value of static friction force is acting on block.·· f=F,,;, .._ =µ,N

l

+

Consider a conveyor belt moving with velocity v A. A small block is gently lowered on it. -+

--+ --+ -+ VBA =VB -VA=O-VA

Velocity of block B relative to surface A is towards left; friction force is opposite to VBA, i.e., towards right. Due to this friction force, the block accelerates

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FORCE ANALYSIS

~

1671

Solution: Force of friction = 0 => N =0

towards right and the belt retards. Finally the block acquires the velocity of belt and moves with it. .. ·1-:-7

J~='~g~_J;~-l~,cl

39

1..>

=>

f=tan0 =>a=gcot0 a

I

A block weighing 20 N rests on a horizontal surface. The: coefficient of static friction between block and surface is 0.40 and the coefficient of kinetic friction is 0.20. ;

How large is the friction force exerted. 011 the block? (b) How great will the friction force be if a horizontal force of

( a)

5 N is exerted on the block?

(c) What is the minimum force that will start the block in: motion? (d) What is the minimum force that will keep the block in. motion once it has been started? (e) If the horizontal force is 10 N, what is thefrictionfprce?

F = ma = Mg case

k.J~.x:g_t;.ti.l!i!l:;;;.--::Q_':-,..._

=

·- .

~~~

·····--··7 A black of weight W rests on a rough horizontal plank. Thel

slake angle of the plank 0 is gradually increased upto 90°. ; Draw two graphs both withe along x-axis. In graph show the: ratio of the normal force to the weight as a function of 0. , In second grapl~ -show ·the ratio of the friction force ta the! weight. Indicate the region of 110 motion and where motion, exists.

Solution

Solution: (a)

I

N

When

- I Fmction

i

=0

From condition of equilibrium, P=F=O (b) First we calculate

mg cos a

_F_lg. 2_1=:~1

= (0.40 X 20)

Till block is static mg sin 0 =

mg

P=F=SN

I

f,; ~ = sine mg

As incline angle is increased, if block does not move friction force has balanced component of weight down the incline In impending state of motion mg sin0 0 = µ,mg cos0 0 tan0 0 = µ,

r.Fx = pmin. - Fkinetic = 0 or Pmin. = F!cin,ti, = µk N = (0.20)(20 N) = 4 N (e) Since P > µ, Nin this case, the block accelerates. From Newton's second law, :r.Fx =P-µkN=ma Therefore F = µk N = 4 N.

40

increased

:

I

(a)_

Fig._2_E.39

(c) When the block just starts to move, it is in impending state. From condition of equilibrium, :r.Fx=P-F=x_=O or .P=F=x_=µ,N=BN (d) When block is in motion, F = µ kN. Minimum force will be required to move the block with constant velocity. From condition of equilibrium,

lc~S~-S¼W,B!iJ

angle of incline is being gradually

----mg

F=,.=µ,N =SN Since P < F max., block is in static equilibrium, i.e.,

. I

..!'!.

mg

- l f - - - - ~ - - - . . . 1 . . . ._ goo

Fig. 2E.41 (b)

[>

-

Th~

A wedge of mass M m~es an angle 0 with the horizon~!. wedge is placed on horizontal frictionless surface. A small' block of mass m is placed on the inclined surface of wedge. , What horizontal force F must be applied to the wedge so that! the force of friction between the block_ and wedge_(s ,._ero ]__ ;

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-

-.

----

_.e

1,'

Anurag Mishra Mechanics 1 with www.puucho.com

-·-·· -- . -

j168

--- ·-····

--------------- - -------------- ---...!...

Impending state

mg

MECHANICS-I ' Tmin=40N

:50N-[I

1----,,L-'--4~

I

- -

,

'/ µsCOS8o ,i---·-----:a,,/ ' I µ,cose 0

-

r

(f,)max•40N (c)

. mkcos00 j Static

T•40N

i

sine motion occurs

50~ f, •10N

e

90°

(d)

Fig. 2E.41 (c)

I.

Fig. 2E.42

When block begins to slip fk =µkmg cose fk - =µk case w

Thus block A remains static Force F can not pull block A 1-····-- -------r-i-

E._f;~~'~'~,P.-'~- i 43 1>

[~>f~'L'gfg~J42!> r

--- -

- -

---,

:Find the acceleration of the block and magnitude and: 'direction offrictional force between block A and table, if block I :A is pulled towarq~ !eft iyith a forq, pf !jO /'{._ 1'

.

And friction force is (10~ N

I

}--X

I

!

;11ie 10 kg block is resting on the horizontal surface when the force 'F is applied to it for 7 second. The variation of 'F with ;time is shown. Calculate the maximum velocity reached by the lblock and the total time 't' during which the block is in1 !motion. The coefficient of static and kinetic friction are both,

;a.so.

' µ•0.8 g•10m/s2

F(N)

100 ...... .

B 4 kg

I

Fig. 2E.42 (a)

L -- --- -

-·-'

Solution: Case (i) If block moves down, maximum possible tension T = 40 N is attained when it moves with constant velocity. In this case N

---··

[

50 N

Fig. 2E.43 (a)

I_ A

0

'---'---'------+ t(s) 4 7 l

/

40N=Tmax'

Solution : Block begins to move when F=µN

40N

= O.Sx lOx 10 = SON 50 N

i . - --- --

From

t

=O to t =4sec

F = 2St

Fig. 2E.42 (b)

Tmax can not over come apposing forces of 90 N,

therefore it is not possible. Case (ii) If F=SON force can pull block A to left, mm1mum tension in string Tmin = 40 N if B moves with constant velocity.

From t=4tot=7sec=40N Block begins to move at t =2 sec. after that F-µN= mdv

dt 2St - SO = 10 dv dt

:41:NF: '

µ

'

'

mg

Fig. 2E.43 (b)

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' - FORCE ANALYSIS

169 V

4

0

2

f dv = f (2.St - 5)dt

or,

V

Concept: Kinetic friction is opposite to relative velocity it opposes relative motion. When horizontal component force is reversed, relative velocity is not_changed therefore, direction of kinetic friction does not change.

-0 =12-~t2 - 5{

Fcoh-' Fsin37°

= (2.5 X 8 - 20) - (5 -10) = 5m/s 4 sec.; block retards due to greater friction V

After t force

=

Stage 2:

v = .Jl 12 m/s

µ,N

a

----+Uinitia! = 5m/s

,._

F cos 3 7° is reversed, block continues along original direction, but due to retardetion created by µ kN and F cos 37° block travels till it stops. -(Fcos37°+µkN) = ma -(20x .:': + (0. 25 x 8) = 2x a or, a= -9m/s 2

- - 0 - F a 40N µ,Na 50N

Fig. 2E.43 (c)

a= 50- 40

5

= lms-2

Displacement of block in this phase

10

0 =v

Velocity of block at t = 7 sec at t = 4sec, v; = 5m/s v 1 =v;+at = 5 - 1 x 3 = 2 m/s

2

v2

-

2as;

s =-

2a (112) 56 =--=-m 2x9 9 Stage 3: Which block returns its a acceleration is: Fcos37°-µkN = mg 2 a= 7m/s Fsin37°

~ LE~ff~J!tl:?l~ .~ 44j;._>

'

A force of 20 N is applied to a block at rest as shown in figure. After the block has moved a distance of Bm to the right the direction of horizontal component of the force F is reversed in· direction. Find the velocity with which block arrives at its. starting point.

Velocity of block when it returns to original position v2 = 2as 56 ) =2x7x(s+ 9

Fco~

~ mg

Fig. 2E.44 (c)

16-.fi

v=--m/s 3

-~7'

-

µ•O.~

-

-

-

.-~

lE?ffl~BL~ ! ~__;>

Fig. 2E.44 (a)

,Find the contact force on the 1 kg block. () ':,

Solution: Stage 1: Motion till force reverses its direction N = mg -Fsin37° Fsin37° = 20-20X~= SN 5

Fcos37°-µkN

= ma

2ox .:':- 0.2Sx 8 = 2x a 5 2 a= 7m/s

• • Ijj . µ,N

'

Vs;;.;

'


acceleration.

. a

Find acceleration of blocks F

f •

-2

.!. x ~ x 4= .'.!; Total displacement= 12-.'.! = 10~ 2

60 N

= 40 N

0

µ 1 =o.s-

a =4 m/s2

Fig. 2E.53 (a)

. --~· Solution: Assuming same acceleration

, µ, = 0.2 .........

Fig. 2E.51 (a)

.~(

Solution: (1) First of all find values of limiting friction at all contact surfaces. CJ, max)

0

60-f=Sx4 or J=l0x4=40 f = 40> 25 hence our assumption is wrong.

fsmax=25

"'l"~~,J~,"

'

= 30

,, 60

•2E • \

(2) Maximum force upper surface of 10 kg can experience is 25 N so it will not more relative to ground. (3) Hence only 5 kg will move.

a1 = 7, Oz = 2.5

Fi!!: 21t,53 (b) t -

-· - -· ·- - - -

~

k:~0:9"''22,~l~ .l~~-•>

aA=3,a 8 =0

Solution:

·

Fig. 2E.51 (c)

25-m

l' !e;?~f~-riiel~. r;-27>

_30~~!;:t

-ill-

:: :

Fig. 2E.54_ ....

A

if they are moving together a 1 = az = a

30 N

smooth~

F-f-30=l0a; f=5a; F-f-30=10x1. · 5' F = 30 + 3/ maximum f is 25. F=30+75=105

,

Fig. 2E.52 (a)

Solution :

\

-

Find m(l,!Cimumforcefor which they can moyf_tgget!Jer..

25____[D--- 40

1.:7 . I 10 25 25••-'-----'1

5

~ 25

Fig. 2E.51 (b)

, µ, = 0.S

a= 30 = 2ms-z 15

Fig. 2E.52 (b)

Fig. 2E.50 (c)

3

MECHANICS_;i-j.

IT] /1 max = 25

Two Block Problem .. - ------.

~ fzmax =0

LJ=:;~~g.fD,~J~ i 55 L-> ,

(1) 10 kg block must move because some force on upper

surface will act on it. (2) B can either move with same velocity and acceleration as A or it can move relative to A. (3) Always assume it moves with A and solve.

,.

as:

_

I Fig. 2E.55 ( b)

--

Solution : For this force both the blocks move in combination acceleration of system

I

I

I.

.I.11_exqmp/e !i!i_fil_lg.frictiol_l_fosce b~tw~~n_ !,locks if F = 6N._

NqfJ! Na B

mAg

I

Fig. 2E.56 (b)

3

I~E:~F~t11-~l.~

_,__,..,,_,.,_._,,""=--. __,.,._---=-~"--'-

! 6

F= 20N

or,

,

Fig. 2E.55 (a) ::u=:,__--,

i ~ F-;

3

.,----µ•0.3,m 6 •3kg,

mh,fir

II

l

.,----µ,

59

b>

!An object is given a quick push up an inclined plane. It slides: :up and then comes back down. It is known that the ratio of ·the ascent time (t up ) to the descent time (t dawn) is equal to the I ·1coefficient of kinetic friction (µ). Find the angle e that the, inclined plane makes with the horizontal Find also the range! 1 ,ofµfor which the situation described is possible. Assume. that; -the_ coefficients_ of static_and_ kjneti~ fric_tion a~e_equal. _i Solution : aup = g sine+ µg case;

= g sine - µg cose L = "I.[g sine+ µg cose]t~;

ad,wn

Till this moment blocks A and B more in combination. (b) Fmax can be obtained by applying Newton's second law on upper block

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-MECHANiC:S:fj '"~---,-'

So, Resultant force = ~ fk + N 2 2

L = .!_[g sin0 - µg cos0] tJ,wn 2

(sin0+µcos0)µ 2 = sin0-µcos0 µ(1+µ2) tan e = '--'--'--c--'(1 - µ 2)

;E

=)(µN)2+N2 =N)1+µ2 = mg cos0~1 + µ 2= 12-F, N

. . ,---,, iJ;;lff½~:EiJ~ I. 62 J.> .

µ -;--·-····------"·

'

'

'A wooden block" slides down the right angle channel as shofvn )in Fig. 2E.69 (a). The channel is inclined at an angle 8 w.r.t. :the horizontal. The,angle a is 45°, i.e., the channel is oriented •symmetrically with the vertical If the coefficient of friction. ,between the block and the channel is µ k, find the acceleration ;of the block.

F n,;~ = : (sin 8 - µ cos 8)

or

(c) Case (i) Impending motion downwards: Block has a tendency to slip downwards and external force just prevents it from sliding. In this case Fis minimum. From _ _ conditions of equilibrium, ____ .

'-

~ I I

I

I I

1·

I

Fnction force -

-In the figure shown, the static friction .coeffici;n; betwe~;~u] contact surfaces is 1/2. What minimum force applied leftward; 'on block 1 will move the system ? Repeat problem if tbe force) is now applied on block 2; _______ •

~~~?~fr i - - - - - - . - - - ·', - - · . - - - - - -

iA man:;of m~§ •eyo kg, stands on a hqrizontal weighing. machine, ofi'!eg{lgible mass, attached to a massless platform P. that slides do111ri ·at 37° incline. The weighing machine read 72 kg. ,:nan is ci/w;;ys~_ a£. r_esP_t _w.r. t. ·weighing machine. \

I '

l'

!l

;:, _.

. "'

•

Fig. 2E.74 (a)

!Calculat,e :

, . ( a) The vertical- acceleration of the man , '(b) The coefficient of kinetic frictionµ between the platform

l_____q_n(IJnriline,_:__ . _ ___ _

~ Fig. 2E.75 (af

Solution: Step 1: Calculate maximum friction force that acts on all the rough surfaces. Step 2: Check the tendency of motion of each block, static friction opposes that. When slipping just begins f, is

, ·1

. l

-~i

·

Solution: Weighing machine measures normal reaction. Draw FBD of man. System of man and platform h& e acceleration at an angle of 37° ax and ay are x and y components of ·acceleration. What is cause of vertical acceleration and horizontal acceleration.

maximum. Note that due to string constraint both the blocks'will b~ at the verge of slipping simultaneously. Case I: f 1,- =0.5x3xl0=15N ' f 2smax = 0.5x2xl0= ION

from FBD of 1 kg F = f1 smax + T + f2 smax = 15 + 5 + 10 = 30 N from FBD of 2 kg zr ". f2,m~ = 10 T=5N

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I FOiCE ANALYSIS

-- -- .. --- . - '" .

181] ..:·===========

Note: _ _:__:__ _:_--_:_·-:..:-:.:·:::.-

First decide whether there is slipping between blocks or not. If blocks have same acceleration then friction force between blocks must be less than µ,N.

