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1

Part One Physical Fundamenatls of Mechanics 1.1

Kinematics

1.1.

Method : 1 (Relative approach) V Motor Boat

V = Relative speed of

Fixed point

raft

raft

6 km

motor boat w.r.t. u

river which is constant

distance = 2 × u

V

Observer on raft see that speed of motor boat is constant because duty of motor boat is constant. Hence if motor boat take 1 hrs in down stream journey then to reach again at raft motor boat will take 1 hrs in upstream journey. Hence Total time in complete journey = 2 hrs. Motion of raft : u = speed of river. Then 2 u = 6  u = 3 km/hr. Ans. 1 hr.

Method : 2 (With frame of ground) Motion of raft : u (1 + t1) = 6 ––– (i)

t1 6 km

Motion of motor boat : (v + u) × 1 – (v – u) t1 = 6  v + u – vt1 + ut1 = 6 

v – vt1 + u (1 + t1) = 6; from (i)  v – vt1 + 6 = 6  t1 = 1 hrs.

put t1 = 1 hr in (i) : u (1 + 1) = 6  u = 3km/hr Ans. Q.1.2: Total distance travel by point is S. Then Time taken in first journey : S t1 =

S S t t 2  V1 2  V2 2  t2 = Time taken in second journey : V1  V2 V0 2 2 2

Mean velocity =

Q.1.3

S t1  t 2

=

S S/2 S  V0 V1  V2

 Mean velocity =

Method : 1 (Graphical Approach) Given tan  =  1   – t  (t  )   tan  Displacement Area of trapzium 2 2     Time  Total time

2V0 (V1  V2 ) Ans. V1  V2  2V0

V t



t 

t   1 –

4 

Ans.

 – t 2

 – t 2

2

Compact ISC Physics (XII)

Method : 2 (Analytical) 2

  – t  1   – t  1   – t  Total displacement (S) = 2   2     2   t  2   2        2

2

2

1   – t  1   – t    – t       t     4 S 2  2  2  2   2   Time taken =  = t =  1 –   

Ans.

S 2m

Q.1.4 (a)

average velocity = 20 s

(b)

2 = 0.1 m/s = 10 cm/s Ans. 20 S

t

1.4

Velocity will be maximum when slope of S (t) curve will be maximum. V=

1.4 – 0.4 m/s = 25 cm/s 14 – 10

0.4

Ans.

10

14

t

S

(c)

Q.1.5

Instanteneous velocity may be equal to mean velocity when slope of line joining final and initial point will be same to slope at point on curve. Ans. : to = 16 s    Velocity of B with respect to A : VB – A  V2 – V1

16 s approx

t

     Position of B with respect to A : rB – A  rB – rA  r2 – r1 v1 A

v2

A r1

VB – A

B r2

B

Particle B will be collide with A if velocity of B with respect A is directed toward observer A.     V2 – V1 r2 – r1      ˆ   Then VB – A || rB – A  V B – A  rˆB – A Then | r2 – r1 | Ans. | V2 – V1 | N

Q.1.6

VW – E = 15 km/hr 60º W

Y

Y VS – E = 30 km/hr E

X

 S

X

3

  VS – E = 30 i; VW – E  15 cos 60 ˆi – 15 sin 60 ˆj     VW – S  (15 cos 60 – 30) ˆi – 15 sin 60 ˆj  – VS – E  VW – E – 15 sin 60  | VW – S |  (15 cos 60 – 30) 2  (15 sin 60) 2  40 km/hr  tan   15 cos 60 – 30

   19º Ans

Method 2 30 km/hr 60º 15 km/hr



   we know VW – S  VW – E – VS – E

VW – S

 2 2 | VW – S |  VW – E  VS – E – 2 VW – E VS – E cos 60  40 km/hr Ans.

15 sin 60      = 19º 30 – 15 cos 60

from figure : tan  =

Person : 1 2 km/hr

Q.1.7

Time to cross the river : t =

d 

d ––––– (i) 2.5 cos 

2.5 km/hr

from figure : sin  =

2 4 3 d   cos  =  Put in (i) : t = ––––– (ii) 2.5 5 5 1.5

Person : B x 2 km/hr d 2.5 km/hr 

Using trigonometry :

x = d tan 1

Time to reach at destination point : t = t1 + t2 Here t1 = Time to cross the river = t2 = Time to walk on bank = from (ii) and (iii) :

d 2.5

x d tan  2 4 d d 4    t=   And tan = –––– (iii) u u 2.5 5 2.5 u 5

d d 4d    u = 3 km/hr 1.5 2.5 5u

Ans.

4

Compact ISC Physics (XII)

d

Q.1.8

tA =

Boat A :

d

time to reach again at same point.

d d  VW V–W

–––– (i) where V = Speed of boat w.r.t. river

W = Speed of water w.r.t. earth.

2d

W

Boat B :

Time to reach again at same point. tB =

d V

tA Also given V =  ––––– (iii) now t B

d d      –    2d 2

2

  –

2

––––– (ii)

V2 – W2



tA  2 – 1  t  B

 2 – 1

= 1.8s

x

d

1.9: Method : 1

Vm – r

d  Time to cross the river: t = | V mr | cos 

 A

Vr – E

d Then Drift (x) = (Vm – r sin  – Vr – E) V cos  Since Vm – r < Vr – E . Hence mr this is not possible that drift will be zero. Hence we should have to minimize dx drift (x). Hence =0 d

d sec2 –

30º

90º

(Vr – E ) d Vm – r 1  sec tan  = 0  sin =   = 30º Vmr Vr – E 2

Angle made by boat with flow velocity of water = 30º + 90º = 120º Ans. Method : 2 (Vector addition method)

Vm – r 

   Vm – E  Vm – r  Vr – E Hence  will taken any value between

Vr – E

 (0 – 180º) hence we can draw a semi circle of radius of length | Vm – r | . Then.

5 C

C2 C3

C1

C4

Vmr 

 A

Vr – E

B

Vr – E

And resultant is given by C1, C2, C3 and C4, ....... Cn. But for minimum drift resultant must be Vmr 1    = 60º Vr – E 2

tangent at semicircle. Then cos  =

Then   = 180 – 60º = 120º

1.10: Relative acceleration of particle (1) w.r.t. (2) = g – g = 0; Relative velocity of particle (1) w.r.t. (2) = V 1 = V0

– 2

=

V02  V02 – 2V0 V0 cos (90 – )

2 (1 – sin ) Where 90 –  is angle b/w two velocity..

g

g

V0 2

1

V0

 = 60º

Since there is no relative acceleration of particle (1) hence its relative velocity does not change w.r.t time then. Distance b/w two particle at time t is : Distance = V0 t

2 (1 – sin )

= 22 m Ans. 3 m/s

1

4 m/s

x (+)

2

Q.1.11 : Method : 1 (Vector)

+ y

Initial velocity in y direction

of both particle zero. Hence vertical velocity of particles (1) at time t : Vy = u + at  Vy = g t; Velocity   of particle (1) at time t = : V1  – 3 ˆi  g t ˆj ; Velocity of particle (2) at time t : V2  4 ˆi  g t ˆj      Since V1  V2  V1 . V2  0 Then – 12 + g2 t2 = 0  t = 0.12 s. Hence distance between two particle will be : Distance = Vrelative × t = (4 – (–3)) ×

0.12 = 7 ×

0.12  2.5 m

Ans.

  Method : 2 (Graphical) : Since V1  V2   +  = 90   = 90 –   tan  = tan (90 – )  tan  = cot  tan  × tan  = 1

––––– (i)

gt gt And tan  =  tan  = 3 4



3 m/s





 V1

 gt

gt

4 m/s

6

Compact ISC Physics (XII)  gt   gt  put in (i):      1  3  4 



t =

0.12

Distance = Vrel × time = 7 0.12  2.5 m

Ans.

1.12 : Method : 1 (Velocity of approach) Since particle A heading to particle B and B to C and C to A. Then position of all particle at t = dt is as figure 2. B

60º

Vdt

V

Vdt

a

a

Vdt

V 60º

60º A

Vdt

C

V

a

Vdt

Vdt

Vdt Vdt

Vdt

Again position of all particle at

t = 2 dt is as figure 3. B



A

30º

P

30º C

A

Again position at t = 3 dt and so on ........ Since at any time all particle travel same distance then at each moment of time, all particle will be at equilateral tringle. Then by symmetry you can say that all particle will be met at centriod of tringle then path of each particle will as : Suppose at any instant of time, distance b/w particle A and centriod P is r. Then, line joing particle A and P make 30º angle dr with side of equilateral tringe then will always constant and A dt dr equal to v cos 30º then – = V cos 30º, (–)ive sign because dt r is dicreasing function. Finally r = 0 while initial r  a / 3 0

a 3

–

dr  v cos 30

a/ 30º a/ 2

t

 dt 0



t =

2a a  A P = sec 30º = 3V 2

a 3

Ans.

3

P

7 Method : 2 (Relative approach) let distance b/w A and B at time t is r then. At any instant of time, rate of decreasement of distance b/w two particle A and B will be constant as shown in figure. 0º s6 co 6 0º V

co s6 0º

B

60º V

1

V

B 11

B

V cos 60º V

60º V

V

V

11

11

V

C

A

V

1

1

A

C

C

A

dr 3V Then = – (V + V cos 60º) = –  dt 2

0

2 dr  – 3V

 a

dt

 dt

3a V

t =



Ans.

0

1.13: Suppose at time t distance b/w A and B is r. Then rate of decreasement of r is : o

– v + u cos 



– dr 

t

t

0 (–v  u cos θ) dt  –l = – vt + u 0 cos  dt.... (i)

t

Now since

0

0

B u

t

t

0 dx  0 (–V cos   u) dt

0 cos  dt =

x

y

cos  dt is not known then to find this integration,

dx we use rate of decreasement of component: Then = – V cos  dt

+ u

– dr = dt

0 = ut – V

ut now put in (1) : –l = –vt + u V

 ut  V 

t

0 cos 

V  

x

r

A

dt 

 V    t= 2  V – u2

W

1.14: With frame of train : With frame of train, train appear in rest



B

then distance b/ there two event is equal to l.

A

With frame of earth : When event (1) will happen. Velocity of train is V = u + at = wt. Since event (2) will be happen after time  then. Distance travelled     u1 t1 + a t21 = wt() + ww  t   Then distance    

b/ two events is =

 – w   t   2   

= 0.24 km.

Ans.

event 1



by headlight (A) = Distance travelled by head light (B) =

A

B event 2 B

A

 w (t +  /2)

8

Compact ISC Physics (XII)



It we want that both event will be happen at same point then velocity of reference frame

0.24 km = 4 m/s Ans. 60

will be Vreference frame =

shaft 2

1.15: (a)

a=1.2 m/s

For observer inside lift at t = 2s. velocity of bolt = 0

Accn of bolt = 10 + 1.2 = 11.2 m/s2. Then assume t time is taken by bolt to reach at floor. s = (b)

  at2  2.7 = × 11.2 t2  t = 0.7s  

2.7m

Velocity of bolt with respect to ground at t = 2 sec.

V =1.2×2 = 2.4 m/s Displacement then S= ut +

Distance : H =  1.3 m

 2  at  S = 2.4 (0.7) – ×10 (.7)2  – 0.7 m  

.  2.4 = 0.288 Distance travelled = 2 × 0.288 + 0.7 2g

H 2.4m/s

Ans. 0.7m

1.16: Method : 1 (Relative velocity) from figure V1

V2

tan  = V  cos = 1

V12

 V22

y V2 V2 1 V2 also tan  =   V y =  1 V  BC =  2 – V 1 1 1 1

  V   – 1 2  Shortest distance = CM = BC cos =  2 V1  1

V12  V22 V1 

0

V1 V2

V1

1 2

V2

2

0

y B  M C

Shortest distance =

|  2 V1 – 1 V2 |

t 

V12

 V22

now t =

V1 1  V2  2 V12  V22

Ans.

AM V12



V22



AB  BM V12



V22



 1 sec   CM tan  V12



V22



V 1  V2  2 V12  V22

9 Method : 2 (Velocity of approach) At shortest distance velocity of approach = 0. Then V1 cos  = V2 sin   tan  = V2 t –   V  1 In tringle A B O : tan  =   – V1 t V2

V22 t – l2 V2 = l1 V1 – V21 t  t =

V2 1

B

  V1    V2 V12

Ans.

 V22

1

A

A V1t

2

(   – V1 t)  (V2 t –   )

And Shortest distance is : Shortest distance =

90–

2

O

B

V12  V22

V2 V2t –  2

 

(   – V1 t) 2  (V2 t –   ) 2

... (1). for shortest

1 – V1t

distance

t=

dl = 0  dt

  V1    V2 V12  V22

 – 2 ( – V1 t) V1  2 (V2 t –   ) V2 = 0  ( – V t) 2  (V t –  ) 2

Ans.

1.17:



1

2



put value of t in (1) :

 

|   V1 –   V2 | Ans.

V12  V22

Method : 1 Time to reach at point D : T =

2

 1 – V1t

Method :3

V1t

V2t

 V1

  V1    V2

V1

V1 V2

m –  tan   sec   . For V V/n

dT  n minimum time : = 0 Then – sec2  + sec dθ V V

1 × tan  = 0  sec  = n tan   sin  = . Then   distance BC = l tan   BC = Ans.  –

m m– tan A

tan

C

B  

 D

10

Compact ISC Physics (XII) Method: 2 (Help of light wave) medium (1)

We know that light travel via that path in which time will be

i

 sin 

  

length BC = l tan =

C

B

sin i speed in medium (1) sin 90  V   less. Then. = sin θ speed in medium (2) sin  V

 medium (2)

 η2 – 1



Ans.



WX

1.18:

2

1m/s

3 1

2

–1m/s

t

6

x

t

Distance

t

1.19: (a)

Mean velocity in irodov is misprint and it is mean speed then mean speed =

(b) Mean velocity = And 

2R τ

Ans. (c) We know             ....(i)

  2 t2      2    ii from   τ2

2 2 (i) and (ii) :    2         now V = RW then τ τ  2 R –0 2 Vf – Vc τ   V= R . Average accleration : =  τ  τ 



πR Ans. τ

Avg. Accn =

2 R τ2

Ans.

A

B R

Time taken = 

11 1.20: This is one dimension motion be cause direction of position vector r is same as constant      vector a . Then r  a t (1 –  t)  a t –  t 2 a (a) (b)

  d r     dv   V  a – 2t a   a cc   – 2  a  V  a (1 – 2 t) dt dt

   At initial position t = 0  r = 0 Now at final position  r = 0 then. a t (1 – t) = 0

t = 0 ro

t = 

Ans.

B At point B velocity will be equal to zero so

Initial A

 that particle will be turn back. Then 0 = a (1 – 2t)

   r = a      a a   4 4

1.21: (a)

t =



 Then position B is : 

  a  1 –     4 . The Distance travelled in up and down journcy is:  a Ans. 2

V = V0 (1 – t/) 

x

0

dx 

t   x then : x = V0 t  t – 2τ   

(b)

Ans.

 t2  t  V0 1 –  dt x = V0  t –   2τ  0 τ  



t

Position at time t is

t=

A

Ans.

B C

Henc we see that velocity will change dirn at t =  because When t >   v =  t   And t <     v = – ive. Then Case I : t <  x V0 t  l – 2τ  Ans.  

Case II : t > 

Total distance travelled = AB + BC

     t  V0  V0  t      = V0   l – 2τ    V0   l – 2τ  – V0 t  l – 2τ   = 2  2 – V0 t  l – 2            t   Distance = V0  – V0 t  l – 2τ   

1.22 : (a) v =  x   x a =

1



when t > 

Ans.

differentiate w. r. t. time :

dv 1 – 1 dx  αx 2 dt 2 dt

 a

1 α – 12 x αx 2 2

1 2 1 2 α Since acceleration is constant; velocity will be : V = u + at  V = α t 2 2

12

Compact ISC Physics (XII)

(b)

S = ut +

1 2 1 at  S = 2 2

=

  s 

 α2   2 

  2 t 



s s 2 s Mean velocity = t  2 s α

t =

Ans.

1.23: Calculation of time : w = – a v ( – i ve sign is used to show deacceleration) 0

dv = – a v  dt



v0

dv v

to

 –



1

a dt



0

t0 =

 V0 2

Ans.

a 0

dv  – a v  Calculation of distance :   – a v  v dx



v

1

x0 2

dv  – a

v0

3

 dx  x

0

0

=

 V0 2 3a

2

1.24 : (a)

(b)

(c)

(d)

1.25: (a)

(b)

 r  at ˆi – bt 2 ˆj  x = at  y = – bt2  y = – b

  dr v  a ˆi – 2bt ˆj  v  a ˆi – 2bt ˆj dt  | a |  2b

x   a2y = – bx2 a 

Ans.

  dv  2 2 2  a   – 2b ˆj  a  – 2b ˆj | v| a  4b t dt

Since direction of acceleration is toward y dirn then angle made by velocity vector with y axis y a is known as angle b/w two vectors then. tan  = Ans. 2bt     x s at ˆi – bt 2 ˆj a Vmean    Vmean  a ˆi – bt ˆj  | Vmean |  a 2  b 2 t 2  t t  x x = at  y = at (1 – t)  y = x 1 –  a   

2bt

Ans.

dv y dx   = a  ay = = a – 2 a  t  v  a iˆ  (a – 2a  t) ˆj  | v |  a 2  (a – 2a t) 2 dt dt   dv    – 2a  ˆj  | a |  2a   | v |  a 1  (1 – 2 t) 2  a  Ans. dt

vx =

13 Method : 1 a (–2a  t)

(c)

acceleration = 2a

Velocity = a

  1 – 2a  (a – 2a t) a .v  1 2 cos  = | a | | v |  2 a  (a – 2a t) 2 (2a ) 2 t = 

Method : 2 – tan 

4



a  – a = a – 2 a t  t =   a – 2a t

Ans.

y  x x = a sin wt y = a (1 – cos wt)   a – 1 = – cos wt ––– (i)  = sin wt ––– (ii) a  

1.26 :

2

2

 dx ˆ dy ˆ x y  i j     – 1  1  x2 + (y – a)2 = a2 Equation of circle of radius a. v  a  a  dt dt       v  a w cos wt ˆi  aw sin wt ˆj  | v |  a w = const.  Uniform circular motion.