2kg f2, 1kg f1, (b)

N1 f2,

2 kg

T

f2,

f1,

20N

10N (c)

Fig. 2E.75

Case II: From FBDof 1 kg T = Fmin

f 2 smax + f 1smax = 10+ 15

= ZT+ f, ,m~ = 2x 25+10= 60N N1

~-~T-;;,

F

Let blocks move together with common acceleration a 6mg - f 2 6mg - 3mg a= =-~-~=g 3m 3m then for upper block T-f1 =ma => 3mg-f1 =·mg => f, = 2mg but f 1 ,;; limiting static friction but here f, is coming out · to be greater than µ,N. Assumption of no relative motion between blocks is incorrect that means there is relative motion. Therefore f 1 is kinetic friction. 3mg-µmg --2g (towards right) m

2T 12,

3mg+µmg-3µmg . -~~~~~~ = g/2 (towards nght) 2m Applying pulley constraint to get acceleration of hand ap

A block of mass m rests on top of a block of mass 2m which ls ·kept on a table. The coefficient of kinetic friction between all ,surfaces ls µ = 1 A massless string ls connected to each mass .and wraps halfway around a massless pulley, as shown. 'Assume that you pull on the pulley with a force of 6 mg. What ls the acceleration of your hand ? F=6mg

µ=1[mJ 'µ=11 2m

-

~

Fig. 2E.76 (a)

Solution : The free body diagrams both the blocks are:

~~. f11

N2~

mg

a2

2

= 5g / 4 (towards right)

r- ~.~p~e} .!:: ..:-1?!;.> A 4 kg block ls placed on top of a. long .12 kg block, which is accelerating along a smooth horizontal table at a= 5.2 m/si: under application of an external constant force. Let minimum; coefficient of friction between the two blocks which Willi prevent the 4 kg block from sliding ls µ, and coefficient of friction between blocks ls only half of this minimum value. of, (i.e., µ/2).Find the amount of heat (in joules) generated due• to sliding between the two blocks during the time in which. 121 kg block moves 10 m starting from reg,

~

= 5.2m/s 12kg 1-I smooth.___,__- - - - ' · - r"l'.'.":J'---'-, a

T N 1 2mg

(b)

+

acceleration of pulley= acceleration of hand am +a2m 2g+g/2 ap =~-2~= 2

11

12

= a1

2

Fig. 2E.77 (a) (c)

Fig. 2E.76

f 1 is force of friction between blocks f 2 is force of friction between block and

___ j

Solution: First assume that blocks have common acceleration, for both block to move together acceleration of 4kg block must be 5.2 m/ s 2

ground.

4kg 1 · I' ,'-"=1--1-2..::kg....=ll:;-b+f a = 5.2 mis~4

from FBD of pulley, we get T = F/2 = 3mg

Fig. 2E.77 (b)

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-

rt:;~·:·~~? b··:;· ·-,~. --.-- ·7 ·

I

a= 5.2m s 2

f= 4x5.2 µ 0 mg = m(S.2m/s 2 )

I.~--. .:--r·, .,.., '· · ,.

· µ 0 = 0.52 Ifµ= .!co.52) = 0.26 the acceleration of 4kg block is 2 . . due to friction . ' ' 2 a 1 = µg = 2.6m/s As there is relative motion.between blocks we apply s,,1 Sre1

L

= - 2,6 m/s 2 , 1 2 = -(-2.6)t 2

Time of motion can be determined from motion oflower block· ~ = .! (52)t 2 = 10 (given) For 12 kg.block 2

sre,

=-Sm work done by friction is given by. w1 =µmgS,., 1 = 0.26x4x lOx (-5) = -52J Heat generated = 52 J ·

Ii+ 12 = 0 b+ /4 = 0 From which we get .a = b = c Applying Newton's Law on.block A Mg-T=Ma on block B T-T1 -µmg= ma on blockC T1 ~µmg= ma solving eqns. (1), (2) and (3), we get

~,,.,~..,--.-rat·-,.,

·, •

·. ,·.:]

.., ~{~· ,

,'

•:,_ \·

• . .: 1 .

w

eC

mC

m

. . •

r:'

an{c\~;;J

B

.

;,

j

!::,::;~::::.: '. ::···,;

M A I ·. : '

:J

ta) ~:~:io·~.$.,·~.t~:!~~.~t?E.i~.-.·.·~~.-.i·.:t~::~~~i.!.·.·~.R.~.~·;~ I .. !, i. 'small: , J . : · ,,; ' , · ·. · , ,;;,,· , '(~) if t/zci Fri~$ of blq~k ;t';:, z;;;ss th~,.; some critical vi:zltii!, the

I ..blocks will not.'accelerate1whefr·relea.sed from" rest.,Write t__.'.etween blocks B qri:d (he support table, bu"ttherel is frict/o~ /zetivee,; qlocks -p:~ncl,\i/denoted; bY,,

.

..:--

_

.....

'' - -----·--··'

- ........ -.-~~,

Solution: During upward acceleration . a1

L0.!.:.

µN

I

4a !

a 1 cos 37q =~

mg

N - mg

= m(

3 ;

µN = m( ~ a1

150N/m

1

\

.I

Fig. 21:.~5 j~).

450N/m

)

1

on solving we get

}

= 15g m/s2 31

Fig. 2E.84

.

Solution : Suppose origin is at the equilibrium position and the direction of increasing x is towards the right. If the blocks are at the origin, the net force on them is zero. If the blocks are a small distance x to the right of the origin, value of the net force on them is -4kx. Applying Newton's second law to the two-block system gives -4kx= 2ma Applying Newton's second law to the lower block gives k(x1 - x)- f = ma where x1 = initial stretch and f is the magnitude of the frictional force. f=k(x 1 +x) The maximum value of x is the amplitude A and the maximum value for f is µ,mg. Thus, µ,mg= k(x 1 + Amrucl· Solving Amax gives A =µ,mg -x =3

k

= a.1 .sin 37' •

N ,

,When the system shown in the diagram is in equilibrium, the ,right spring is stretched by 1 cm. The coefficient of static: ' I ::Jriction between the blocks is 0.3. There is no friction between[ \the bottom block and the supporting surface. The force) ,constants of the springs are lS0N/mand 450N/m (refer Fig,; 2E.84). The blocks have equal mass of 2 kg each. : Find the maximum amplitude (in cm) of the oscillations of1 ·the system shown in the figure that does not allow the top: 'block to slide on the botto111. ;

max

~a

/

Concept: When lifting arms accelerate up, caus.e ofj ,acceleration a1 cos37° is friction µN. And resultant force up1 :is .(N -c 111g):,1chich causes acce!er;it!~n,J{i.sj.n}J_0,. _. -

-- - -

- - --·· ''f

4a~

a2 cos37°=5

··71

37°

82

+

a2 sin

37°=,

1

·

I

Fig. 2E.85 (c)

FBD when Arm is in Deceleration . Concept: During deceleration direction of friction force ,is towards left. Student is advised to ponder over a simple question.

I.-----

- .

''Which force is cause of component of acceleration a 2 cos37° parallel to surface."

1

mg -N = 'In the manufacturing process disks are moved from level A to: B by the ·lifting arms shown. The arms start from level A withi lno initial velocity, moves first with a constant acceleration a,! 'as shown and then with a constant deceleration a 2 and comes: ,to step level B. Knowing the coefficient friction between disks '.and the arm is 0.30, determine the largest allawable. :acceleration a 1 and the largest allawable deceleration a 2 of ·the disks are not to slide. '

Which on solving given

m( 3;

2

)

a2 = lSg m/ s 2 4a

'·-------..---···-···-~ ~

6-,~~~.~~~-~-·~ "-

- ---

--- - --- ·- - --

- --- -------·--- -·

- 1

;In the Fig. 2E.86 (a) shown a constant force Fis applied on, :lower /:,lock, just large enough to make this block sliding.outi from between the upper block and the table. Determine the i 'force F at this instant and acceleration of each block. Take: g_= l()_m/s 2_. • • •.. ______ - - - - - - - - · - · · _______ ...

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11as

, / , f,

>'

r:;1·;;,d ~2 (11)(~2>

a,

dt= r [ -sm . 0 -1+cos-J de • de "] ;

, Fig. 2.51 (b) ----~--·--

-2+ -- .

dt

dt

di

i

= roo[ (- sin0) i + (cose)j]

... (1)

[,

When a,. is )n direction of motiol! i.e. parallel to velocibvector speed of object i'!creases. :, Centripetal acceleration chrmges direction of ,;elocity vector. ,_ 1_,: ' ' ' ' When ta11gential acceleration is opposite ·to veJoi:ilX vector speed ·of_ o_bject decreases. · · · .. : · Note that aligular velocity> vector, position vecton: cmd tangential ac;celeratio/1 vector are -rtormal to each other. . ·· Total acceleration of particle a(t) is ; j


, ; ', Solution : Forces acting on water are weight mg and

........

'

the force of pail on water Fp . Fp may be termed reaction of pail on water. Same force will be exerted on pail by·water. ~·,,,...,.··---- .

.~·--c,:- .·- i ·Y

t-=-~~

r,:-"1

~-t

:ottoms lis what ~e ·se!1s~ )Vhen we speak offeeling our weight. Wlien !the airpla11e:biinks, the seat has. to exert enough normal force ito·balance weiglit and to accelerate· the plane. •

-·--

>•·

---·

-

•

!

-+

L,

! ,.

0

0

-+

w

• (aj.

(b)

~1 I

'

L__C_. - - - - --- ' - · · · ·- · - -

8b:;;

:-7

MECHANICS-I s·

'

1

___ }

Motorcycle Stunt ~· f

_,

_, N

_,

N

w --; I

w:

Fig. 2.68

Fig 2.68 is a free-body diagram for the motorcycle and rider, modeled as a single particle.

Concept: Static friction, exerted by the cylinder walls· on the motorcycle tires, balances-the weight of cycle and rider. 'The normalforce acting on the tires causes the centripetal· iaccel_':ration of cycle and rider. If the rider tries the stunt at too low a speed, the normal force will be correspondingly small, and the maximum possible friction will be too small to balance the weight. (On a straight wall, there is no horizontal acceleration, no normal force arises no matter what the speed, and the stunt cannot be done.) The minimum speed for the stunt is that for which maximum friction can just balance the weight. Vertical Components Horizontal Components 'I.Fy = 0 'I.Fx = Max f~Mg=O N=Mv 2 /R At the minimum speed, friction is at its limit; µ,N = fmax =Mg.Thusµ, Mv~;n/R = Mg; so:

Non-uniform Circular Motion on Horizontal Plane Let us consider that a particle of mass 'm' is moving in a horizontal circle of radius 'r' with velocity' v' and tangenti;tl acceleration We will solve problem in reference frame of car. To oppose the tendency of skiilding of the particle (body) in the direction of net force F,er, a static frictional force F, is developed as shown in the Fig. 2.69 . To avoid skidding,

a,.

rnv 2

\

-·-~~ ' · -

· - - - - ._ - - · · ~.,u,.

r

bank angle

-+

w·

Fig. 2.69'

·(c) , Fig. 2_.67 , --

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FORCE ANALYSIS i.e.,

- - - - - - - ' - ' - - · - - - ____________ ------~-'-----'-'---'---'1~9.:-111 F,

=

(m;2r+(ma,)2

. mv 2 LFx = N sm 0 = max = - -

... (1)

r

Since F, ' : N

;

......

... (2)

de = (~) and tangential acc. of particle about dt

2

Ca =Fsine ' m

ar=Ra=(F:::'e}

y

f

... (1)

a=F:e

t

~

a=dro= d[2~]=2d2e d2e=~ dt dt dt 2 ' dt 2 2

1---1mg

'

Fig. 2E.94

Solution : In this case normal reaction of surface provides centripetal force and friction force prevents the man from sliding vertically. From Newton's second law, mv 2 Lf'x =N = - -

... (1)

r

Lf'y

where

=f

- mg = 0

Fig. 2E.95 (b)

... (2)

d 2e Fsine -=-dt2 2mR

2

f = µN = µmv

r µmv 2 From eqns. (2) and (3), - - = mg r or

C

µ = rg = v2

gr

(2nr/T) 2

,.,(3)

From eqn. (1) de = ~ = _! x (F cose)1/ dt 2 2 mR

= 0.5

(:~r ::(~r (:~r

la~~R~P Ie i_~5,:y A particle Pis moving on a circle under the action of only one, force acting always towards fixed faint O on the:

p

=¼F:e

... (4)

~ = Fsine x 4mR = 2 tane. 2mR Fcos8 1···-- . . . . .

r-c.

ks~q_!TI__p!~J -~:.---96 l > '

.

IA car is moving in a circular path of radius 50 m, on a flat., !rough horizontal ground. The mass of the car is 1000 kg. Ata• ;certain moment, when the speed of the car is 5 m/s, the driver' s 2• Find the value of ,is increasing speed at the rate of 1 •sta_ti,frictio!J.. on tyres at this moment, in Newtons,

m/

Fig. 2E.95 (a)

2

d 2e

- ·:

. ,, F'md ratw . oif -d2e & (de) czrcumJerence. . dt 2 dt

... (3)

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- -- .-- - -- -- --- -MECHAN·1cs:1 i --·-----------··- ·---- __.

1202

~-------------·-·--,.-··

. 5m

···IIJ

:......._____~-----··.-· B

'

Fig. 2E.103 (a)

'(a) Tangential acceleration of the block. (b) Speed of the block at time t. (c) Time when tension in _rope becomes zero.

Fig. 2E.102 (a)

Solution: (a) Tangential acceleration is the retardation produced by the friction a= -Jim= -µmg/m a, = -0.2x 10 = -2m/s 2 dv (b) - = a, =-2 1 ••.•------·:· ...... dt

Solution: Radial direction: T1 sin60°+T2 sin60°= mco 2 r

11:......... .

(T1 +T2 )sin600= mco 2 Lsin60°

= mco 2 L

Vertical direction : T1 cos60°-T2 cos60°= mg T1 -T2 = 2mg adding eqns. (1) and (2), 2T1 = 2mg + mco 2 L mco 2L T1 =mg+--

... (1)

... (2)

60°

10

0

•/

5m

p1Qm/s: .

·'·

: ·~········--------------·· ·,

·

I

'.

Fig. 2E.103 (b)_

v-10=-2t V = 10-2t (c) Tension in the rope will become zero when centripetal acceleration becomes zero i.e., when speed becomes zero v=0 => 10-2t=O => t=Ssec.

L.s,dq~p,,e;__[W41~ ·A ball of mass M is swing around in a circle around on a lighti ·spring which has spring constant k The ball describes a :horizontal circle a distance h above the floor. The stretched spring has a length I and makes an angle ewith the vertical as, ,shown in Fig. 2E.104 (a). Neglect air resistance. :

. 2

_____

C

f dv = -2 f dt

Fig. 2E.102 {b)

T1 + T2

V

60°!