(a)

Distance travelled in time  is = (aw)  Ans.

a

(b)

Since motion is uniform circular motion. Hence only radial acceleration is present then Angle b/w velocity vector and accn will be   Ans.

(0, a)

1.27 : y = ax – bx2 differentiate w.r.t. time :

dy dx dx a – b 2x Vy = a Vx – 2bx Vx dt dt dt

At x = 0 Vy = a Vx ––– (ii) differentiate equation (i) w.r.t. time :

dVy dt

a

–––– (i)

dVx dVx – 2bx – 2bVx2 dt dt

ay = a ax – 2bx ax – 2b Vx2 At x = 0 Given ax = 0 and ay = w ay = | – 2b Vx2 | w = 2b Vx2 –– (iii)

Speed at origin will be : V =

Vx2  Vy2 =

w a2 w w  V = (1 + a2) 2b 2b 2b

Ans.

g 1.28 :

(a)

V

0

    1 1 s  u t  a t 2  s  V0 t  g t 2 2 2

Ans. 

14

Compact ISC Physics (XII)   s  v   (b)  t

  1  v   v0  g t 2

V

–g Ans.

 –

     2 u sin  – v0 . g   g  – v 0 . g  2 we know T =  u sin  =   v   v  0 g g 2  g  g     v0 . g     v   v 0 – g g2 y

1.29 : (a)

2 V0 sin  1 S = 0 in y direction  0 = (V0 sin ) T – g T2 T = g 2

V0 sin 

V0 

(b)

At maximum height final velocity in y direction = 0 Vy 2 = uy 2 + 2 aH

02 = (V0 sin )2 + 2 (–g) H H =

R=

(c)

x

 2V0 sin   V02 sin 2    Range R = V0 cos     g 2g  

V02 sin 2  2V02 sin  cos  V02 sin 2   when H = R  tan  = 4  = tan– 1 (4) g 2g g

1 x = (v0 cos ) t  y = (v0 sin ) t – g t2  y = v0 sin  2 gx 2 y = x tan  – Ans. 2u 2 cos2 

  1   x x  – g   v0 cos   2  v0 cos      

V0 V2  R0 = g cos  normal acceleration

Radius of currature =

 

A V0

g

V0 cos 

g

RA =



V02

2

cos  g

Ans.

1.30 : w = g sin  andw = g cos Here  is first dicreasing and then increasing and projection of total accn on velocity vector  will be (–)ive then. wv = w1 .vˆ = – g sin 

wn = g cos 

w

  wn g

=w w  in gs 

V w

2

V0

2

(d)

V0 cos 

wV

V

t



15 y

h 2g

 sin



h 2g

M

2gh



V0 = 2gh

s co

O sin



1.31 :

h 2g



C



Just before collision

Just after collision x

Time to collide with incline T =

2 uy g cos 



2 2gh cos  g cos 



2 2g g

  Length MO = (velocity in MO) × T =  2gh cos  sin   2gh cos  sin      2 2gh cos  sin  ) 2 2gh  MO   Range OC = (MO) sin  = = R = 8 h sin  g cos  cos 

1.32: We know R =

Ans.

u 2 sin 2θ ()  sin 2  5.1 × 103 =  1 = 32.5º. Also we know that range g g

 usin  g

will be same for 2 = 90 – 31.5 = 59.5º. Then time of flight will be: T1 =   240 sin 31.5 T1 = = 24.3s = 0.41min 10

m/s 240

Ans.

   240 sin 59.5 T2 = = 42.3s = 0.69 min Ans. 10

1.33:

3

5.1×10 m

Method: 1   º     º. For both particles x and y co-ordinate must be same then. Particle (1): y = x tan 1 –

Particle (ii): y = x tan 2 –

From (i) and (ii) :

g x2 2 V02

2

/s 250 m

.... (i)

cos 1 g x2

2 V02 cos 2  2

x tan 1 – g x 2

60º .... (ii)

2V02 cos 2 1

= x tan 2 – gt

= v0

v0 m/s = 250

45º= 

2V02 cos 2  2

16

Compact ISC Physics (XII)

x=

 V02 sin (1 –  2 ) cos 1 cos  2 x .... (iii)  Time for particle (1) : t1 = V cos θ 2 2 g cos 1 – cos  2 0 1

x x  1 1  TIme for particle (ii) : t2 = V cos θ  Then t = V  cos  – cos   .... (iv) 0  1 2 0 2

Put value of x on (iv) : t =

2 V0  sin (θ1 – θ 2 )    g  cos θ1  cos θ 2  = 11s

Ans.

Method: 2 Particle (1) :

x = V0 cos (t + t) .... (i)  y = V0 sin (t + t) –

Particle (2) : x = V0 cos (t).... (iii)  y = V0 sin (t) –

(iii) and (iv): t =

2V0  sin (θ1 – θ 2 )    g  cos θ1  cos θ 2  = 11s

 g (t + t)2 .... (ii) 

 2 gt .... (iv)  From (i), (ii), 

Ans. y

v0

V0 1.34 : (a) tan = ay  Time to reach at hight y : t = y/V0



vx = ay

y

dx Also = ay  dt

x t dx dx  aV0 t dt = a [V0 t]  dt 0 0





0

aV0 t 2 a V0  y   x=   2 2  V0 

(b)

ay = 0

ax =



 a  x =   2 V0

 2 y  

Ans.

dVx a dy  = aVy = aV0  anet = aV0 dt dt

= aV0 cos  a =

a V0 (ay) V02  (ay) 2



x

x

a2 y 1  (ay/V0 ) 2

Tangential acceleration = a Ans.

Radial accn or normal accn : (an) = a V0 sin  an = a V0

   a y  V0  



Ans.

17 a  x = at .... (i) Vy = (a t) b bx

1.35: (a) tan  = y

0

dy  ab

Then y =

(b)

t

0

t dt

ab  x    2 a

ax = 0



ab t 2 y 2



anet



ay =

Vy bx= v

y

 b  y =   x2  2a 

 dVy dt

 b

an

Ans.

x

dx = ab  anet = ab  Then normal accn : an = anet cos dt

anet = ab Then an = ab  a   xb     1    R= b   a  

a = vx

a a 2  (bx) 2

 R =

V2 (a 2  b 2 x 2 ) 3/2   an a2 b

/ 

Ans.

1.36:

Method: 1 (Work-Energy)    Tangential accn is given by : | w  |  a .  = a cos 

Suppose at time t particl e at poseti on A and in smal l time dt particl e displacement by ds.



Work done : dw = Ft ds = (ma cos) ds = ma (ds cos)  dw =

(ds) cos  ma

 ds cos

Using work energy theorem : W = max 

 mV2 = max  



V =

ds





a x

0

ma (ds cos) = ma

a x

Ans.

18

Compact ISC Physics (XII) Method : 2 (Kinematics)   Tangential accn : at = a .  = a cos. Also since we know d| v | that tangential accn is rate of change of speed then. dt v dv = a cos  = a cos  v dv = a (ds) cos = a dx ds v

0

v dv 

x

0

a dx 

v2 = ax  v = 

ds V 

dx

Ans.

a x

v

1.37: Angle travel by particle:  = 2n. Speed of particle : v = Rw = at w =

at dθ   R dt t2 =

2n

t

at

 ds   R dt 0

0

at 2 2R

t2 =



  nR  a

R

  nR 2 a 2 4 n R a2 t2 Now : Radial accn : ar = V R  = a R a R

= 4na Tangential accn : at = Total accn = a

–

dv V2  dt R



dt

R

 V0

–

0

R

V

 tR V0

 V 1

dv v2 = –v a 2t  a 2r = at  at = ds R

dv  v

S

 0

ar

(–) sign because

V0 V0 t



R

–v dv =

Where ds = distance travel by particle in time dt. –v dv = V

a 2  (4 na) 2

Ans.

t

– V –2 dV 

V0

V –1

a 2r  a 2t =

V2 Since at is rate of change of speed then. R

V



V is decreasing

anet =

dv = a  Total accn = dt

1  (4n) 2 = 0.8 m/s2

1.38: (a) At any time t : at = ar =

(b)

2n =

a

ds R – n

V S  V0 R

Then V = V0 e–S/R  at =

v2 dt R v2 ds R V 2 e – 2S/R v2  0 R R

at Given at = ar at t = 0 u = V0

19

anet =

V02 2 e – 2S/R V2  R R

2

1.39:

 anet =

V02 2 R e 2S/R



V2 R

2

Ans.

Method : 1 

2

v = a s Squaring both side : v = as

at

ar

a Compare with : v2 = u2 + 2 accS  u = 0 acc = 2 Hence motion is constant magnitude tangential accn. Then. At

as V2 = R R

any time t : at = a/2  ar =

at 

Since we know that velocity vector and tangential accn is parallel then. a

as

2S

r tan   a  R a/2  R t

Ans.

ar Method : 2 dv ds dv  a  a = a = av  2 t dt dt dt

v2 = as  differiate w.r.t time : 2v = 2 ar = V

R

 as

R



1.40 : (a)  = a sin wt now

tan  =

ar

at



2S R

 = 0    t = 0

d  a w cos wt  = ± a   t =  2 , 3 2 , V = dt

Ans. y

wt a sin = 

x

dv  a w 2 sin wt Then at = – aw2 sin wt a = dt

ar =

V 2 a 2 w 2 cos 2 wt   anet = R R

anet = aw2

at t = 0 

sin 2 wt 

a2 R2

cos 4 wt

a 2t  a 2r

–––– (i)

a2w 2  anet = aw2 Ans.  = 0  at t =  2 or 3 2  anet = R

20

Compact ISC Physics (XII)

(b)

da net  0  cos wt = R 2 a put in equation(i) dt

For minimum value of acceleration 

2

amin = aw

R  1–    2a 



 R    = a 1 –   2a





  = a 1 –

Speed of particle when distance is S. V2 = 2aS –––– (i)

1.41:

S=

R2 2 a2

Ans.

(Because a = const.)

2S 1 2 at  t2 = a 2

a =

a 4

t wn = b

2

4bS2  2S  Then wn = b    2 a  a

Radius of curvature : R =

w=

Net accn :

a 2T

V 2 2aSa 2 a3   R = wn 2bS 4bS2

 w 2n

1.42: (a) y = ax2 diff. w.r.t. time :

 4bS2 a   a2  2

 w =

   

Ans.

2

Ans.

dy dx  2ax  Vy = 2ax Vx dt dt

y

Again diff. w.r.t. time : ay = 2ax ax + 2aVx2 x

At x = 0  Vy = 0  Then Vx = v  ay = 2aV2 Since speed is const. its tangential acceleration will be zero. Then 2 dV V2 1 V 2  ax = = 0  anet = 2 a V  R =  R = a 2 n 2av dt 2a

(b)

x2 a

2



y2 b

2

2x Vx

= 1 diff. w.r.t. time :

Again diff. w.r.t. time :

At x = 0 and y = b :

2x a x a2 2 Vx2 a2

a





2

2 Vx2 a2 2 ay b







2y Vy b2

2y a y b2 2 Vy2 b2

Ans.

0



2Vy2 b2

0 V0

0

21 2 V02

As shown in figure : Vy = 0 and Vx = V0 

a

2



2 ay b

 0  ay =

b

Since speed is Const. tangential accn = ax = 0 at; t = 0  anet = V02

Radius of curvature : R =

 R =

an

V02 b a

1.43 :

Given

2

V02

 R =

a2

a2 b

b a

2

V02

V02

Ans.

d d(2) d 2 = w = const. Then = 2w = const. dt dt dt

A 

Hence angular velocity of point A w.r.t. point P is



constnat then. WP = 2W and velocity will be perpendicular to position of A w.r.t. P then V = R WP = 2RW Since WP = const.

dWP 0 dt

O

 P

Ans.

Then  Tangential accn = 0

Radial accn = R WP2 = 4 RW2 = anet  Direction is toward the centre. 1.44 :

 = at2 

Ans.

d dw = w = 2at  =  = 2a dt dt

Tangential acceleration : at = R = 2aR  Radial acceleration 2

2

V

2 2

: ar = RW = R (2at) = 4 a t R

at ar 

And. v = RW = R 2at  2at = v R –––– (i) Now anet =

a 2t  a 2r = 2aR

from (i) : anet = 1.45 :

1 (2at 2 ) 2

v 1  (2at 2 ) 2 = 0.7 m/s Ans. t

l 1 Since acceleration is constant. l = at2 and a V 2 Vt V = at Then l = –––––– (i) and also. angular acceleration is const. then 2 1 1  =  t2 and w = at then  = wt and since particle taken n turn in its journey.. 2 2

22

Compact ISC Physics (XII) = 2  n (I)

1.46 :

 V 2v  (II) : 2 W  W = 

 = at – bt3  w =

Wavg

Ans.

d = a – 3bt2 when body is in rest then w = 0  a – 3bt2 = 0 dt

a  in = 0 and final = t (a – bt2) = 3b

t =

a

final –  in = = t

Win = a and Wfinal

And.  = 1.47 :

1 wt –––– (ii) 2

Now 2 =

3b a

2a 3 



2a 3

a

 2a  3b  3 

Ans.

3b  a  Wfinal – Win  = a – 3b  3b  = 0   avg =   t

dw = – 6bt = – 6b dt

a   = 2 3b

3ab

a a

w

t

 0dw   0 dt

 w =



3ab Ans.

3b

Ans. y

Given  =  = at To find tangential accn : (a t) : at = R = Rat To find radial accn : (ar) :  =



at

dw  ar = Rw2 dt

ar x

at 2 R a2 t4 at dt   ar = 0 2 4 t

R

at 

Then

ar R a2 t4 tan  = a  tan  =  t = 3 4R a t t

4 tan   7 s Ans. a

ar

1.48 :

Given

 –W

 

W    = angular accn.   = k

dW k W d

W  k = const. 0

 (–)ive because W is decreasing. –

 W0

1

W

 2

dw 

 k d 0

23 3

2 W0 2 3k

  ––– (i)

 = Angular displacement.  Also  = k 0

– dW = k

W dt 

w

–1

W

2

dw 

0



W

–1 2

t

k dt  2W0 0 k

 t ––– (ii)

3

Avg. angular velocity : Wavg

1.49: (a) Given W = W0 – a –––– (i). At d Also = W0 – a  dt

 n

(b)



0

2 W0 2

 = = = t

t = 0

d  W0 – a



1

  2 W0 2  3 k  k   

  = 0



W0 3



t

–1 dt  n (W0 – a) 0 a

W0 – a W0  – at   = (1 – e – at ) W0 a

Ans.

W = W0 

 t 0

Ans.

 W0 – at  Put value of  in equation (i) : W = W0 – a  a (1 – e )  W = W0 e–at Ans.   y

dW 1.50: Given  = 0 cos    = W = 0 cos   d w

0

w dw  0



0

W = ± 2  0 sin 

cos  d   Ans.

w  2  0 sin 

0

W2 = 0 sin   2

 – 0

   0 cos 

– 2 0

w  – 2  0 sin 

1.51: (a) Instanteneous axis of rotation is passing through that point which velocity is zero always. If we observe carefully its

y A

point must be at line joining AB. Then. at time t :



VP – 0 = RW = Rt = yt

B

V

x

24

Compact ISC Physics (XII) y

x = vt ––– (i) Velocity of point P :

VP = 0 y= (b)

 v  v  v = ty x = v  y  t = y  

v2 x

P (x, y)

yt

put in (i)

V

O

x

Ans.

Velocity of point O = 0 y

VO – P = Rw  VP = u + at = wt O

yw = RW

wy yw = wt  t = w

Then

(x, y) x P

 wy  1 2 1  x= at = w  2 2  w 

2

1 w2y2 2 w

 x =

Ans.

1.52 : Since there is no slipping at ground. VC = 0  V = RW where VC = velocity of contact point. To find distance, we have to find speed of a particle of rim then.  = wt  Time to one complets v 2  v 2  2 v v cos ( – ) = 2 v sin /2 = 2V sin (wt)/2  Again

journey = 2/w VP =

s ds = 2 v sin wt/2  ds  2 v dt 0



2 w

0

sin wt/2 dt  S =

8v = 8 R w

Ans.

y

V

x

w

v 

P

C

1.53: (a) Acceleration of point C : aC = w  Velocity of point

 v

A

C at time t : VC = u + at = wt Angular accn of ball about its centre:

w w  R R R  Angular velocity of ball at w time t : w = w0  t = t  Velocity of point A R  =

aC



w.r.t. centre C : VA – C

R C 

B

O

= RW = w t ˆi  VC – E = w t ˆi  VA – E = 2 w t

Ans.