,::·..··.·.·.::·"·..--+-.C............... .......... .

mg (02

> 2g L

=>

/illlllllll/111/JIIIIIIUIIIIJ/J

Fig. 2E.104 (a)

... (2)

co>Ff

~~mpJ~f 103 )> 25

~-block of~ass ~-res~ on a hor~o~taiflo;r (~- = 0.2). It; lis attached by a 5 m long horizontal rope to a peg fixed on, ,floor. The block is pushed along the ground with an initial.

\;L~f~

(a) In terms of only the given quantities, what is the· magnitude of the force F that the spring exerts on the mass M? ,Cb) In terms of F, k and i what is the natural length 10 of the: spring, i.e., the length of the spring when it is not: stretched? '(c) In terms ofF, l,M and 0, what is the speed v of the ball? (d) At same instant aftime, the spring breaks. The ball moves' a horizontal distance x before it hits the floor. In terms al, l .v, Ii, ,ind g, what is x_?

of 10 ~/s so that it mov~ in a circle_ aroun_d ~he_ ~eg. j

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l_Fo~c~_~N_A~s~ -- · ·--- · ·__ .. __ -·--~ .::_:_ ~~-

-- ------

Solution: Concept: When a particle moves in a circle, ,perpendicular forces along y-axis balance out. Towards of horizontal circle centripetal acceleration acts therefore that must be a resultant force. Note that along the length of spring forces are not balanced because this direction has component of acceleration. ~~

friction between the shoes and the drum is µ, find the power required in watt to tum the governor shaft.

Solution Centripetal force for rotation of brake shoe comes from normal reaction between brake shoe and drum. N = mrro 2 = mr(2rrf) 2

! \._

+.

mg

Friction force

(c)

(b)

(F)

Fig. 2E.104

(b)

F,p

=>

=

F sp

= µmr(2rrf)2

10 = 1 - ~ Kcose 2 . mv F,p sme = - r -

Mg case

P=2Fv = 2 x [µmr(21tf) 2 ] x r(21tf)

= l6mµ1t3 f3r2 r

~

,--~,. '-................... ;r=l~_;_~~---·

,

Fig. 2E.104 (d)

mg . mv 2 --sme=-cose I sine

--~

---

- · - - ----

A particle suspended from the ceiling by inextensible light string is moving along a horizontal circle of radius 1.5 m as shown. The string traces a cone of height 2 m. The string breaks and the particle finally hits the floor (which is zy plane 5.76 m below the circle) at point P. Find the distance OP.

I _

glsin 2 e cose

h X

1 2 = -gt 2

=>

t=f!

~ 1.5m

.-, ,./

/

= Vt

x=vf! l.,S:~R\J\.i:?J?

--r--,..

~~gm~,!,~ }106 I >

v=.1=--(d)

Fig. 2E.105 (b)

Power required to overcome friction force on both the brake shoes

= K!il

Iii = F,p = ~ K Kcose 10 = l - Iii

(c)

--

Fig. 2E.105 (a)

(~·.·.·.·-.-, -~~---5t --~···· ,.,

F,p cos0 = Mg

- - - - - - - ~------ -- ---

acceleration

"\:T··.0,

(a)

·---- ..

component of

+y-axis tTcosB

8

-

y

!1057>

Q

112m 5.76m X

.! -------p[_..-····· Fig. 2E.106 (a)

The essential elements of one form of simple speed governor are as shown : to a vertical shaft a horizontal rod is mounted symmetrically and on the horizontal rod are freely sliding' brake shoes, ' . When the shaft turns at a frequency of rotation f the brakel shoes press against the inner surface of a stationary, cylindrical brake drum. If the brake shoes are each of mass m! and their thickness dimension is negligible compared to the inner radius of the brake drum rand the coefficien~ of sliding'

Solution : Let the string breaks when the particle is 1.5 m right of point O and direction of its velocity v is along y-axis. . mv 2 Tsme=--

r

and

Tcos0=mg --v = .Jgrtan0 Now time to reach the floor,

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. ·-

. -. - --

______ M_ECHANICS-1_i

I _____ -- ------- ---- . - - - -

t =

~

=;

Before it hits the floor, l'>.y = vc:t,---c: = ~2h 2 rtane r where cane= h1

T 0

--,-

v=~=Sm/s

(b) Tangential component of force =k(3:)sine dv 9kR m-=dt 25 Rate of change in speed dv 9kR -=-dt 25m

Fig. 2E.106 (b) ,

l'>.y=~

=

2xl44x(l.5)2 25 . 2 18 =-m=3.6m .

Fig. 2E.107 (b)

·=;

24m/s 2

5

Its position from 0, when it hits the floor = LS i + 3.6] OP= ~(1.5) 2 + (3.6) 2 = 3.9 m

A bead of mass m = 300 gm moves in gravity free region. .along a smooth fixed ring of radius R = 2 m. The bead is 'attached to a spring having natural length R and.spring, constant k = 10 N / m The other end of spring is connected to;

An inclined plane of angle a is fixed onto a horb:c,ntal. tum-table, with its line of greatest slope in same plane as a 'diameter of tum-table. A small block is placed on the inclined plane a distance r from the axis of rotation of the tum-table and the coefficient of friction between the block and the inclined plane is µ. The tum-table along with incline plane, spins about its axis with constant minimum angular velocity

~-

'

:

6

r

:

,.__,.;

a fixed point O on the ring. AB= R_ Line OB is diameter of

!

-5

!

1ring:

Fig. 2E.108 (a) Fig. 2E.107 (a)

_Find (a) Speed of bead at A if normal reaction on bead due to ,ring at A is zero. i(b) The rate of chqnge_ in SJJ.eec:( qt this irJ5.tanf.

Solution:

Concept: Spring force has component in radial as well _as tangential dir_ection. (a)

Elongation in spring= (2R) 2

-(

6 :

r

-R

(a) Draw a free body diagramfor the block from reference of. ground, showing the force that act on it. (b) Find an expression for the minimum angular velocity, oo,, to prevent the block from sliding down the plane, in terms of g, r, µ and the angle of the plane a. (c) Now a block of same mass but having coefficient of friction (with inclined plane) 2µ is kept instead of the original block. Find ratio of friction force acting between block. and incline now to the friction force acting in part (b).

Solution: (a) N

k

=SR_R=3R 5 5 Radial component of spring force =

k(3R) case= 12kR 5 25

As normal reaction is zero

12kR mv 2 --=-25 R

.

Fig. 2E.10B_(b),•

Concept: Not force along vertical axis is zero and along radial axis provides centripetal acceleration vertical axis. . .

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[ioRCE ANALYSIS .

~~~~ Ct. stone is ia~~ched upward at '45° with speed v~. -fl beej

(b)

0

[follows the tra;ecto,y of the stone a_t a constant speed equal to/ \the initial speed of the stone. ·· . ; (a) Find the mdius of curvature at the top point of thei • •. . .· 'I I traJectory. · · . · . . i '.Cb) What is t/ie acceleration of the bee atthe top po/nt of, the1 L_. trajectorxLf.QrJ/Je store,J1_egle_c_t.thLair.J:.i.~

t

a-

·ji

is hrn,g by fu>mg 24. Blocks A & C starts from rest & inoves to the right with acceleration aA =12tm/s 2 & ac = 3m/s 2 • Here''t' is in seconds. the time when block B again comes to rest is :

.I . A.

,.....+

I,

,_ -:

25. In order to raise a mass of 100 kg a man 60 kg fasterts a rope to it passed the rope.over a smooth pulley. He climbs the rope with acceleration Sg/4relative to rope. The tension in the rope is: (g =.10m/s 2 ) ·. , (a) 928 N (b) 1218 N (c) 1432 N (d) 642 N . 26. A ball is held at rest in position A by two light cords. The horizontal cord is now cut and the ball swings to the position B. What is the ratio of the tension in the cord in position B to that in position A? . (b), 1/2 (a) 3/4 (c) 3

(d) 1

27. In the shown figure two beads slide along a smooth horizontal rod as shown in figure. The relation- between v and v O in the shown position will be : (a) v = v 0 cote (b) v,;, v 0 sine (c) v = v 0 ,tan8 (d) v = v 0 case 28. Two masses each equal to m· ,~,--·--.-;..,+; are constrained to move only · >,..f · ' ., · · along x-axis. Initially they m ' '· m x are at (-a, 0) and (+a, 0). (-a, O) i. (~. of . They are connected by a light string. A force F is applied at the origin along y-axis resulting into motion of. masses towards each other. The accel~ration of each mass when position of masses at any instant becomes (-x,0)_and (+x, 0)is given by: F.Ja 2 -x 2 . Fx

y~d.

(a)

(c)

m

---

(c) 2·s

2m.Ja2 -x2 (d)

_!_

~

,2mV~

29 . All surfaces shown in figure are smooth. System is released with the spring unstretched. In equilibrium, compression in the spring'will be :

I

3 (b) -s

2 1 (d) -s 2

(b)

X

F x m:J 0 2_x2

[---'---·--·-~II (a) 1 s

j

~! ·I

1 (a)

2mg

k

mg

(c)

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..J2k

(b) (M +m)g

..J2k (d) mg k

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I FORCEANA~iL:s.L'..;..,.---'-"~30. Find

the ' maximum · ('."".

/ ~;~fi~!~~~ l~~~~~:

.

-- ""-

kufu!::::e~~

with acceleration 'a'. All the surfaces ate smooth : ma 2ma (a) (b) 2k . k ma 4ma (c) (d) k k 31. A block of mass M is sliding down the plane. Coefficient of ·static friction is µ, and kinetic friction is 0 -- , µ k. Then friction force acting on the , =block is : (a) (F+Mg)sin8 . (b) µk{F +Mg)cos8 (c) µ,Mg cos0 (d) (Mg +F)tan8 32. The displacement time curve of a particle is shown in the figure. The external force acting on the particle is : ' a. I-~ , ,o~---~--.. (a) Acting at the beginning 0 Tim~~, part of motion (b) Zero (c) Not .zero (d) None of these 33. A block of mass 'M' is slipping down on a rough inclined of inclination a with horizontal with a constant velocity. The magnitude and direction of total reaction from the inclined plane on the block is : (a) Mg sin a down the inclined (b) less than Mg sin a down the inclined (c) Mg upwards· (d) Mg down wards 34. A block of mass 0.1 kg is held against a wall by· applying a horizontal force of SN on the block. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is : (a) 2.5 N (b) 0.98 N (c) 4.9 N (d) 0.49 N 35. A body of mass Mis kept on a rough horizontal surface (friction coefficient= µ). A person is trying to pull the body by.applying a horizontal force but the body is not moving. The force by the surface on the body is F where: · 2 (b) Mg ,,;p,,;Mg~l+µ (a) F = mg

·. _•.-·. [E

lI . lj

(c) F =µMg

(d) Mg?. F?. Mg~l-µ 2

36. A spring of force-constant kis cut into•two pieces such ,, that one piece is double.the length of the. other. Then the long piece will have a force-constant· of :

00 ~k

(b) ~k

3

2

(c) 3 k (d) 6k ' . ' 37. In the arrangement shown in figure -----, tlie wall is smooth and friction coefficient between the blocks is µ =0.1. A horizontal force F =1000 N is applied on the 2 kg block.The wrong statement is : (a) The normal interaction force 1:>etween the blocks· i~"lOOON. (b) The friction force between the blocks is zero. (c) Both the blocks accelerate -downward with acceleration g m/ s 2

(d) Both the blocks remain at rest -r--,~38. 1\vo blocks are kept on an inclined plane and tied to each other with a mass-less string. Coefficient of friction between m1 and inclined plane is µ 1 & that between m 2 & the inclined is µ 2 . Then: (a) The tension in the string is zero if µ 1 > µ 2 (b) The tension in the string is zero ifµ 1 < µ 2 (c) Tension in the string is always zero irrespective of µ, &µ2 (d) None of these 39. A block kept on an inclined surface, just begins to slide if the inclination is 30°. The block is replaced by another block B and it is just begins to slide if the inclination is 40°, then : (a) Mass of A > mass of B (b) Mass of A< mass of B (c) Mass of A =mass of B (d) All the three are possible 40. A force of 100 N is applied on a block of mass 3kg as shown below. ·The coefficient of . friction between wall and the I .. 1 F = 100N block is 1/ 4. The friction force Fixed vertical'wan: __ _ acting on the block is :

,:--

"d

.,' '~-~·-·_. h..,,,, . 0

l

(a) 15 N downwards (c) 20 N downwards

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_

(b) 25 N upwards (d) 20 N upwards

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:,., ..

- ' ......

41. An insect surface crawlsveryup hemispherical slowlya

1.• ME~HANICS-1

·

iw· _ ·-. I (X

·• _

•

.(see the figure). The coefficient of f'r!cti~n betwe_en the insect and · ):lie surface is 1/3: If the line joi1Jing the :centre of the hemispherical surface to the insect · -makes· an" angle a with the vertical, the · maximum possible value of a is given by : (b) tan a= 3 (a) cot a=' 3 (c) seca = 3 · (d) coseca = 3 ;ri,1ocl(of mass 2 kg is held. at rest. against a rough vertical wall by passing a horizontal (normal) force of 45 N, Coefficient of friction between wall and the block is equal to 0.5 .. ·Now a horizontal force of 15 N (tangential to wall) is also applied on the .block. Then the block will :. . . M~;e horizontally with acceleration of 5,m/s2 . ' (b) ·.. Start to move with an acceleration of magnitude · . • :· .L25 '!'( s2 (c) .Remain stationary (d) 'Start to .move horizontally with acceleration .,• .gte~t~r than 5 m/ s2

46. A stationary bcidy of mass m is slowly lowered onto a rough massive platform moving at a constant velocity v O = 4 m/s. The distance the body will slide with is : · respect to the platform µ· =

:o.~

(a) (b) (c) (d)

(a) (b) (c)

J.'i. Mg.

42 mg

0CM + m) + m )g

(d) (~(M+m) 2 +M 2 )g

~:~.

~

'

'

(b) 30° (a) ·_Zerp ·. (d) 60° 45~· · 45. The',force· F1 required to just moving a body up an incljned plane is double the force F2 · required to just preyertt 'the body from sliding down the plane. The coefficient of friction isµ. The inclination 0 of the plane

.

is·: ' '

ca)

'

truJ..:1

·

µ

(c) tan~' 2µ ,. •'

(b) t an -1 -µ

(d)

tan-1

2 3µ

''

•

Smooth surface

--;:=-=· ::;;=;-::-::------7 m =10kg F , . · ·

1 m2:::::15kg

, 1_ µ·::o 0.1 between the blocks , (µ:-coefficientoffrl,ction)

F'·' .. (Smooth ground) '

WeSI

Eas\

. (a) m1 experiences frictional force towards west only iJ; F1 > F2 (b) If F1 '# F2 then it is possible to keep the system in equilibrium certain suitable values of F1 &F2

for

(c) · If the system is to remain in equilibrium then F1 must be equal to F2 & F2 :s-10 N ·

.!i = !-!,_, m1

m2

then frictional force betwe'e~ the

blocks is zero 48. Consider the system as shown. The wall is smooth, but the · surface of block A & B in contact is rough. the friction force on B due ... to A is equilibrium is: (a) Zero (b) Upwards (c) Downwards (d) The system cannot remain in equilibrium 49. Given mA = 30 kg, mB = 10 kg, . m, = 20 kg. Between A&B µ 1 = 0.3, ~ A ,. F between B&C µ 2 = 0.2 & between B · • C & gronnd µ 3 = O.L The least . c · horizontal force F to start motion of · ·· · any part of the system cif three blocks resting upon one another as shown below is: (Take g = 10m/s 2 ) (a) 90 N

(b) 80 N

(c)' 60 N (d) 150 N SO. The coefficient of friction between the block A of mass m & block B of mass 2m is µ. . There is no friction between blockB & the inclined plane. If

I ~ ·

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v0 =-4m/s1

rnw·.

m

cc1

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44. The pulleys -and strings shown in tit!! ,figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle e should be :

i

I\ " ''.. Platform

1,m,J,,,;;;;;,t;,,, I

r-·

i m1 , then acceleration of the weights i~ :

hanging icleal string. The maximum possible tension in the string is 1000N. --The minimum time taken by the man to reach -upto the pulley : (a)

m

!~-~ \

60. A plank of mass 3 m is .placed on a rough inclined plane and a man of mass m walks down the board. 1f the coefficient of friction between the board and inclined plane isµ = 0.5, the minimum .acceleration of does not slide is : (a) 8 m/s 2 (b) 4m/s 2

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0.2

l ; l;;;;, ~kgl;?

(5 - 2t)N

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l I

30°

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~-~- -· ....,.._ __ I

(b) 2n + 1

2n 2n-l (c) 2n,- l (d) 2n 2n+l · 2n+l '• 62., A· wedge of mass 2 m and a cube of :1.' mass m are shown in figure. Between ":' ' cube and. wedge, there is -no friction. . The minimum coefficient of friction 45° between. wedge and ground ·so that ~ wedge does not move 'is : caJ 0.20 CbJ 0.25 (c) 0.10 (d} 0.50 63. The figure shows a block 'A' resting on a rough horizontal surface with µ = 0.2 A man of mass 50 kg standing on the ground surface starts climbing the

- J

. (b) 3 N

!

:

sn+l

il

67. In the above question 66, if the same acceleration is towards right the frictional force exerted by wedge on th,;, block will be : (Coefficient of friction between wedge & block = ../3/2}

61. A small block slides without friction down an inclined plane starting . from rest. Let Sn be the distance

~ n -.1 to .t =·n. Then ..!!.E_ is

4

· (d) Zero 66. The acceleration of small block m with respect to ground is (all the surface are ·smooth) : (a) g . ,(b} g/2 (c) · Zero · (d) .fig

(d) 3 m/s 2

.

'jto~-J ·

. r,L-----~ . - _,,_ '"--·· ,. ...·___I,

--·---·---· ~ ~ - - · · - · - - - - - ~ -

m1 +m2 ,

. :

¢•··'"· - · - - · - --.-

·""r

(a) 2 N (c) 1 N

(c) .Cm2 - m1Jg- f ·

(a) 2n - 1

20m

(b) 1

,------.

m1 +m2

traveled from time t

i

(c) .Ji (d) none of these 64. In the above question 63 distance between the man · and the block' A', when man reaches the pulley is :(a) 10 m · (b) 2 m (c) 20 m (d) None bf these 65. The force acting on the block is give1_1 by F = 5 - 2t. The frictional force acting:1m the block after time t = 2 seconds will be : (µ = 0.2)

(a) f :- (m2 - -m1) g

(c} 6m/s 2

,;r-~· : 50kg

(c) 2mg

(b) 3mg 2 (d) mg 2

68, A· block of mass 'm' is held stationary against a rough wall by applying a force F as shown. Which one of the following statement is incorrect ? ' (a) Friction force f = mg (b) Normal reaction N = F (c) F will not produce a torque (d) N will not produce any torque

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69. Two blocks A and B of masses and m, respectively, are connected by a massless inextensive string. The whole system is suspended by ·a massless spring as shown in the figure. The magnitude of acceleration of A and B, immediately after the string is cut, are respectively : (a) g,g/2 (b) g/2,g (c) g,g. (d) g/2,g/2 70. Two particles of mass m each are tied at the ends of a light 1 string of length 2a. The whole O system is kept on a frictionless nJ p '',m horizontal surface with the string held tight so that each : Jc a >Jc ., a >j mass is at a distance 'a' from the center P (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force F. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes 2x, is :

o-

(a)

a

F 2m )

(c)

(b)

2m )

0 2 -x2

F x 2m a

F

(d)

r

00

0 2 -x2

F )a 2 -x 2 2m X

(b). mro~a

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,., :I +=-.er

i

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.

r-.·- ---- ~--1

, , h-..!

. Jc· I l [ . . i

(d) mroro 0 a

73. A particle of mass m1 is fastened to one end of a massless string and another particle of mass m2 is ,fastened to the middle point of the same string. The other end of the string being fastened to a fixed point on a smooth horizontal table. The particles are then projected, so that the two particles and the string are always in t:lie same straight line and describe

-- ~I

r----·

{d) : ~

.J

____t

----------'

particle moves along on a road with constant speed at all points as shown in figure. The normal reaction of the road on the particle is : (a) Same at all points (b) Maximum at point B (c) Maximum at point C (d) Maximum at point E 72. A particle of mass m rotates about Z-axis in a circle of radius a with a uniform angular speed ro. It is viewed from a frame rotating about the same Z-axis with a uniform angular speed ro O• The centrifugal force on the particle is : (c) m(

released from rest from point A ii~-A~ :_:·::~--~::_:~-~----·~;;-. inside a· smooth hemisphere bowl \ _,.,___, , 8 as shown. The ratio (x) of magnitude of centripetal force & normal reaction on the particle at any point B varies withe as:

X

71. A

(a) mro 2a

horizontal circles. Then, the ratio of tensions in the two parts of the string is : (a) m,/(m1 + m 2 ) (b) (m,. + m 2 )/m1 (c) (2m 1 + m 2 )/2m1 (d) 2m 1 /(m 1 + m 2 ) 74. A small particle of mass 'm' is