W

25 A

wt y

wt x

A

y C

Point B :

wt

B



x

wt

VB–C = Rw = wt (– j)  VC–E = wt i  VB–E = wt (i – j)  VB–E = wt Ans. Point O : VO–C = – wt i  VC–E = wt i  VO–E = O Ans. (b)

Acceleration Calculations : A

wt

C

wt

at

0

y

ar C

Point A: x

aC–E = wi .... (i)  aA–C = at i – ar j  Where at = tangential acceleration. ar = radial acceleration. at = R = w  ar =

v2

R

(i) + (ii) : aA–E = 2 wi –

 (wt) 2

R



w2 t2 w2 t2  aA–C = wi – j .... (ii) R R

w2 t2 j  aA–E = 2w R

 wt 2      2R   



Ans.

26

Compact ISC Physics (XII) y

  Point B : a C–E = wi .... (i)  a B–C

 v2  = (–R) ˆj +  R 

 ˆ i =  C R

 B

w2 t2 ˆ – wˆj – i .... (ii) R  a

(i) + (ii) :

B

x

 w2 t2  = w – R 

  ˆ ˆ  i + w j  aB = w 

  P oint O: a C–E = w ˆi .... (i)  a

–w ˆi –

O–C

 wt 2      2R   

 v2 = (–R) ˆi +   R 



Ans.

y

ˆ  j =  

w2 t2 ˆ j .... (ii) R

C

 w2 t2 w2 t2 a O–E = – j  aO–E = Ans. R R

(i) + (ii) :

x

0

A

2V

n

1.54 : Velocity of point A = 2v  Velocity of point B = v   acc of point A = v

2

R

A

Again

R

V

B v

R

v

2V

Radius of curvature =

2

v

2  accn of point B = v

(speed) 2 (2V)2  RA = = 4R V 2 /R normal acceleration 2

2

v Again an = cos 45º = R

v2 2R

 RB =

(V 2 )

v R



v2

 2 2R

V

2R 1.55 :

Method : 1 (axis is rotating):

w1–2 =

w 12



w 22

 w 1–0 = w1 ˆi 

B

45º

V V 2 x

an

  w 1–0 = w2 ˆj  w 1–2 = w1 ˆi – w2 ˆj

 dw 1– 2 dj d ˆi d ˆi    1–2 = = w1 – w2 direction of is toward y axis and dt dt dt dt

this is rate of change of dirn of x axis.

27 y

y

w2

w2 x

0

w1 x

0

Method : 2 (axis is not rotating)    w 1–C = w1 cos  ˆi + w1 sin  kˆ   w C–0 = w2 j  w 1–0 = w1  cos  ˆi + w1 sin  kˆ + w2 ˆj  | w 1– 0 | 

w12

 w 22

w2

  w2

z

1 C w1

d  = w2   = – w1 w2 sin  ˆi + w1 w2 cos  kˆ dt

 |  | = w1 w2. Here x axis is directly attached with observer. Then

d ˆi | d ˆi | dθ  = – w2 kˆ  | d ˆi | 1  dθ  = w2 dt dt dt

But

x

0

  d   d   d w 1– 0  = – w1 sin   dt  ˆi + w1 cos   dt  kˆ dt    

And

y

y y axis

z

1 unit d 1 unit

passing

j di i

d ˆj  = 0 Because with frame of this observed dirn of axis does not rotating then.  1–2 = – dt

w1 w2 kˆ 

 1–2 = w1 w2

 dw  1.56 : w = at ˆi + b t2 ˆj    = a ˆi + 2b t ˆj (a) dt

 (b) |  |  at

cos  

 2ab t  1    a 



b  1  t  a 

  w .    w   = w cos   cos   w

a 2 t  2b2 t 3 a2 t 2  b2 t 4

 | w |  at

a 2  4b2 t 2

   = 17º

Ans





28

Compact ISC Physics (XII) y

1.57: Method : 1 (axis are moving): It we see carefully then x axis is moving while y direction is not rotating. Angular velocity of disc with respect to centre M  V x is : WM – O  ˆi .... (i) Angular velocity of centre R

 

0

 V ˆ j  OM = R cot  M w.r.t. origen. WM – O  OM

1  tan 2  

p

y axis

 V ˆ [ i  tan  ˆj ] quantity it follow vector addition law WD – O  R



R

v

 V  WM –O  tan  ˆj . Since angular velocity is vector R

 V | WD – O |  R

M

V Ans. R cos 

y axis

j di

d

i z

   V  d iˆ d ˆj  dw β  Again Angular acceleration () is :  β  R  dt  tan  dt  dt  

And since y dirn is constant  V β    R

d ˆj d ˆi = 0  But  | dˆi | = d dt dt

d ˆi d  dt dt

= wM O 

dˆi V  – tan  kˆ dt R

2 2 V     tan kˆ  β  – V tan kˆ  | β |  V tan  R  2 2   R R

Method : 2 (axis are not rotating) angular velocity of disc w.r.t. M : WD–M =

V cos  ˆi + R

y

V sin  ˆj angular velocity of M w.r.f. O R  V ˆ V j tan  ˆj angular velocity of disc : WM – O  R cot  R   wrt O: W D–O = W

D–M

  + W M –0  W

D–O

V = cos  ˆi R 



x

0 



D M

V

z

w0 = 

 V V V  V  V  sin  ˆj   tan  ˆj  | WM – O |   cos     sin     tan   R R R  R  R 





V R

V Ans. R cos 

29   dw D – O –V V  dθ   dθ  dθ αD– O   sin θ   ˆi  cos θ   ˆj  0 and is angular velocity of M w.r.t. dt R R  dt   dt  dt

0:

d/dt =

 – V2 V2 V2 V ˆj | α ˆi + tan   α D – O  sin  tan  cos  tan  | = D – O R R2 R2 R2

tan  Ans. 1.58 :

Method : 1 : (Direction of Co-ordinate axis is fixed)

At time t line AB rotate by  angle then

y

 WP – A =

0

w0 cos  ˆi + w0 sin  kˆ    t ˆj  | WP – A | = w 20 cos 2   w 20 sin 2    20 t 2 

 | WP – A | = w0

 t     0   w0 

x

A

 w0



Now

z

  d  ˆ  d WP – A d  k   0 ˆj   – W 0 sin  ( iˆ )  W 0 cos    dt dt  dt 

 = –W0 0t sin  ˆi + W0 0 t cos  kˆ + 0 ˆj  |  |   |  |   0 1  w 20 t 2

w0

w 20  02 t 2   20 

Ans.

Method : 2 (x axis is moving) : Here we take line AB along

  t 2  x dirn. WP – A  w 0 iˆ  0 t ˆj  | WP – A |  w 0 1  w2

y 0



w0 P

0

  dw P – A d ˆi d ˆj    w0  0 t  0 ˆj  dt dt dt

But Ans.

d ˆj 0 z Here dt

di ds ˆ d iˆ  k   0 t kˆ   = w0 t kˆ +  ˆj =  t kˆ  Then dt dt dt

x

A



1  |  |   0 1  w 02 t 2 2





30

Compact ISC Physics (XII) F

1.2 The F undamentals Equation of Dynamcis 1.59: Where F is force due to air which will be constant then. mg – F = mw .... (1). When m mass is taken then F – (m – m) g + (m – m) w .... (ii) From (i) and (ii) : m =

m – m

2 mw Ans. w

F

w

m

mg

(m – m) g

a

1.60 : Net pulling force = (  m) a  m0g – k m1 g – k m2

m2

m1

 m 0 – k (m1  m 2 )  g = (m1 + m2 + m0) a  a =  m  m  m  g Ans. 1 2 0  

km2g

km1g m0

F. B. D of m2 : a T

m2

m0 g

k m2 g T – k m2 g = m2 a. Put value of a then

(1  k) m 0 T = m  m  m m2 g 0 1 2

Ans.