75. A particle of mass' m' oscillates along the horizontal diameter AB inside a smooth spherical shell of radius R. At any instant KE. of the particle is K. Then force applied by particle on the shell at this instant is : K (b) 2K

~@i1··-~-

A--- --------.;......... ••· B

i

_____ ·;_ _ _ _ _j

(a)

R

(c)

R

3K R

K 2R 76. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration a, is varying with time t as a, = k 2 rt~ where k is a constant. The power delivered to the 'particle by the forces acting on it is: 2 2 2 2 (a) 21tink r t (b) mk r t (c) (rrik 4 r 2ts)/3 (d) Zero (c)

77. A long ·horizontal rod has a bead which can slide along its length and is initially pl~ced at a distance L from one end A ofthe rod. The rod·is set in angular motion about A with a constant angular acceleration, a. If the coefficient of friction between the rod arid bead is µ, and gravity is neglected, then the time after which the bead starts slipping is : (a)

fa

(c)

../µa.

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(b)

µ

../ri.

(d) infinitesimal

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78. · In gravity-free space, a particle is in constant with the

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·1

witho1;1t fri_ction thr~ugh it. B is . r.;· .. . hii,er- surface• of. a hallow cylinder and ~aves in a circular path aJo!'g the surface .. There is some friction Dunng. •CA the moaon from. A, to C if . · ·+::: .B .•.. '""" . •,d ru,,i.""1 ""'""' ·between the particle· and the -surface, The.retardation will: •' . : '' ·::· of the particle is: ... : .:· · · . (a} Alw~ys be in contact witlr the ... ' .,:,.~:~:-'.· . 9, · Ca) Zero · . .' inner wall of the tube (b) Independent of_the velocity .(b) Always be in contact with the outer wall of the Cc) Proportional to its velocity , .tube : · _ · . . -. ,. '. Cd) .Proportional tQ'the square of its vel99ty Ci)- Initially be in contact w:it:li. the inner wall and later 'with the outer wall 79. A curved ·section of a road is banked for a speed v. If there is no friction Between· the road and the tyres (d) Initially be in contact with the outer wall and later · then: · with the inner Wall , 85. A particle is ~i>ving iri. ·a: 'circle ~t radius R in such a Ca) .a car. moving with speed v ·does noFslip on the .road way .thilt any instant 'the normal. and tangential components ·a( the acceleration '!i:e· equal. If its speed (b) a caris more likely to.slip on the road at speeds at t ·=· 0 is ·u 0 ,. :the tim_e taken to complete the first higher than ·v, than iit speeds lower .than v revolution is : Cc) a car is mqre likely to slip on the road at speeds Ca) R/u (b) u 0 /R 0 ,, _lower thaji v, t!i~. at _speeds. hlgher than v ·,- , Cd} ,a ·car can remain stationary on. the road without Cc) ~(1--e"2lt) Cd) ~e-2' slipping . -. ' .' _ · Uo 'Uo 80. In' a, circular :mbti~n- of ~ particle the tangential 86. A ·particle P is inoving in a circli ~f radius r with a acce)eration of the particle is given by 2t m/s 2 • uniform speed u. C is the center of the circle and AB is diameter. The angular velocity of P about A and C are The radius of the circle described is 4 m. The particle is · in the ratio·: initially at rest. Til)l.e• after Which-total acceleration of (a) ~ : .2 the' particle makes,45° with'radial acceieration is : (b) 2 : 1 Ca): sec · ... -· .r,'. ,_ (b),'2 sec (c} 1: 3 Cd) 3 : i 87. A small body of mass m can Cc}"3 sec -Cd)' '4 sec 81 .. A partide travels along the arc ofa circle bfradius Its s_lide without friction along 1 ' ] a trough bentwhlch is iri the . ·1 _· • • . ; ''.. ,j' · · speed depends on the distance. travelled l as v = a.ff., 7 where 'a' is a -constant. The angle a; between the form of a semi-circular arc . of radius R At what height h i · ~ h ; · vectors of total, accejeration. and the velocity of the particle is : , ,_ will the body be at rest with " · ~----1 1 respect to the trough; jf the trough rotates with (a)_.g = tan- C2l/r) - . Cb),. a= ~os- (2Zjr) ·., angular velocity OJ about a vertical axis.: uniform Cc} a; =·sin-1 C2Z/r) · ·(d), ·a= cot" 1{2Z/r) Ca) R · (D) R -· 2,g 82. p~~le of ·mas~ m is atta~~ed to· ~ne end of a string OJ2. oflength 1while the othe'r end' is fixed to point Ii (h < l) ' (c) 2,g· -Cd)· R-.L · · above a horizontal table. The particle is made to 2 . QJ2 OJ . revolve ·in a circle ion the table so as to make p 88. A car moves, along a horizontal circular road of radius r revolutions per second. The maximum value of p, if with constant speed v. The coefficient of friction the particle is to be in i:ontaciwith the table, is : CZ > h) between the wheels and the road is µ. Which .of the (a}-• 2rc.Jifi · Cb)' ..jg/h following statement is not true ? ; -. '. ' . - '.I: ' Cc) 2rc..jh/g Cd) -, ..jh/g Ca) The car slips if v > .Jµii · 2rc Cb) Th~ car slips ifµ < (v 2 /rg~ 83. A. sto~e is thrown horizontally with a velocity of 10 mfs

1~_· .- .·.-

·

at

a,·=

.i

r.

W·

0~ ·_· ·. .

i

R_;

Cc) The car slips ifµ_>(v 2 /rg) at t = 0. · The radius · of curvature of the stone's trajectory at t = 3 s is : [Take g = 10 m/ s 2 J Cd) _The- car slips at a lower speed if it moves with ·cal 1oJio m . Cbl 100 m some tangential acceleration, than if it moves at constant speed Cc) 10oJio ~ Cd) 1000 m 89. A smi>otli liollow cone whose vertical angle is 2a; with 84. T)Ie narrow tube· AC forms a quarter circles in a its axis vertical and vertex downwards revolves about vertical plane. A ball B has an area of cross-section i~ axis 11 tl!ne p~r seconds. A Particle is placed on the can _move slightly smaller than that of the tube and www.puucho.com ;· . '' ·, . . ,,

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',

,·· '.,

•• .. '

.

•.

inner surface of cone so that· it rotates with same speed. The ,radius of rotation for the particle is : (a)gcota/4rr 2 T] 2 '·, (b) ·gshl'a/4rr 2 Tj 2 (c) 4rr 2 TJ~/g

(d)

g/ 4rr 2 TJ,2 sin a

90. A particle is kept fixed ~n a turntable rotating uniformly.As seen frpm the. ground the partjcle goes in a circle, its speed is 20 cnys & acceleration is 20 cm/s 2 • The particle is now shifted to a new po~itiol) to make the radius half of the original.yalue. The new values of · the speed & acceleration will. be : (a) cm/s,·'10 cin/s 2 , (b) '10, ~/s, 80 cm/s 2

'10

(c) 40 emfs, 10

cm/s

2

(d) 40 cm/s, 40. cm/s\::,

'91. A particle· of mass in is suspended from a fixed point O by a: string of length' l. At t ~ 0, it is displaced from . its equilibrium position and released. . The graph which shows the variation of the tension T in the string with time t is : (a)

.

(b)

~o

· ~. . . .1.. ,:,-,:· rn e

I

1-.

:.

' ,

f,~r, ,; ·~.· :;• .· .•..·._·J· :j ,I_,_

··'

1· ---·-.

:,,,

-·-L.:J

~~-1·;r~,· r~I .,.

(c) l : , . ~ ( d )

It

(b) 10 se{ (d)'-'s tee'·,

(a) 20_,sec (c) 40 sec

~

96.

'~;;7

!1~:S~~cl~o:ti;ll~s :~::_ • figure'. .The , · · approximate· ~ · ./" = 01 variatio~ · of· direction . ·of · · t :· . ·· ·.::'.,;,·{. •, i' j resultant acceleration . as·· B ·····;,; >\J:~:.J particle µioves 'from A to B is : .. ' . . (a) clo~se ,. (b) anticiockwise '. , (c) · direction· does not changes.' (d). no~e o~ the~e ,., .· . . . . 97. In the above question 96 the net acceleration of particl~ is h«;>rizont'al only at 8 (8 is acute angle made by string.~th liµe OB)':. ' ··

=

•., .• ('1)· '(1) ./3 · .·' .'J3 . ·c· ).. .i) ~;(·.1r,;:J. r,; . . , .

' ·. (a) cps-' .

(b{si~~' .

. ~'( (~)_:rr: -rr:. -.sm · .. ;c-.,-cos_. 2. . '!3: . . ,_.,,.,.2 ',,: . ....,3 '· 98. Two similar trains are moving along the equatorial line with same.speep_but in·«;>pposite dire~tion. Then: (a) they'.will exert 'equaH9rce ,6n rails ,: :-, (b) they ~ :not exert 'any f~rce. as they are on equ!'ltorial line · · ·. , (c) on~. of them will exert.'zero ·force. (d) both exert.different forces _... ' . . ·. •. 99. Two b~· of'mass m and 2111 are attached with strings of length 2L and L respectively They ,are released from horizontal position. Find ratio tensions in the.string when the accelerati9n ·of b9th, is only ,in vertical direction: · . · · · ·. _. ' ·. ·· (a) 5 · . , · '(b). 5 c

92. A rod of length Lis pivoted at one end is rotated with a uniform angular velocity in a horizontal plane. Let T1 &T2 be the tensions atthe pointL/4and 3L/4away from the pivoted ends. (a) T1 > T2 (b) T2 > T1 (c) T, ~ T2 (d) The relation between T1 &T2 depends on whether the rod rotates clockwise or anticlockwise The driver of a car-travelling at speed V suddenly sees 93. ci:). 2 ,.s,·:·_ . ca) a wall at a distance r directly infront of him. To avoid collision. He should : · 100. Indicate.the direction offrictional f~r~e·6l' a car which . is movhlg along. ,the.· ctJrv~d. path- with .. ,non-zerf> (a) apply the brakes tangential acceleration; ih a,nti-clock' directioJi': . . . (b) tum the car simply away from the wa:11 (c) do any of the above options . (a) · (d) none of these .·,.-< _:I · . 94. A body is undergoing uniform. circular motion then which of the following quantity is constant : (d) . (a) velocity (b) acceleration • (c) force (d) kinetic energy 'il A particle is resting on an inverted cone as shown. It 95. . ·' is attached to cone by a thread of length 20 String is_ given remains parallel to slope of cone. The cone www.puucho.com

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Anurag Mishra Mechanics 1 with www.puucho.com

I 220 j

M_ECH~N.lCS-1 ] ---~-Cc-------~------ -------c:,-------:.....-....J.

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10 I. If a particle starts from A along the curved circular path shown in figure with tangential acceleration 'a'. Then acceleration at B in magnitude is : r·--···-s-··_ . 7

l' C'\!. .:f':.__ .•

103. A simple pendulum is oscillating without damping. When the displacement of the bob is less 'than

maximum, its acceleration vector in:

.c•

(a) 2a~1+1t 2

(b) a~l +1t 2

(c) a~1t 2 -1

(d) a1t~l + 1t 2

(a)

102. A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height.The speed with which the block enters the tracks is the same in all cases. At the highest point of the track, the normal reaction is maximums in :

':

,:

: -

L-,--." -,- ,_ · - - - - ~

¾,~

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(b)

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[ FORCE ANA~YSIS ···-

----

2 __ --~~~~ ,t~-~~-~~~-~ltern~ti~=-~~~~~~ 1. A particle stays at rest as seen in a frame. We can

conclude that : (a) Resultant force on the particle is zero (b) The frame may be inertial but the resultant force on the particle is zero (c) The frame is inertial (d) The frame may be non-inertial but there is a non-zero resultant force 2. A particle is found to be at rest when seen from a frame S1 and moving with a constant velocity when seen from another frame S 2. Select the possible options : (a) Both the frames are non-inertial (b) S 1 is inertial and S 2 is non-inertial (c) Both the frames are inertial (d) S1 is non-inertial and S 2 is inertial 3. Figure shows a heavy block kept on a frictionless surfaces and being pulled by two ropes of equal mass m. At t =0, , · ·· - - · ····-- - ·1 the force on the left rope is ' . ~ .i . 1 m m Fj withdrawn but the force on the 2@JN _ ! . right end continues to act. Let F1 and F2 be the magnitudes of the forces acting on the block by the right rope and the left rope on the block respectively, then : fort < 0 (a) F1 =F2 =F + mg (b) F1 = F, F2 = F fort > 0 for t < 0 (c) F1 = F2 = F fort> 0 (d) F1 < F, F2 =F 4. The force exerted by the floor of an elevator on the foot of a person standing there is more than the weight of the person if the elevator is : (a) going up and speeding up (b) going down and slowing down (c) going up and slowing down (d) going down and speeding up 5. If the tension in the cable supporting an elevator is equal to the weight of the elevator, the elevator may be: (a) going down with increasing speed (b) going up with uniform speed (c) going up with increasing speed (d) going down with uniform speed 6. A particle is observed from two frames S1 and S 2. The frame S 2 moves with respect to S1 with an acceleration a. Let F1 and F2 be the pseudo forces on the particle

-~or~ect -~

when seen from S1 and S 2 respectively. Which of the followings are not possible ? (a) F1 ,;, 0, Fz = 0 (b) F, 0, F2 _o (c) F1 = 0, F2 ,;, 0 (d) F1 = 0, F2 = 0 7. In the arrangement shown pulley r- ······ and thread are mass less. Mass of plate is 20 kg and that of boy is 30 I . ·: kg. ' : ' ' ' Then: . . 1 1~1a _i -.--- ---·-"~ ~. (a) If normal reaction on the boy is equal to weight of the boy then the force applied on the rope by the boy is (lS0g/7) newton (b) If the boy applies no force on the string then the normal reaction on him is 30 g. (c) If the system is in equilibrium then the boy is applying 125 newton force on the rope (d) None of the above 8. A smooth ring of mass m can slide on a fixed horizontal rod. A string tied m to the ring passes over a fixed pulley B and carries a block C of mass 2m as shown below. As the ring starts ' sliding:

*

*

:LI -

f h . . 2g case . (a) The acce1eranon o t e nng 1s --"--'-1+2cos2 0 (b) The acceleration of the block is

2g

1+2cos 2 0 . m 'the stnng . .1s - -2mg -"-c e tension ()Th 1 + 2cos 2 0 (d) If the block descends with velocity v then the ring slides with velocity v cos8. 9. A block of mass mis kept on an inclined plane of mass 2m and inclination a to horizontal. If the whole system is accelerated such that the block does not slip on the wedge then: (a) The normal reaction acting on 2m due to m is mg sec8 (b) For the block m to remain at rest with respect to wedge a force F = 3mg tan a must be applied on 2m · (c) The normal reaction acting on 2m due to m is mg sece (d) Pseudo force acting on m with respect to ground is mg tan a towards west

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:,~:1\·>f·: l--~-,:~22,.,:.'__·,f;~~~:~~~:>,~~4 . "~ J-!:=

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·.:.· ;~~~./: }'::~~~-1~~,f. ·i,, ~._:-~ves wlth:some·acceleraticin ~) _th~)iiO:tet~~ sti;i ha;:y_ w!ill~--the h~~~er .... man·moveswitlisciineaccelebition:'. , · cd:'rhe light~r:in~n _ls stati\)hazy,~hile,the heavier -_,: . •m·.an.·,m,,9.."e.'Yi,\li, some,,ac.~e\e,ra.t_io_h .. .-·-'.. , _ . td)< The twii' men move. wiih acceleration: cif the· same .. ·(, \•: -~,amJf.U~e ll! O~po~it,-. ... _.~~w f'

;,::.~:-,~-~,

, . (b).

The maximum force, which the man can exert on the wall is the.maximum frictional force which exists between his feet and the floor Cc) 'rhe man can never exerts a force on ,the wall °&hich exceeds his weight ,.(di .. The.man cannot be in equilibrium since, he is · exerting a net force on the wall · 15. A block of mass mis placed on a smooth wedge of

' ··

:~~;=~*re~~~~~!Jt:a:~:~e~s::~~~~~~~t:: from .the .grou.nd: ·. ' . . ' '

··· · · ··

Cb) ,(M + 171sin8)g -_ (d)• .('M_ +msin8)gcose · M+m· '

(~) (M'.t- mig (c) . Mg

16. A block of mass m is placed on ,a smooth Wedge of inclination e m,·th the horizontal..J'he ~hole system is accelerated so, that the_ block does not .slip on the wedge. Theforce exerted by the· wedge on the block has a magnitude:

Ca) mg/cose . Cb) mgcose (c)· mg (d). mgtan0, 17. In ,arrangement .shown below, the thre~d ,pulley :and spring, ~e .. all massless and there is no friction

r,

:::t;~:~.

I~eadsi;ri:~ctin~ in4 ~ m.,·r,J'-;,-'--\_;"·1 cut then just after thread· is cuf: ·, '

=·

.

(~) 1s:1;1Js\0::', (c) 40 rfi/s

2

. :/, :_': (b) 27:s:n;s . ,... -~

' '.

7.S·m/s 2 : .: •

r - ::·. _; : (d)

', '.· i ·,;l : ,.

. .

2

' ' ,·, ' ' '• ''

4

' -

f -. . l:-' ,/• , , '

13. !Ii.th~ figur,e; the puµey·P·mov:es· td -7>~ the right-with 1i·const1!-'!t s~eed u. T_he ~ -"' dOw,11\\'.ar~, speed M, 1ps VA, and tl}e' -. speed of B to,the tjght is vB: , ,· L , , AJ

(J)

(b)

VB

=h+~~'· a 3 > a2 · · ,', · (bl. a,·= a2,a2 =.a,' (c) a 1 =a2 =a 3 - , .. ', (d);:a1>a 2 ,a 2 :aa3 24. A man has falleli'into ~'ditch of i -_· ~ ·· ·, width d and two 'of liis· trienas' ,ar~ · , · '• " - -1 ~lowly' pulling him ciut using a light rope and. ·two fixed pulleys as . · shown in. figure .. Indicate ·the ' . · ·- correct statements : (assume, both the friends !ipply equal forces of equal magn/l}lde) ·' (a) The force exerted by both the friends·deqeases as the man move up · ·, · _, . · , mg 2 (b) The force iipplied by'each friend is , h ;-~ d~ + 4h 4 when the man is at depth of h -~c) The force exerted by both the friends inc~~ases as ·· the man moves up ·

r· : . , ·. Jj

(df The force applied b~ ea~ fri~nd is 25.

7, .Jd

rd th~ figure shown m,.~ 1 'kg; m

(b) .~

.

.

(d) g/3i' · 23. In the figure the block A, B and C of mass m each, have accelerations a 1 , a 2 &a 3 respectively. F,.&Fi are external. · forces of magnitude · '2 ·, mg - a_nd mg

. . ..

-

.

. .

E·I··;l /

·

.

~

F J

: '..~-,

m : , -\ 1

• ,

m ,

/>.{ ['~ ~mg B zm_

~ 1

S

~

I'

.

r

""'"" ,.i

_•ram.

Afu= F - "

is applied to2pulley (t is in second) then (g=1Dm/s

):

·

,

:

2~.

' ,

~~j• : . ;; .

.

iI ·

m

(c) 1 is lifted off the ground at t = l_O sec(d) both blocks are lifted off simultaneously 26. In the following figure all . surfaces are smooth. The 0

. . I -----'-..c

I

. ..

' ," .

~~- .,~"""''"" ''. (a) acceleration ofwedgeis_greater then g sine 2

(c) acceleration of mis g '. (d) acceleration· of wedg~ is g sine 27. In above question 26, the normal .fdrce acting,between: (a) wedge and incline plan~ i~ Mg cose · . ,, (b) m and wedge is mg cos~· (c) m and wedge is•zero · ·. . · ( d) m ~nd wedge is mg siri 8: : 28. In the figure shown .."'.i '." 5 f ~ kg, m 2 =10_kg & fnctt f (b) FN > f (c) F > FN (d) (FN-f)endiculaf to it which is alw;iys perpendicular to velocity ofj>4rticle. The motion is taking place in a plane it follows that : (a) vela~!~ i~ constant (b j accel~riition is constant (c) KinetiJ'~~etgy is constant (d) ii lllOVes in circuiar path 66._ A parti~le 9f mass m moves in a.conservative force field along' aifis where the potential energy U varies with position coordinate x as U = U0 (1- cos ax),U0 and a · being positive constants. Which of the following statement is true regarding its motion. Its total energy is U O and ~tarts from X = 0. (a) !i]e 'cceleration is constant (b) It's speed is maximum at the initial position.

70.

x

(c) It's maximum x coordinate is~ 2a

rd) It's maximum kinetic energy is U0

.

71.

(c) '.'cceleration of both the masses is same (d) ·"\!cceleration of both the masses is upward ' ·, Which of the following is / are incorrect: (a) If net normal force on a surface is zero, friction. will be z¢ro. '(b) Value ofstatic fii~tion is given byµ ,N. (c) Static friction oppo~es relative motion between two surfaces is contact. (d) Kinetic friction reduces velocity of an object. A spring block system is . placed on a rough ?orizontdaltfloor.dTh: bhlock ( .. 1s pu 11 e owar s ng t to ~---·--~-give spring some _elongation and released. Then: . (a) the bloc1' may s~op before the spring attains its natural length (b) the block m\lst stop with spring having some compression (c) the block may stop with spring having some compre_ssion (d) it is not possible that the block stops at mean ·position In the above situation the block will have maximum velocity when: (a) the spring force becomes zero (b) the frictional force becomes zero (c) the net force becomes zero (d) the acceleration of block becomes zero A book leans against a crate on a table. Neither is moving. Which '. :1 of Lhe following statements ! concerning this situation is/are incorrect ? (a) The force of the book on the crate is less than that of crate on the book (b) Although there is no friction acting on the crate, there must be friction acting on the book or else it will fall

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(c) The net force acting on the book is zero (d) The direction of the frictional force acting on the book is in the same direction as the frictional acting on the crate 72. An iron sphere weighing 10 N rests in a V shaped smooth trough whose sides an angle of 60° as shown in the figure. Jhen the reaction forces are:

73, In the sy~tem shown in the figure m1 > m2'. System is

held at rest by thread BC. Just after the ·thread' BC is , burnt: r, . . ' .

form

· 14 ..

li;

~·

G ___ ____1~---'~____

(a) initial acceleration of m2 will be upwards (b) magnitucje of initial acceleration of both blocks

(a) RA= ION andR 8 = 0 in case (i) (b) RA = l0N andR 3 = ION in case (ii) 20 dR 10 N , . . ("') (c) RA=. .f:3Nan 8 = ../3 m case m

will be equal to ( mi -. m2 ) g · m1 + m2 (c) 'initial acceleration of m1 will be equal to zero (d) magnitude of initial acceleration of two blocks will be non-zero and unequal.

60°

60~

.c_(ii-'-i)_-.__...,

(d) RA = l0N andR8 = 10.N in

all the three cases .

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: 230

Compreh~nsion Based Problems

I

-::

s.S'A;l(E

--,

-"·

3. The tension on side ofheavier\nass will be:

1 ...: ~,~---

pA

(a) m1g (c) '2m 2g

(d) 2m 1g 3 3 4. The tension c;m side of lighter ll_lass will be:·

Effect of friction between pulley and thread : In ideal cases i.e., when pulley and strings are massless and ,no friction exists at any contact surface, then tension in the string is constant throughout its length. But consider a, massless pulley and massless string but friction exists: ,between pulley and string With coefficientµ. Then tension :at the two .ends of the pulley will be different. As .shown in. figure, consider an element of string :

2

2

µdN i 8 8 Tcosd2 ~ ' (T+dT)casd2 , :

,··· e d;e___ ~_T+dT

.

m,

--···:· de· .

de.

.

· 2

,_

t

2

2

2

·., ._

.?.Ji~~

twq,

2 .

2

JliTdT = l"oµ. de T2

2

2·

=> In

·(T.~ J= µ1t

=}

T

~

= eµ:c

T,

!Suppose coefficient of friction between the string· and'

. 1 . pulleyis µ. = - .

'-·-· · · - · - •. , lt.

1. What should be the ratio of heavier mass to lighter mass for no motion ? (b)

I

.

2

2

.____ __ ,------ - ,-- - - ~I ' ,._

de de]· dT=µ [ T·-+O+T=µTease

eL

,·r-rr'~fi;t

'• ,-

37'

de [r sm-. , de + dT ·Sm-+ · . de T sm~ . de] dT -cos-=µ

(c)

.,

3

2

dTcos de= µ[er+ dt) sin do+ Tsin de]

(a) e

4m1g

, Cons!,. Jl:4

"." . ·~

... (ii)

From eqn. (i) and (ii) t = t'

t~

B

_j [. ~: gs:J

Time to reach B = t'

=>

C

tTI

,£, •t,.

N = 600 - T

=>

,

... (i) ... (ii)

T = 600 = 150 N 4

5. [c]

~~ : 2 tN

Both blocks will moves together 21 = 3x a a= 7m/s 2

=> T=lx7=7N Net external force on block A = 7 N 6. [a]. 1 Tension at all points will be F => rope is not moving, acceleration will be 0 F-T=0 F=T 8.· [c] T~20w Reading in spring balance =T/g T -4 tension in thread connected to spring is in Here system equilibrium and T = 20 x g . 20g 20 read mg=-= g

[~L

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\;FO! RCE ANALYSIS ,~'J.,; "'"'-'-··~-'----~=··'-----'-~'-"'..,,,~_....,,--9. [c] Tension T = !Oxg Also T'=T=_lOxg reading in both the spring T' T = - = - = 10kg

g

.

g

10 .. [a] Magnitude of F1 and F2 may be equal or may not be but their direction cannot be same because F1 is accelerating and F2 is decelerating.. 11. [a] Deceleration of body A

~

dA=(MAg+f) MA

Similarly dB =·MBg + f MB 2 Now, v = 0=u 2 +2ah u = same for both bodies u2 u2 ~ hA=-= 2dA 2(g +_L__) . MA u2 u2 hB=-= 2dB

15. [a] T=mg ... (i) [. . . Mg= zr case ... (ii) '.·T~e. \a , , T... e. T .~i" . . . . From eqn. (i} and (ii) 1 · . Mg ·: . •, mg Mg= 2mgcose I~--~-· . ,•-- ,, , ~---, M = 2mcose (as case< 1) MMB hA > hB 12. [b] When cable is ·cut down then chamber will fall freely under gravity, wedge and block both will also fall freely under gravity. :. acceleration of both will be g ,J. :. block will remain at top of wedge 13. [a]

with time velocity will increase since initial velocity was 0. 'a' is decreasing also after certain time a=O mg v=k when velocity = mg then a= 0 and. ball moves with k constant velocity. 18. [a]

While going upward a = F - mg·

m

h =.!..at

2

2

~

t1 =

T'=m 1g T' . m2g--=m 2 a

/2h v~

T'

2

m

t2 ~

= ~

f-;,-

... (i)

2

While moving downward , F+mg a=---

.

-m 3 g

= m 3a

... (ii)

Form eqn. (i) and (ii) as

a'> a

a=

t1 > t2

(m 2

-

m 3 )g

m,2 +m3

'

14. [c] In both cases initial relative velocity of elevator = 0 and g,J. · :. time will be same

Putting value of ci

( m m·)

T'=2m 2 l+ 3 - 2 g . m3 +mz

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2m 2 x2m 3

m1g =.

m2 +m3 4 . 1 1

23. [b]

g

Mg-T=Ma T-mg-=ma (M-m)g=(M+m)a (M-m)g a=

-=-+~1

m2

m3

19. [b] mg-T=ma

3

given

Tmax

=4mg

a.

=f4

mm

(!"f + m)

r'

for minimum value of acceleration 'T' should be max.

.

:r

l'

m(M+m)g T =mg+----~ M+m T= 2mMg M+m

..' .

~ ..

.n,g

T-2mMg =2mg M

Total downward force on pulley =2 T

20. [a]

=4 mg.

24. [d] Block B will come to rest when V!!locity of block A velocity of block B cc}

J~12t dt

. '..(i)

6t 2

Now let m kg sand is put (M+m)g-B=(M+m)xf . 6

cc}

t

... (ii) 25. [b]

From ~qn. (i) and (ii), m =~ M 5 21. [b]

· ~

cc}

...

...Fnet = 0

~

.

T2=W2+N2

=0, 0.5 =0.5sec

(,~:~;IlJ· T

And get

N cose =W sine N=Wtane

22. [b]

.

... (i)

a+ a'= Sg 4 T-100g=100a' Solve eqn. (i), (ii) and (iii)

T+W+N=0

also, W and N are at right angles also

= 3t

T--600= 60a

~

...

=J~3dt .

t .:___ _ ~]

System is in equilibrium :::::}

t

cc}

=

... (ii) ... (iii)

= l 9 soo ~ 1218 N 16

.

26. [a]

Tcose = mg

... For. moving with constant velocity F = 0 ...F+mg=0 _, ... A is in x-direction . ... :. For net.force to be 0, F should be in +ve y-direction ...... Now v xA_= mg .

... (i)

j

rT-!8 -. -r:,~ . -· . -. - -.·.l - •• .. r

net

, ·- :• . A

[____~ ·.

cc}

''

,

B

..

. .

/.

m~2

T'-mg cose = --·= 0

r T'= mgcose

vAsine = mg

. .

v=·~ A sine

;

For min. v, sine should be.maximum

cc}

v= mg A

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2

•

= cos 0 = ( ~ )

T'

3

T

4

2

... (ii)

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1

IS-'-'-~"------~-"""'-'-----~~''""·::1""·t1"':;c..?... I.• .... · _ _.....:..,_•..•i~-'~}~'-'--dv y--_.,l--_. a=-=0 I o ..

i...:..FO...:R...:C_E_AN...:A._LYS;.,.___

27. [a] As shown in figure vsine = Vo case V = Vo Cote 28. [b]

'j

0;58

.-

I~_ ij.

, ,.

sEf

dt

F=0

=}

33. [c] -+ -+ Finclin~d + Fgravity -+

~ F inclined

=0 --+

-+

= - Fgravity = - Mg

34. [b] Since the block is held held against a wall, the coefficient of friction will be equal to the weight of the block. Hence µ=mg = (0.1 kg) (9.8 !IlS-2) = 0.98N

I

'-'·--2Tcose = F For any mass Tsine=ma T sine

.a=--=

35. [c]

N = Mg & Fp,non

F sine

m 2cosem F Fx a= -tan8 = ----,=== 2m 2m.J a 2 - x2

29. [d]

flmax =µMg Clearly the magnitude of net force acting on the block . from the horizontal surface is F = ~f2 +N2 = ~f2 +M2g2

---

But=} =} =}

o,;;J,;;µMg o,;;J2 ,;;µ2M2g2

M2g2,;; f2 +M2g'2,;; M2g2 +µ 2M2g2

Mg,;; ~f2 +M2g2 ,;; Mg~l+µ 2

36. [b] The force constant is inversely proportional to length. If the length 1 of the spring is cut into x and 1- x such that x=2(l-x)

then x = 21/_3 From the inverse relation, we can write:

I

30. [b]

k1 l l 3 -=-=-=-

dv .mv dx = (ma - Toe)

J;mvdv= J;cma-kx)dx

k

~a,

I

0

kx2

2

gm/ and

2ma

k

21/3

37. [b] Since the blocks cannot accelerate in horizontal direction therefore the nom1al interaction force ber,,reen the blocks as well as between 5 kg block and the wall is F = 1000 N. Again both the blocks accelerate downward with acceleration s2

0=max-2 X=--

X

therefore the relative acceleration between the blocks is zero. Hence the friction force between the blocks is zero.

.

31. [b]

f, = µkN

38. [d]

=µdF+Mg)cos8 32. [b] Slope of displacement-time graph gives v~locity which is constant here v = constant

If a block is released on ·an inclined plane of inclination 8 and having friction coefficientµ with the block then the acceleration' a' of the block is (assuming tan 0 > µ)