1.61 : (a) Pulling force F = m1 g sin  + m2 g sin  – k1 m1 g cos  – k2 m2 g cos . Then acceleration a is : a =

g sin  (m1  m2) – g cos  (k1 m1  k 2 m 2 ) .... (i) m1  m 2

F. B. D of m1 : N = Normal force between two blocks. m2 g sin  + N + k1 m1 g cos  = m1 a .... (ii) put value of (a) in (ii) : N =

2

m

g

a m2

g m1

m

1

si

n

Ans.



os gc

os N 1g c m k1



m2 s k2 co

in gs





m1

m1 

(b)

g m1 k1

si n

(k1 – k 2 ) m1 m 2 cos α m1  m 2



For minimum value of acceleration will be zero and friction force will at maxm value then 0 = [g sin  (m1 + m2) – g cos  (k1 m1 + k2 m2)] / m1 +m2  tan  =

k1 m1  k 2 m 2 m1  m 2

31

1 2 at where V = final velocity in this situation Vf = 0 2

1.62 : Upward Journey : We know S = Vt – Then  =

1 (g sin  + µg cos ) t2 2

–––– (i) =

0

f

v



t = Time taken in upward journey. 

v

a

 a=

Downward journey : We know S = ut +



+

µ

)

1 2 at u = initial 2

u=

(i)

and u = 0  now :

a=

(ii)

 2 – 1   g sin   ug cos   1      =  g sin  – ug cos   2   µ =  2  tan    1    

0

v

1 (g sin  – µg cos ) (t)2 –– (ii) 2



velocity. Then  =

(g

s in

 os gc

(g

) a cos g µ – si n 

Ans.

m2 1.63 : (a) Starts coming down. m2g > m1 g sin  + fmax m2g > m1 g sin  + k m1 g cos   m 1 m2 > sin  + k cos  (b) m1 g sin  > m2 g + k m1 g cos   g sin  > m + k cos   1 m2 m2 m1 < g sin  – k cos  (c) At rest : Friction will be static : g sin  – k cos  < m1
m1 g sin Block m 2 has tendency to

m2 Given m   . 1

ax

2 m g  0.66 m1g Then 3 1

in

 Here m2 g = m1  g =

m

and m1 g sin  = m1 g sin 30º = m1 g/2  0.5 m1g

1

g

co

1.64: To find tendency of sliding check value of m2g

move down ward. Then. Pulling force = m2g – m1 g sin  – k m1 g cos  acceleration a =

a =

g [ – sin  – k cos ]  1

m 2 g – m1 g sin  – k m1 g cos  m1  m 2

Ans. fr = k m 2 g

1.65 : (a) Before no sliding b/w m1 and m2 :

m2 m1

F = at

F at F.B.D. of System : Acceleration of both block will be same. w1 = w2 = w = m  m  m  m 1 2 1 2 fr w But friction b/w m1 and m2 will be static then. m1 m2 F = at m1 m1 at k m 2 g (m1  m 2 ) at fr = m1 [w1] = m  m < k m2 g  t <  w1 = w2 = m  m ––– (i) a m 1 2 1 1 2

If

t > k m2 g m2

km2 g

F = at w2

F.B.D. of m1 :

Assume to =

(m1  m 2 ) . Sleeping b/w two block will be start then F.B.D. of m2 : a m1

m1

  w2 = k m2 g

at – km 2 g m2

  w1 =

k m2 g m1

–––(ii) Ans.

––––– (iii) Ans.

k m 2 g (m1  m 2 )  t > to : Slope of w2 > a m1

w

w2

 a  Slope of w1 = w2 because of Slope of w2 =  m  from w =w  2   a t (ii) [After sliding]  Slope of w1 =  m  m  from (i) [Before sliding] 2  1 1

0

Slope w1 = 0 from (iii) [After sliding]

w1 2

t

33 1.66 :

F.B.D. of block : ma = mg sin  – k mg cos   a = g sin  – kg cos  Time to reach at bottom : s = ut +

1 2 1 at  sec  = (g sin  – kg cos ) t2 2 2

A km gc

R k = coefficient of friction

os 

mg









t2 =

s in

2  sec  2  t = g (sin  – k cos ) g [sin  cos  – k cos2 ]

2 g [sin  cos  – k cos 2  ]

–– (i)

To minimise t  (sin  cos  – k cos2 ) will be minimum also. Assume. x = sin  cos  – k cos2   Now

dx = 0 – sin  sin  + cos2  + 2 k cos  sin  = 0 d

cos 2 = – k sin 2  tan 2 = – 1 k

Ans.

Put value of k :  = 49º  Put value of  in equation (i) : 2  2.10

tmin = 1.67 :

10 [sin 49º cos 49º – 0.14 cos 2 49]

 1.0 s.

Ans.

When block pull up Direction of friction will be downward. F.B.D. of block : N = mg cos  – T sin   fr = k (mg cos  – T sin ). At just sliding:

m k



N

T

 sin g m 

sin

 



T

mg sin   k mg cos   Tmin   cos   k sin  T

T cos  = mg sin  + k (mg cos  – T sin ) T =

T

s co



m µN mg cos  mg

(cos  + k sin )  min  minimum value of cos  + k sin  = 1 k 2  Tmin =

mg (sin   k cos ) 1  k2

34

Compact ISC Physics (XII)

1.68 : (a) At time of breaking off the plane vertical component of  F must be equal to weight mg. Then F sin  = mg = at sin

F

V

 0

m dV  a cos  V

(b)

 0

t1



m

µ=0

at cos  dv  m dt

mV 1 m2 g 2 m g 2 cos  mV t2   1   V = a cos α 2 a 2 sin 2  a cos α 2 2 a sin 2 

t dt 

0

m dV  a cos 

at



 t = mg a sin  . Motion equation of block : a1 = Accelration of block. F cos  = m a1  a1 =

=

t

 t dt



0

Ans.

mV t2  a cos α 2

1.69 : Motion equation :  = aS 

mg cos  = m a1 a1 = 3

F

mg g cos  = cos 3m 3

acceleration of block of mass m. a1 = v

s





=

3 g/ m

 m

g g v dv  cos as ds    a1 = cos as  3 3 0 0

v2 2

v

s

 0

g sin as 2g  v2 = sin as  v = 3 a 3a 0 s

t





2g a cos  2 sin as  v = t  3a 2m

3 2 3 a cos  ds a ds  t 2 dt  s = a cos  t  s = m g cos   cos  t 2  2m dt 2 m 2 m x3 6 a 2 sin 3  0 0

1.70 :

Ans.

Method : 1

Motion of mass 2 m : T – k2mg = 2 mw  T = 2

2m k2mg

mw + 2kmg

w T

k

T

m

a k mg

k

Motion of motor : T – kmg = ma  2 mw + 2 k mg – k mg = ma  a = 2 w – kg Acceleration of 2 m w.r.t. m : a2m – m = 2w – kg + w = 3w – kg  Suppose in time t both will be met then.

 =

1 (3w – kg) t2 2



t =

2 3w – kg

Ans.

35 Method : 2 On system : Tension will be internal force hence k2 mg – k mg = – 2 mw + ma  a = 2w – kg  a2m – m = 3 w – kg then  =

1.71: For observer inside elevator : a1–2 =

But in form of vector :

a1–car

2 3w – kg

1 (3w – kg) t2  t = 2

m 2 g – m1 g  m 2 w 0 – m1 w 0 (m 2 – m1) (g  w 0 ) = m1  m 2 m1  m 2

  (m 2 – m1) (g – w 0 ) = Ans. m1  m 2

shaft 2T

(m 2 – m1) (g – w 0 ) Acceleration of m1 w. r. t. shaft : a1–shaft = a1–car + acar = m1  m 2 + w0    a1–shaft =

Ans.

T

(m 2 – m1) m1 g  m1 m 2 w 0 Ans. m1  m 2

m1

m2 m1 g

m2g

2 m1 m 2 (g  w 0 ) Tension in string : T – m1 g = m1 a1–shft   T = m1  m 2 Force applied by pulley on ceiling = 2 T = 4 m1 m 2   2T = m  m (g – w 0 ) 1 2

w0

T

T

4 m1 m 2 (g  w 0 )  as vector form: m1  m 2

Ans. a/2 T

1.72: Motion of body (2) :  mg – T/2 =  ma  2 mg – T m 

= 2 ma .... (i) Motion of body (1): T – mg sin  = m a/2 .... (ii) from (1)

2g [2 – sin  ] and (ii) :   Ans. [  1]  a1  a 2  1.73 : Equationof motion : T = m0  2  .... (i) m1g  

s mg

 in

T/2 a

 m

mg a1 + a2 2

m0

T

= T/2 = m1 a1 .... (ii) m2 g – T/2 = m2 a2 .... (iii) T/2

4 m1 m 2  m 0 (m1 – m 2 ). g from (i), (ii) and (iii) : a1 =  m m  m (m  m ) 1 2 0 1 2

m1 mg

Ans.

a1

T/2

m2 m2g

a2

36

Compact ISC Physics (XII)

1.74: Motion equation on system: Mg – mg = M a1 + m a2 .... (i) Motion equation of m : fr – mg = ma2 .... (ii) Since length of rod is  then.   t2 .... (iii). From (i), (ii) and (iii)

: fr =

m m  (M – m) t

2

1 (a + a2) 2 1 a1

Ans.