a = ..!. (mg sin 8 - µmg cos8) = g (sin8 - µ cos8) m

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.- d

Hence greater the value of µ lesser is the value of acceleration irrespective of mass of the block.

39. [d] A block begins to slide on an inclined plane ifµ = tan 9 irrespective of mass of the block, where µ = coefficient of friction and' 9 = angle of inclined plane with horizo1_1tal. 40. [c]

fl~ax =µN =

(¾)(lO~)

N~Fsin30 30 Fcos 30 3g

25,.J3

=--Newton 2

Since 1_let force (excluding friction) acting on the block · is 20 N upwards therefore f = 20 N downwards. 41.

[af

·

·

For tile insect to be at equilibrium Ffr = mg sina or µN = mg sin a or I+ (mg coscx) = mg sin a. Hence, · · cota = 1/µ = 3. 42. [b]

.flmax = µN = (0.5)

(45) = 22.5 newton. Since magnitude of net external force except friction is 25 N, therefore, . f = 22.5 N . lal=.25-22.5 = 1.25 m/s2.

and

2

43. [d] Tension in the ·string, T = Mg . Ther~ two forces acting on the pulley. The force T acting horizontally and the force (M + m) g acting vertically· downward. The resultant of these force is ( ~CM +m) 2 +m 2 )(gl.

'are

44. [c] If T is the tension in the string, then T = mg (for outer masses) 2f cos9 = :/2 mg (for inner masses) 2(mg)cos0=:/2~ · or cos9 = 'lj:/2. => 9=45° 45. [d] Let m be the mas of the body.

F1 = mg sin9 + µmg cos9

... (i)

F2 +µmg cos9 = mg sin9 . . :(µ) => mg sin9 +µmg cos9 = 2(mg sin9-µmg cos9) => 3µ cos9 = sin 9 => 9 = .tan-1 3µ 46. [a] With respect to platform the initial velocity of the body of mass mis 4 m/ s2 towards left and it starts retarding at the rate of a= 2m/s 2 Using v 2 =u 2 +2as we get: 0 2 = 4 2 + 2(-2)(s)

=> s = 4meter. 47. [c] If F1 & F2 are not zero then friction force on m1 acts west wards & on m2 acts east wards. For m; to be in equilibrium F1 - ·f = 0 For m2 to be in equilibrium F2-f=O => F1=F2=f But f S 10 N. Hence F1 = F2 & F2 S 10 N 48. [d] Consider A and Bas a system. There is no vertical force in upward direction to support their weight. Therefore, the system cannot' remain in· equilibrium. 49. [c] Limiting force of friction between A and B is F1 =µ1mAg=90N Limiting force of friction between B and C is F2 =µ 2(mA +mB)g = BON Limiting force of friction between C and grou~ci is F3= µ 3 (mA + mc)g = 60 N As F is gradually increased the force of friction between A and B will increase. When F = 60 N block A will exert a horizontal force of 60 N on C. Hence C will be on the point of motion. Hence the least value of Fis 60 N. 50. [b] The acceleration of blocks down the incline will be g sin 9. Horizontal component of this acceleration is .·,,,,: Nl aH = aco.s9 and vertical component a, ';'asin9 f.',"'' \ •... aH = acos9 = asin9cos9 \mg and av =asin 2 9 For body A: Mg-N=ma, . 2 or, N = mg - mg sin 9 = mg cos 2 9 and, µN 2' maH µmg cos 2 9 2' mg sin9cos9

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µ;:: tan0 0=tan-1 (µ) _

or,

51. [a] Horizontal acceleration of the system is

F

F

a=-----=-

2m+m+2m Sm L_et N be the normal reaction of the system is

Tcos45°= ma

or,

2F N=2ma=5

Now B will slide downwards of

T =.fi. ma mg-Tcos45°= ma mg-ma=ma a= g/2 T.=mg_

µN;:: mBg

so,

µ(~);::mg p;:: 5mg

2µ

52. [b] Friction force between A and B(=µmg) will accelerate B and retard A till slipping is stopped between the two and since mass of both are equal acceleration of B = retardation of A =µg

. .fj_

55. [d] Extension in the spring= AB -R = 2R cos 30°-R =(.J3-l)R

V1=Vo-µgt v 2 =µgt

and Hence the correct graph is B. When the slipping is ceased the· common velocity of both blocks becomes • v 0 /2. 53. [a]

Free body diagram (RB.D.) of the block (shown by a dot) is. shown in figure. For vertical equilibrium of the block, . F N = mg +Fsin60°= ,J3g +-./32

So, spring force = kx c-./3 + l)mg c-./3 ..:1)R = 2mg R Free body diagram of bead is : N =(F + mg)cos30° = (2mg + mg)

2

Tangential force

f+- N. -:-:~--· ·1 . '· . •

I;f" 1·,

\.fert1ca1

··-· .: .

.

r: :

. F cos60'_ •

; : ,' : Horizontal

.!1'19 + Fsin 60° ,.. : •. "

2

· 56. [a]

For no motion, force ·of friction •

.J3 = 3-./3 mg

... (i)

I

:

=F sin 30° - mg sin 30° =(2mg + mg)si~30'= m{ I>

i

••.,. ~.,.

Tangential acceleration = g /2 57. [b] When the inclination of the slant side ·reaches the angle of friction, sand will betin to slid do~. So, for maximum heightµ= tan0 = · .

I

J;::Fcos60° µN;:: Fcos60° or F g;::or 2 or F ~ 2g or 20 N Therefore, maximum value of F is 20 N

R

or, 58. [d] T=Mg

54. [d] Just after the release B moves downwards and A moves horizontally leftwards with same acceleration say a.As shown in the free body diagram of both A and

.

N =60g-Tsin60° Also, T cos 60° =µN Solving these three equations M = 32.15 kg

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· MECH~~ 59. [c] For m2 mass

= m2 a T1 =T2 + f.

m2 g-T1 Also,

... (i) . .. (ii) ... (iii)

T2 - m1g = m1 a . From above three equations

T should be maximum 1000-SOxlO = SOa a= 10m/sec2 1 . . 2 Now 10=-xlOxt

=> 64. [b]

(m 2 -m 1 )g·-J a=~~-m1 +m2

f~

2

t = .Jz sec.

fmax = 0.2xN 1000- 0.2x SOg = SOa . 400 a=-=18

60. [a] .

so

' d=-xl8x(v2) 1 r;:; 2 Now =18 2 :. distance between man and ~lock= 20-18 = 2 m

65. [a] fmax = lOx 0.2 = 2N Initial force = SN > 2N block will move with acceleration

For equilibrium of 3mg sin37°= f + 2mg cos37° => f=2m For man, mg sin37°+f = ma 6m+2m=ma a=8m/s 2 61. [c]

a=S-21:-fmax S-2t-2 1 dv -=3-2t dt v=3t-t.2

S =~[2n:...l] n 2

a

Sn+1 =-[2(n+l)-ll 2

so,

~

l

N'.

~2t) '

10

(•: at t = 0, V = 0)

v=O t_= 0,3sec :. at t = 2 sec block is moving :. Jmax will_ act i.e., frictional force acting = 2 N

[2n - l] [2n + 1] '

Sn+l

,--B~-

=>

66. [a] Small block m will fall vertically as no external force is acting on it.

67. [a]

=>

N'=mgcose N'= 2mg + mg cos 2 8 = 2mg+ mg= Smg

2 fmax

=µN:=

2

S~ xµ

= mg case X sine= mg

N=.J3mg

2

1 µ =-= 0.20

fmax =.J3x ~mg= ¾mg

s

63. [c] fmax = SOg X 0.2 = 10g T-SOg = SOa For minimum time acceleration of man should be maximum

·1·~·,,,N',,1 .. .· ·.· •.·, .i""'. T'-' f

I.'.__.

50g.

:. block will not slide 3 Since f= mg _ mg =mg Ne =}·NE is maximum. ·: RE 0, then ball will be in contact to lower surface and if N < Oit will be in contact with upper surface.

= vr = ( a;zJ

Angle between a 0 et and v is same as angle between a.et & a,· : a tan ex== ___f_ a, a= tan-

1

N >0

,., .·_·,- 8~--_ ·a·_·(2"'

e ,' . ,, .

r•

=}

cos 8 > ~ 3

[from eqn. (i)],

net,

- ' ·

.

conservation

mgR(l - cos8) = -1 mv.2

dv a dl av a2 a--------' dt 2./f. dt a.ff. 2·

, a,

Using energy between A & B.

= a./f_ 2

=}

'

mv 2 mgcos8=-r

Let, when particle is at angular position 8, then distance travelled = 1.

and·

v =30m/s' • Jy/'

81. [a]

But

"

,. . '

" ·--·-•-....:... .. '' .....--t,_..:,.;...._ L

(X' •

•

, .• ,;:v.. a,

85. [e] (~)

Given

82. [d]

N+Tcos8= mg

... (i)

=}

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dv v2 = dt R V dv t l J-dt uoV oR -

f2=

... (i)

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[ FORq ANALYSIS =?

R (__!__· Uo

.!) =

t

... (ii)

Again from eqn. (i) dv ds v 2 -·-=ds dt R 2 V dV '"' ds

J

=?

====>

""---;;- o·R t

v

90. [a]

e-2•

~

I

r@II

2(!~) (!:) 1:2 =

2.roA=COc

N sine= mro 2 r

... (i)

--=-

R-h

=?

.

mv 2 T-mgcose = - r

h

g

h=R-_L

=?

mv 2

- - > µmg r

as well as mv both increases. Hence graph will be r (d), (c) is not acceptable because at t = O, T ;t 0. (b) is not acceptable because the variation is not linear.

dm=(7)dx

Hence (a) and (b) are both true. Again If there is tangential acceleration then for slipping : µmg=m V

r

when mass is released from displaced position, 0 starts decreasing and v starts increasing. As a result mg cos0

> .JµriTrue

V

=?

mv 2 =mgcos8+--

92. [a] Considering an element oflength dx at distance x from axis of friction.

(02

88. [c]

Car slips if

T

2

g

=?

=?

At any angular position 8

=?

and 2 ro r tan8=-

=20

91. [d]

=

N cos8 = mg

2

R=20cm

=?

co= l rad/s At new position R = 10 cm So, v =Rro =lOcm/s And acceleration = R 2ro = 10 cm/ s2

II . A•'

. 87. [d]

=?

-=20 Rro

)

I

26 =

COA COc

v2

... (iii)

Uo

=?

=(gcotet)

R

86. [a]

=?

x

=21t11X)

4it2112

= Uoe21t

=~ (1 -

(·:v

v2

=?

dv v 2 =:> v - = ds R

- J

from (ii) and (iii),

=?

tanet= gx

=?

V

K

at x = L, T = 0 (T for tension) (T + dT)

2

2 )1/4 =.Jµri l - _a_ ( µ 2g2

·: (d} is also true. 89. [a]

m

T

X

0

LL

2

= JdT= J-xro dx

= -nt2 ( x22

I

=?

T

=?

T = -mro (x2 -L2) = mro (L2 -x2) 2L 2L

2

2

=?

.

N

93. [a] Let 'F' be force of friction in each case for stopping car by applying brakes

sina

.!

N cos mg

i

2

mv

2

N cosu = - X

and N sinu = mg

mv 2

~ F. r

(i.e., work done by friction should be .

greater than kinetic energy) 2 p;,, mv =? 2r

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... (i)

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MECHANI~ 2

For turning the car F :c, mv

N = mg - mv

2

... (ii) r The required force is less in case of applying brakes. 94. [d] . Direction of speed is changing so velocity is changing => acceleration and force are also changing. 95. [a]. Length of thread = I T Mass= m 0 N ~ 0 in limiting case Tcose = mg Tsin0 = mrOJ 2 rng • rOJ 2 => tan0=--

g

99. [d] At position B acceleration is only vertical. For particle 1. Let velocity at B·= v from energy conservation 1 2 -M1v =MgL1 2 M v2 Also atB T1 -M 1g = -1-

=>

.J3:,; !xsin60°xOJ

=>

10

From eqn. (i) and (ii),

T1

OJ :c, 10

Conserving energy at points AandP

Net acceleration at 2

mv =-

cos0 =

R

~

=>

B = ~a

2

+ a 2 it 2

= a~l + 1t 2

102. [a] At the highest point, we will have Mg +N = mv 2 /r

= 2mgcos0 T = 3mg cos0 when particle is only horizontally accelerated at this moment => T cose = mg => 3 mg cos0 x cos0 = mg

=>

2

R

= 2gcos0

· => T-mgcos0

= 3Mg

ic2 Normal acceleration at B = .....!!.. = a 1t

2

2

R

= 3M1g

vf = a!tR

97. [a]

-

= M 1g + 2M1g

100. [c] As car is moving in anticlockwise direction and have , tangerrtial acceleration .(swell as radial acceleration :. Friction component should be along tangential and radial direction 101. [ b] 2 2 1tR VB "' V0 + 2a X

When particle is at point A acceleration g ,J, Point , B acceleration is towards ot :. acceleration varies as i.e., clockwise

v2

T1

T M m 1 -1 = -1= - = Tz M 2 2m 2

96. [a]

mgxRcos0=~mv

... (ii)

Similarly for particle 2 : T2

=> t = 20sec

OJ= ext

... (i)

L,

For block to leave contact e :c, 60° 2

will be different

R

1 ) e =· cos-1( .J3

Hence, minimum the curvature r, the maximum is the normal reaction. 103. [c]

a

Net acceleration of the bob in position B has two components. ->

98. [d] Since earth is also rotating Therefore, both will have different velocity w.r.t. centre of earth as they are moving in different directions

~-.,',

(i) an = radial acceleratioh''(towards ' '~ .' BA) ~--~,.~( (ii) a, = tangential acceleration (perpendicular to BA) Therefore, direction of (c). .

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ais correctly shown in option

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FORCE ANALYSIS

·251

2. [a, c]

Particle is not accelerated as seen from both the frames. :cc; frames are not accelerated w.r.t. each other :cc; either both are inertial or both are non-inertial but moving with same acceleration. 3. [c]

.

·- -· ·

, F ~ F,

1;- -

j'

=;

- ··----

for

t 0 system accelerates :cc; F-F2 =ma>0 F2 ~ F1 >F 4. [a, b]

N-mg=ma N=mg+ma N>mg

=;

.

F2~FI

F = F1 = F2

:cc;

N 4 T = S0xlO 4 T = 125 N

=;If

For t < 0 system is in equilibrium

I --

= 70g

N

N=30g 30g+T-30g=30a 3T- 50g = 20a 4T = 10a 50g a=-70 T = 150g 7

If boy

applies no force on rope T :cc; free fall will be there

"i ir'

a

·•.·.

!L_ ·___•· .mg

if a is +ve i.e., elevator speeds up while going up or speeds down while going down.

8. [a, c] Tease . ~

.

~mgl

-x2+y2=h2 -2xvx + 2yvy =;

T = w.r.t. elevator

=0

V =-y-

V

x

cosB

=;

Vy= Vx COS0

=;

ay = ax cosB

a=O It can move only when with uniform speed

= aring x cosB T = 2m x ablock

S2 is accelerated w.r.t. S1 =; relative acceleration of the twci frames is not zero :cc; minimum one of the frame is non-inertial at least one of F1 and F2 * 0

7. [a, c] For equilibrium N+T=30g 3T-N=20g 4T = 50g. N = 30g- SOg

4

... (i) ... (ii) ... (iii)

ablock

6. [d]

F1 = F2 = 0

I

··--··- - .. . - __ J

r

5. [b, d]

is not possible.

=0

2mg T cos0

= m X aring

From eqn. (i), (ii) and (iii) 2g cosB and T = Zmg aring = 1+2cos2 8 1·+ 2cos 2 8 9. [a, b, c] When block does not slip mg =N coscx N = mg seccx Since block m does not slip on block 2m :. both can be taken as on~ system N'=3mg

Normal reaction on 2m by ground Also from figure 1

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M~CHANl~~-1 .] .

14, [a, b]

, ~-N:cosa

Nsina

'

For equilibrium on man, net force on him should be zero. Also as shown in figure m=f, =µN

,

.',, .mg·

.·

N sin a= ma, N = mgseca ~ a=gtana And from figure 2 ,, F=3ma '.F ~ F=3mxgtana F= 3mgtana 10. [a, b] H2+x2=y2

Differentiating 2xxvx = 2yxvy

•

N'

'

'

/'/'.,

.

3mg,

15. [a] Since small block m is not moving w.r.t. wedge :. Both can be considered as a single system which is accelerated horizontally N=(M+m)g 16. [a].

... (i)

N

.

(M·~·;;g,

N= mg cos0

Vear= y xvblock

17. [c, d]

X

~x2 +H2

1/x

= Vear = ----Vbtock X

X

--a==== --block V ' 2 2

~

V x1

vx +H

Differentiating eqn. (i) again, 2 2 xax+vx=Yay+Vy given ~

a;=acar=O v2X -v2y --~=ablock

y

v2H2 ~

-(H_2_+_x_2_)~31~2 = ablock

In equilibrium acceleration of each block is zero. ~ kx 2 =(m1 +m 2 -m 3 )g Just after .the string is burnt only T = 0 and no other force is changed ~ acceleration of m1 = m2 = m 3 is zero kx 2 -m4 g · and acceleration of m4 = ,m4

12. [b] Let acceleration of pulley·is a

T ~":_i_ - :T }9s· --- ,

. . . . . 50 .