M m

mg

Mg

1.75:  = 100 cm – T + mg = m a1 .... (i) + 2T mg = m (i) and (ii) : a1 =  =

 (–  2) g  4

a1 .... (ii) from 2

3 a1 . Suppose t time is taken then 2

arel =

T

T

 (  4) 1 arel t2  t =  (2 – ) g 2

1.76 : Motion equations T – mg = ma

a2

Ans.

2T nm

a1 2

.... (i)   mg – 2T =  m a/2 ....

a1

 m

nmg mg

2g ( – 2) (ii) from (i) and (ii) : a = . When body  4 (1) travel h distance then in same time body (2) travel 2h distance in upward dirn using constraint relation. Nowvelocity of body (2) Just a 2

before string slack.

2T

V2 = 2a (2h). It body (2) travel x distance again then V2 = 2a (2h) = 2gx a

T

h

mg

2

x =

mg

2 ah 2 ah 6h  h  g . Total hight from ground : 4 = g  4

Ans.

1.77: F. B. D. of N sin  = m a2 .... (i) F. B. D. of rod : mg – N cos  = m a1 .... (ii) Constraints: a2 sin  = a1 cos  .... (iii) From (i), (ii) and (iii) :

a1 =

g 1   cot 2 

a2 =

g tan    cot 

a1 N

a1



a2



m 

a2

N

 a2

a1

 mg



37 1.78: F. B. D. of m : mg – KN – T = m a2 .... (i) N = m a1 .... (ii) F. B. D. of wedge : a1 = a2.... (iv) From (i), (ii), (iii) and (iv) : a1 =

mg g  M 2m  M  km 2k  m KN

a1 M

T

T

KN

m

a1

N

a2

Ans.

m a2

mg mg

1.79: F.B.D of bodies on frame of wedge: Since system is stationary on frame of wedge hence : mg = kma + ma + kmg  a =

g (1 – k) 1 k

Ans. ma

ma 1 m a

kmg

ma

m

2m

kma

N

m

mg

mg

1.80: F.B.D. of block (2) with frame of wedge: At maximum accn w : fr1 will be maxm then fr1 = k [mg cos  + mw sin ]. Since block is under rest with frame of wedge, then equilibrium equation of block along incline. k (mg cos  + mw sin ) + mg sin  = mw cos   w =

g (1  k cot ) cot  – k

m 2

fr 

mw 1



fr1

w

mw

 

mg

mg

1.81 :

Method : 1

F.B.D. of System a2 = accn of bar w.r.t. incline. Since no force on system in horizontal direction then O = ma1 + m [a1 – a2 cos ] –––– (1) F.B.D. of bar w.r.t. wedge : m2 a1 cos  + m2 g sin  = m2 a2 –––– (ii) from (1) and (2) : a1 =

m 2 g sin  cos  m1  m 2 sin 2 

Ans.

38

Compact ISC Physics (XII) m2

m2

m1

2

m1

a2

1 

a1

m2

m2a1



m2g

Method : 2 F.B.D. of bar : with frame of wedge, bar has zero accn in perpendicular to incline then. N + m2 a1 sin  = m2g cos  –––– (i) F.B.D. of wedge : N sin  = m1 a1 –––– (ii) from (i) and (ii) : a1 =

m 2 g cos  sin 

N

m1

2

m1  m 2 sin 

m2a1

 

a1

90 – 

m 2g

N



N

Method : 3 F.B.D. of bar with frame of ground observes : Motion

a1

equation in perpendicular dirn of incline then : m2g cos  –

a2



N = m2 a1 sin  ––– (i)



m2g

v

1.84: v = 360 km/hr =

360  1000  100 m s 3600

C v B R A

At point A : N – mg = g+

mv 2 70 [100]2 mv 2  N = mg + = 70 g + = 70 R 500 R

70  100  100 = 70 g + 140 g = 210 g = 2.1 kN 500

At point B : N =

v

Ans.

mv 2 ––– (i) put value of m, V and R N = 1.5 kN R

N

mg

Ans.

N

mg 2

2

At point C : N + mg = mv /R N = mv /R – mg ––– (ii) put value of V, m, R in (ii) N = 0.7 kN Ans.

N

39 1.85: Tangential acceleration (at) : mg sin  = m at  at = g sin 

m

Radial acceleration (ar) : Energy conservation : mg  cos 

1 = mv2  v = 2 anet =

a

t

2

a

r

2

2 g  cos   ar = v  = 2 g cos  

 mg

v

 g sin 2   4 cos 2  anet = g 1  3 cos 2 

Now : T – mg cos  = (b)

T



2

mv 2 = 2 mg cos   T = 3 mg cos  

Ans.

Component of velocity in y dirn: vy = v sin  vy =

y dirn.



2 g  cos  sin   v2y = 2 g  cos  sin2 

For vy maximum v2y will be maximum. Then x = cos  sin2 

will

be maximum. dx d = 0 = 2 cos2  sin  – sin3  = 0  2 cos2  = sin2   tan  =

= 2 g  (c)

1 3



2  sin  =

1

2

cos  =

3

3

 v2y

 1  2 4g  = mg v max  y  T = 3 mg cos  = 3 mg  3  3 3  3

If no acceliration in y direction then. ar cos  = at sin   2 g cos2

3



1  = g sin2   tan  =

2  cos  =



v

3

Ans.

ar 

Ans.

1.86 : Extreme position : Only tangential acceleration is prerent at extreme

at



position. at = g sin  Lowest Position : Energy conservation : mg [ –  cos ] =  v2 = 2 g  (1 – cos )  ar =

1 m v2 2

v2 = 2 g (1 – cos ) 



A/C condition : ar = at  g sin  = 2 g (1 – cos )  sin  + 2 cos  = 2  cos  = 3 5

  = 53º



Ans. v

mg

40

Compact ISC Physics (XII) A

1.87: Particle will break off sphere when normal reaction will be zero.

mv 2  v2 = Rg cos . Energy conservation : mgR R 1 1 (1 – cos ) = mv2  mgR (1 – cos ) = mRg cos   2 2 mg cos  =

cos   2

1 – cos  =

 2 2 Rg Rg  3   v =   3

cos  =

A R 

v

R  mg

2 3 cos   = 1  v2 = 3 2

Ans.

1.88: Top view : Spring force = F =    N = normal reaction on sleeve. Since no acceleration in tangential direction. N sin  =    cos  ––– (i). Equation in radial direction: N cos  x   cos  (cos ) +    sin  sin  m = m ( + ) cosec  (w2)  x   = m  + m   w2   = Ans. x – mw 2

+    sin  = m r w2 = m ( + ) cosec  w2 from (i):



w

l

w

r N

  

 v



r  1.89: k = k0 = 1 – R  friction coefficient. Suppose cyclist at radius of r. then friction provide centripital   r mv 2  force to motion on circular path. Then where kmg = friction force. k0 1 –  mg   v2 r  R

 r2  dx r2 = k0 g  r –  . To maximum value of v : x = r– will be maxm. Then = 0  dr R R   R R  k0 g R 2r 1 2 1 –  0  r = R  Vmax  k 0 g  –   k 0 gR  Vmax = Ans. 2 2 4 4   R 2

r O R

kmg

mv2 /r

41 1.90: Tangential and radial both acceleration is only provided by friction because friction is acting as external force. Then maximum value of friction = k mg.

v2/R

Velocity of car after d distance travel. v2 = 2 w d. Then ar = Radial

R

2

2 w d v  acceleration =   at = Tangential acceleration = w   R R 2

anet =

 2w  d  4d 2    w 2  w  1  2  Fnet = m w  R  R

 4d 2  2  R  w 1 2 2  Squaring : 2  = k g  d =  2 R  

 kg  w  

4d2

1

R2

w

= k mg

2

  –1  

 

1.91 : y = a sin x α  k = friction coefficient. Centripital force m v2 = and centripital force will be provided by friction. R m v2  k mg  v2  k Rg. At limiting condition : R

Ans. y

x

For v maximum  R will be minimum. And we know : speed

y

is constant also we are secing that radius of curvature will be minimum at maximum point of curve then. ay =

d2 y dt 2



v

v2 a

ay

2

 

sin x α . At maximum value of curve : sin x α = 1  ay =

v2 v 2 α2 R = a  2 v a y

v2 a α2

2 R = 

a

x

. Then v2  k

α2 g   v   kg a a

Ans.