T·

~

= [ (m1 + m 2 ) - (m 3 _+ m4 )Jg m4

OON

'

,· .- .t_.1_t l-,,a ,-

100 • ,: •

T-50 = S(a+ a') T-100 ='l0(d-a) zr = soo From eqn. (i), (ii) and (iii)

R .

T

.

T

... (i) ... (ii) ... (iii)

d=ss 2 13. [a, d] Clearly if'B is stationary and pulley moves then block · will rise. VB =u+vA aB = 0+aA

18. [c] At first B will move downward and C towards tight with a constant acceleration and·v, =at· The· moment when B touches ground A will lift up. Now as C is moving toward tight A will rise and string between BC will become loose. Therefore block C decelerates with a constant deceleration due to the tension generated in string between A and C. At a certain moment v c = 0 (after this A moves downward). C again accelerates in the opposite direction upto the moment A reaches the ground. 19. [b, c] Just after BP is cut . For block A-no force has changed :. acceleration of m1 = 0 for m 2 downward force is being reduced :. m 2 will move upwards

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I FORCE ANALYSIS .

'253]

20. [a, b, c] at . Acce1eranon = m i.e., a straight line passing through origin dv at -=dt m at 2

1. c=J:--+•tl

In

2mg-mg = ma1 a, =g 2mg-T = 2ma 2

znd case :

T-mg=ma 2 a2

v=2m

at m

23. [a] In 1st case:

Parabola

=! 3

In 3 rd case : mg +.mg -T = mag

t

V=-X-

T-mg = mag

2

t . v = acce 1eranon x2

21. [b, c]

+--'t;,,-,,~-+ T2 sln 8

AO--o

T2 •

mg

T2 cos a

-,=~I

24. [b, c] 2h

tane=d d . cose=zF' d2 +h2

mg

T2 sine= mg

T2 case= mg T1 sincx = T2 sine T1 coscx = T2 case+ mg From eqn. (i) and· (ii) tane = 1 e = 45° =} T2 = ..f2.mg From eqn. (iii), (iv) and (v) ·

... (i) ... (ii) ... (iii) ... (iv)

R 4

sine= .

... (v)

T

In 2nd

ma=2mg-mg a=g case : T-ing=ma' 2mg-T = 2ma'

a'=! ' 3 a-·a'= 2g 3

~ s i n e+Tsln8

T

slowly

T

Teas 8

-

,

. mg,

TC0s,8 .

.

= ...!1!!L_

mgRd2 T=-h +2xh 4

= mg.J d 2 + 4h 2 ·

2tancx = 1 = tane

22. [b, c] In 1st case:

.

2sine as man moves upward e becomes small sine decreases =} T increase

2

=}

- - __'!2_ga__ _...1

4

.as man moves 2Tsine = mg

tancx = mg 2mg 1 tan ex= Ti= ~ = mg-./s smcx T, ..f2. = T2 X -.J5

T

4h

... (vi)

25. [a, b, c]

[]

2t-2T=0xa =}

T lift

=t

{1-)2t . ~.T =_t

TMT

For m1 to. off. 10 T=mg=lO So t=lOsec Similarly for 2 kg block aN = 20 sec 26. [c, d] The acceleration of mass' m' and' M' along the inclined plane is g sine so the contact force between them is zero. So mass 'm' will fall freely with acceleration g and acceleration of wedge will be g sin_9.

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I 254· ·'.

MECHANICS-I

'.·

27. [a; b] As discussed in question No. 26 contact force between 'm' and 'M' will be zero. So contact force between wedge and inclined wedge will be Mg cos0. 28. [b, d] · (i).Let the force F be applied on m1 and both the blocks · accelerate without any relative acceleration. fm,';. =0.lxSxl0= SN F-J=Sa adding: F = 15a a=F/15

J=10(;s)

Hence

3f

F-=2

3 15 Fmax = - fmax =~newton 2 2 Hence (b) is correct and (c) is wrong. (ii) Let the force F be applied on m2 and both the blocks accelerate without any relative acceleration. f =Sa ] F-f=l0a adding: F = 15 a

=>

If~~

F a=-

~

J=s(:S)

f = mg sin0 - mg 2

1

30° s; 0 + 45°; sin-

=µmgcos0, .::

4

1

( --)

2../2

.

. -l (2../2 ~) < 0 < 90° +sm __

Hence (b) is correct curve between 0 and friction force. 30_. [a, c] N =Fcos0+Mg ... (i) ... (ii) fmax =µN = µ(Fcos8+Mg) To just push the block Fsin0 = fmax => Fsin0=µ(Fcos0+Mg)

~ -j-

=> => => => =>

F=

~ _ _ M9,c._

µMg sin0-µ cos0

sin0-µ cos0 >. 0 tan0>µ tan0 > tan(tan-1 µ) 0>tan-1 µ

Hence the block can be pushed forward only if 0 > tan- 1-µ.·

15

=>

Hence

J

=>

F=3f

=> Fmax = 15 newton Hence .Ca) is wrong and (d) is correct. 29. [b] At 0 = 30°, mg sin0 = mg/2 which is equal and opposite to external force. Hence at this moment friction force is zero. As 0 starts increasing from 30°, the mg·sin0 component starts increasing. Here

Again as 0 decreases sin0 .decreases while. cos 0 increases, therefore, sin 0 - µ cos0 decreases.

Hence

µMg increases. sin0-µcos0

31. [a, d] The free body diagram of blocks A and B is as sho'IVll below.

( mg sin_0 - m;). will be compensated by opposite

!! f2 q_N_" c, . T, "

I

mg

'

•

j

I

,•

friction force until · mg sin0 ~ mg < µmg cos0 2

sin0-.! < µ cos0 2

N1 =mg N2=2mg+N 1 =3mg

sin0 -µ cos0 < .!

.

1

2

f2=µN,=µmg J/=µN2 =3µmg Tr= f2 =µmg . F =fr+ f2 +T1 = Sµmg

sin0- cos0 < -

2

../2 (cos.'.: sin0 . 4

sin.'.: cosa) < .! 4 2

1

sin(0+%)
NS

25

=-

CO=

9

-

9

Passage-7 1.. [b] N cos8

.

v

.... '.

= l0xlO

trolley velocity= -v/2 vre1 = 3v/2 Fmax =µgm => amax = µg and Vmax(rel) = 3v/2·=·9m/s Vmax = 6m/s · 6x 3 t = v/a.= - - = 1.8s 10

= 10/3

= 6m/s Vy = 3m/s KE= KE 8 + KE 7 1 · 2 KE; = - X 50 X (6) v

•

~sin8

tane = rg 3

1. [b] If velocity of girl w.r.t. ground = v,

•

v2

4

= lOm/sec

2. [a]

mv 2 Nsine=-r Ncose = mg

2

v

Passage-11

...,,.,, 0,8

12- f = 2x 0.2xco x 0.8

3

A ,mg

When car just topples, contact at B will be no more i.e., NB =0 M9ment about A is just zero 2 mv 2 ·=> mgx-=--xl 2· r => gxr=v 2 => 10xl0=v 2

=>

..

•

>rriv2 r

1m B

co= VJ rad/sec

=>

·NA

Ne

from eqn. (i) and (ii) 2 25 (0

10

3. [d]

2· -N + 12 = 2x rco 2 x 0.8 3

=>

. + f cose = mv- = ::..:..:.c..::.::...:. 50x100) X cose N sme

(N cose - f sine= mg = 50 x 10) sine => f = 500cose- 500sine = 500x 0.8-500x 0.6 f = 500x 0.2 = 100N

,.,,_ ~ .·. _, •.

4

5--J3 m/ sec

.

.· mg

:•. ..'

8

2

and

..

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KEr

1

=-

2

. 2 x lO0x (3)

KE= 1350J

Anurag Mishra Mechanics 1 with www.puucho.com

ni -=5 M

3. [d] 2

3 Dr

·=>

Dr+Dg Dr . 1

-=12 3

=>

1 3

--

4, (A)

=> Dr =4m

4. [c]

When vrel = 9m/s =;> vg = 6m/s Max. retardation = µg = (10/3) m/ s 2 Minimum time= v/a = 6/(10/3) =18/10 = 9/Ss

(B)

5. [c]

Force.on trolley 7 lm/s 2 x.IOOkg = IOON This also the force on girl by newton's 3 rd law

6. [b] D•

2

-=Dr l =:,

D•

=>

Dr+Dg

Dg .2 2 => - = ~ , 12 3 3

(C)

Dg = Bm (in earth frame)

= 9m/s => vg = 6m/s v 2 -u 2 =2as 2 =:, 6 - O= 2 x ax 8 v,el

=:-

5. =:,

a= 36/16

9/4=2.25m/s 2

~ ~. c~~ ~ -

.

~ ~Ma~~h!!l!/;I!'!!t~le~.!~!!1~\

P = k = constant F = k = constant Fv= k ma=k mav=k m(v:}

a_> µg

~

F.

mm.

A

a mg). 12. No slipping any where. Net force is centripetal as v = CO!Jstant.

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:; ,K;: ~~~, , r

·,, •

,c,.b·.L;:i:i~t,

\

WORK AND ENERGY/

WORK DONE (i) Work Done by a Constant Force The work done on a body by a constant force is the product of the force in the direction of motion and the magnitude of displacement. __, __, W =Fscos0 = F· s

Examples: 1. Consider a block sliding over a fixed horizontal surface. The work done by the force of gravity and the reaction of the surface will he zero, because force of gravity and the reaction act perpendicular to the displacement. N ~

F sine

-2-+ direction

rmmrmmrrrlmr of motion F cos B

mg

Fig. 3.2 s Block displaced by an external force

~ r=go• 5

W=O

F

0

0

s

F

F

Sign of work depends an angle between force and displacement

Fig. 3.1

2. Consider a body moving in a circle with constant speed. At --> . every point of the circular path, goo s the centripetal force and the 1 ' displacement are mutually perpendicular (Fig. 3.3). So, the work done by the centripetal Fig.3.3 force is zero. 3. The tension in the string of a simple pendulum is always perpendicular to displacement. Which place along arc (Fig. 3.4). So, work done by the , tension is zero.

t

. . :r

Case I : When 0 = 90°, then W = Fscos90°= 0 So, work done by a force is zero if the body is displaced in a direction perpendicular to the direction of the force.

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Fig. 3.4

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!266

______ MECHA~t~s-17

Work done by a force is zero if the body suffers no displacement on the application of a force. A person carrying a load on his head and standing at a · given place does no work. Work done by a force is said to be positive if the applied force has a component in the direction of the displacement. Examples of Positive Work: 1. When a horse pulls a cart, .the force applied by horse and the displacement of cart are in the same direction. 2. When a body is lifted vertically, the lifting force and the displacement act in the same direction during lifting.

-> 5

1

i__ _

Positive work __ Fig. 3.5 · ____

i

_j

3. When a spring is stretched, by an external force both the stretching the external force and the displacement act in the same direction. Work done by a force is said to be negative if the applied force has component in a direction opposite to that of the displacement. Examples for Negative Work : 1. When brakes are applied to a moving vehicle, the braking force and the displacement act in opposite directions. 2. When a body is dragged along a rough surface, the frictional force acts in a direction opposite to that of the displacement. . 3. When a body is lifted, gravitational force acts vertically downwards while the displacement is in the vertically upwards direction.

r-· ·- ------ ----- --·--- --- -- ---1

(a)

(b)

(c)

(d)

Fig. 3.7

In (a), 0 = 0°, cos0 = 1 (maximum value). So, work done is maximum. In (b), 0 < 90°, cos0 is positive. Therefore, W is positive. In (c), 0 = 90°, cos0 is zero.Wis zero. In (d), 0 > 90°, cos8 is negative. W is negative. 1. Work is defined for an interval or displacement. 2. Work done by a force during a displacement is independent of type of motion i.e., whether it moves with constant velocity, constant acceleration or retardation etc. 3. Work by a force is independent of time during a given displacement. Work will be same for same displacement whether the time taken is small or large. 4. When several forces act on a body, work done by a force for a particular displacement is independent of · other forces. 5. A real force is independent of reference frame. Whereas displacement.depends on reference frame so work done by a force is reference frame dependent. Unit of Work In SI i.e., International System of units, the unit of work is joule (abbreviated as J). One joule of work is said to be done when a force of one newton displaces a body through one metre in its own direction. ljoule = 1 newton x 1 metre= 1 kg x 1 rn/s 2

= 1 kg ms-2

t

Work done by a force when an object is displaced along a general path .. .

->

5

system is scalar product of F and differential change in the

Negative work b •..

·-·-

·-·· _

->

The differential work done dW by any force F on a

->

->

.

position vector dr of point of application of the force

Fig. 3.6 _____ _

Fig. 3.7 shows four situations in which a force acts on a box while the box slides rightward a distance d across a frictionless floor. The magnitudes of the forces are identical, their orientations are as shown.

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IWORK AND ENERG'i'. +

The work done on the system by the force component Fx as the system moves from X; to x f is the area under curve between X; and x f. W = Fxdx+ fyYJ Fydy + f:l Fzdz

J:I '

'

'Fx

'

X Flg,3.8

_ _ _ _ _ _ _ _ _ _J

........

= F-dr

dW

--+

A

A

Each term is the area under the curve of the graph of that force versus the corresponding coordinate.

,.,_

dr =dxi+dyj+dzk f

= fi

W

•

•

ke~«~il_C~

,.

(Fxi +Fyj +FzK) -(dxi+dy j+dzk)

!An objecUsdisplaced;:;~::~:~ vector;t1 =·~2 i+3);

=ff Fx dx + ff Fydy +ff Fzdz I

,

l

l

If force F is constant, W = Fx dx+Fy f; dy +F,f; dz

to ;t2 =(4):f 6fc)m under aforce• the. work do[lej)y_this iorce.

J;

Solution : W

=Fx(Xf -X;)+Fy(yf -y;)+F,(zf -Z;)

....

........ = F-b.r ---+

or

W,0 , .1 =

-+

f.1-iF,-dr + f. --+

l

---+

'1

f

-+

--+

F2 · dr + J.1 F, 3 -dr + ... An object'is displaced from point A(2m, 3m, 4m) to il point under 'ci _constant .force

IF~

i.~

IB(lm, 2tii1 3m)

3) +4 k)N. Find.the work done by this force in this process., . . , . (2

Solution :

-+

--+

r,

(2 i+ 3j +4k)-(dxi+ dy j + dzk)

= [2x + 3y + 4z]Clm2m3mJ (2m3m4ml

;(·.

······•·.·..

;

W=f} F-dr

3 ·= J(lm2n\ ml (2n\3m4mJ

perform any work., - - - - - - - - - , , - - ,

=-9J

...

~.

··- ......

__ Fig. 3.9 (a)

:2 (3x2dx + 2y dy) = [x3 + y 2J(f ~?

= 83J

ds along curved path; therefore centripetal force does not

..

....

= f

....

F,

2

'1

---+

f--+

1

f_; (3x i + 2.Y.i) - (dx i + dy j + dz k)

.

= F, + F2 + F3 f.... .... = Ji Ftotal • dr

=W1 +W2 +W3 + ... Total work done on the system is work done by the total force or algebraic, scalar sum of the work done by individual forces. + When a particle moves along a curved path, the work is done by tangential forces only. W=fF,ds Centripetal force is perpendicular to small displacement

I

---+

....

---+

Wtotal

= f! F- dr r, =

where b. r is displacement of system. + When more than one force acts on system,

F,ota1

1--+

F= (3x 2i + 2Y.i)N. Find

.(

Illustration for Work Done · (i) The Fig. 3.10 ~hows a smooth circular path ofradius R in the vertical plane which is quarter of a circle. A block of mass m is taken from position A to B under the action of a constant force F that is always directed horizontally.

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'•'"

~-,·::,-,·,. tr·/:

i:.·

:,i

,r;-----··"·7--:·7 ?-------~'--,- 13 · ::· I

- - - - ~ - - ~ - - - - --·-..

l

--~;.;/4...l

Fl

.

"

MECHANf($-1

'·1

.{:.~h~-~---~-~·~·~=-~·- ~ "-:,~.;-:; •, - - -

: ''

tRl,

I [

~'-'::+F

A Fig,.3.10 (a)i ,,

...... = fFdscos0

WR= JF-ds or

· W = J:Fdx

(dscos0 = dx)

or W =FR As the block moves from A to B, the displacement of the block in the direction of force is equal to radius R. -

--:a·-~ ,R

!: , •: ;l: ..

... ...

dW = F-ds = Fdscose

Thus

~ : :.:. r dsj(! d~

=F(Rda)cos (~-%)

dX-,.,, ,

~--Fl~g.. 3.10

.

. or

(E]...:_ .,

dW = FR (~os.C: + sin.