1.92: F.B.D. of differential element of length dl. Equation of motion : 2T sin  – N cos  = (dm) Rw2 ––– (i)  N sin  = (dm) g put value of N in (i) : 2 T  – (dm) g cot  = (dm) Rw2 2 T w  mg

N  T

T

 = very small angle. sin  =  N T 



T (dm) g

42

Compact ISC Physics (XII)

 –

m m m m g cot  m R w 2  (R 2 ) g cot  = (R 2 ) Rw2  T – T = [Rw2 2R 2R 2 2 2

mg + g cot ]  T = 2

1.93 :

 Rw 2  cot    g  

Ans.

m2 m1 = 0   T1 – m1 g = m1 a ––– (i)  m2 g – T2 = m2 a ––– (ii) Relation between T1 and T2 : (T + dT) sin d 2 + T sin d 2 = dN a

Td = dN  dfr = µ dN = µ Td ––––– (i)

T2

Since  is very small. sin d  d  cos d  1

m2

In horizontal direction : (T + dT) cos d 2 – T cos d 2 = dfr

m2g

T2

dT = dfr ––– (i)  dT = µ T d 

 T1

 µ d 0

m2 m1

y dfr

T + dT

d

T

(b)

x

g –a g – a        ga g  a 

0   = 0  eµ = 0  µ =

m1g

   n (T2 – T1 ) = µ 

T2 µ from (iii) : T  e  1 Then eµ =

m1



dN

T2 = T1 eµ ––– (iii) T2 m 2 (g – a) from (i) and (ii) : T  m (g  a) 1 1

dT  T

T1

––– (iv). Since pulley is fixed : Before skiding a =

1 n 0 

g – a  0 g – a [n – 0 ] from equaiton (iv) : en0 =   g  a     g  a  a = g   n0  

Ans.

43 y

Hence velocity along y axis is not responsible for circular

1.94:

motion only velocity along Z-axi s i s responsibl e.

d2 x

1.95: x = a sin wt  a x =

dt 2

2

y = b cos wt  ay =

d y

V0

v0

m VZ2 m V02 cos 2  Then N = VZ = V0 cos  N = R R

O

 R

x



Z 2

= – aw sin wt 

 = – bw2 cos wt  a net  a x ˆi  ay ˆj

dt 2

w

y

 2 2 = – a w sin wt ˆi – b w cos wt ˆj  F  – mw 2 [a sin wt ˆi  b cos wt ˆj]     r  x ˆi  y ˆj  a sin wt ˆi  b cos wt ˆj  F  – m r w 2

x

 where r = position vector of particle. F = mw2 (a sin wt) 2  (b cos wt) 2

 F = mw2 

x 2  y2

1.96 :

(a)

Ans. Method : 1 (Impluse equation)

 We know  P = Impluse in time t =

t

 F dt 0

t

=

 – mg dt

  P = – mg t

Ans.

0

V0

g

90+

V0 m

(b)

g





 2 V0 sin  2 V0 sin       P  mg  = V g cos (90 + )   – V0 . g = V0 V . g 0 0 g g      – V0 . g V0 . g g sin   = 2 m V0 sin   V0 sin  =   P   2m Ans. g g

T=

Method : 2 (Kinematic)

  y V0  V0 cos  ˆi  V0 sin  ˆj  Vf  V0 cos  iˆ  (V0 sin  – gt) ˆj     where Vf = final vel oci ty vecto r t hen  P  m Vf – m V0    P  – m g t ˆj Ans. (a)

V0 

x

44

Compact ISC Physics (XII)

(b)

T=

 2 V0 sin  2 V0 sin  ˆ  P –mg j    P = – 2 m V0 sin  ˆj g g

From method : 1 :

    2 m V0 . g – V0 . g ˆj V0 sin  =   P  g g

Ans.

m

1.97 :

  F  a t (T – t)

(a)

(b)

   V  d V  a    2   w  [ t – t ]  dV  a F  m w  w = linear acceleration. dt m m 0   3   a   t 2 3   V  –   m V  a   Pf Ans.  m  2 3  6



 V

 a w  [ t – t 2 ]  m S



dS 

0

 0

t t      t dt – t 2 dt     0 0 





t   a  t  a  t 2 t3    t dt – t 2 dt  dV  V   –     m  m  2 3   0 0 





 3   2   a   3 .   4  4 4 a  t t  dt – dt   S  –   S  a   S = a  Ans.  m  2 3  m  6 12  m 12 12 m 0 0 





  F0 dV sin wt  1.98: F  F0 sin wt  a =  m dt S

t





V

 0

F dV  0 m

t

 0

t

sin wt dt  V = – F0 cos wt mw 0

F F 1 F0 dS  0 [1 – cos wt] dt  S = 0  t – sin wt   [1 – cos wt]  V = mw mw  w mw  0 0

S =

F0 mw 2

[tw – sin wt]

distance

distance will be increasing function w.r.t. time then dS = (+)ive or > 0  w – w cos wt > 0  dt

cos wt < 1  0 < wt <   2 < wt < 3  4 < wt < 5









wt

45 0

dv F  a  0 cos wt  1.99: F = F0 cos wt  dt m π sin wt = 0  wt =   t = w 0

Now :



dV 

0

S=

– F0 mw



F0 mw

0

F0 m

t

 cos wt dt 0 = 0

cos wt  V =

0

s

t





F F0 ds  0 sin wt dt  si n wt .... (i )  m w mw 0 0

s =

2F0

velocity will be maximum when sin wt = 1

m w2

Ans. V

t





dV – r dV –r  dt  V  1.100 : (a) F = – rV   n V V m dt m V0 0

V = V0 e–r/m t  V will be zero when t  

(b)

F0 sin wt  mw

Ans.

t

(1 – cos wt) at t = π w

2

 Vmax =

F0 m



dV 

V

s





V V0

 –

r t m

Ans.

r dv r r r v dV  – v ds a= – V = v    V – V0 = – S  V = V0 – S m ds m m m V0 0

Total distance travel by particle is S1 then final velocity = 0  0 = V0 – r m S1  S1 =

(c)

Ans.

V0 m V0 r = V0 – S2  S2 = r  m t2 =

1.101 :

m V0 r

  –1     

–––– (i)  Also

S V ( – 1) m n  –––– (i)  = 2  0 t2  n  r

– kv 2 F  v2  F = – kv2  a =  m V

V

 V0

dv v2

V0 

= V0 e –

Ans.



–k m

t

 dt



h

0

–1 – kt  V0 – V  m 1 1 kt    –      V V  k  t ––– (i) V V m V0 V m 0   0

V0

V

r

mt2

46

Compact ISC Physics (XII)

– kv 2 dv – kv 2 dx  v Also a =   v dv =  m dx m

k=

V  V0 – V  h m  n V put in (i) : t =  V V V h 0 0  n 0 

V

 V0

m dv  –k V

h

V V0 = – kh

 dx  m n 0

Ans. V

1.102: Suppose at time t distance travel is x then F = mg sin  – ax mg cos   mw = mg sin  – ax mg cos  where w = acceleration of block. w = g sin  – ax g cos  .... (i) v = g

m

in gs

K



c mg

 os

k = ax



sin  – ax g cos   xm

0



v dv = g sin  – a x g cos   dx

v dv 

0

 (g sin α – axg cos α) dx  0 = (g sin ) x – 0

2 g sin  2 a x 2 g cos  x = a g cos  x = a 2

tan . Velocity will be maximum when dv dt = 0 = acceleration  g sin  = ax g cos  1 x = a

v max



t an   no w

ya tan 

v dv 

0

1  2 g cos   1   tan   –  tan   a 2 a    

 (g sin α – a x g cos α) dx

 Vmax 



0



 Vmax =

g sin 2 α g  sin  tan  a cos α a

= g sin 

Ans.

m

K

1.103: K = friction coefficient  Block start sliding when.



F = at

F = at1 = k m g  t1 = k m g a . Assuming over time start from block start sliding then. Suppose acceleration of block is w then. mw = at – kmg = a (t1 + t) – kmg = a t  w v

a =  t  m

 0

dV 

a m

t



(t d (t)  V 

0

a a S= (t)3 = (t – t1)3 6m 6m

a t 2  2m

s



ds 

0

a 2m

t

 t

2

d (t) 

0

t = t – t1

Ans. t=0

t = t1

t=t

47 v=0 2

V0

h

1.104 :

Upward journey :

V0

mg

Fnet = mg + kv2  a = g + 0

v dv mg  kv 2 kv 2   – ds m m

h

mv dv

 mg  kv 2

F = Kv

 –

v0

 ds



h=

0

 mv dv = (mg – kv2) ds

KV02  mg m n .... (i) 2k mg 2

Downward Journey : F net = mg – kv2 v



h

mv dv

 mg – kv 2



0



ds  h 

0

v0

 v = 1

KV02

v

kv mg

 a =

mg – kv 2 m

m KV02  mg m KV02  mg m mg n n  n  2k mg 2k mg 2k mg – kv 2

Ans. mg

 1.105 :(a) Position vector of particle : r = r cos  ˆi + r sin  ˆj   At time t :  = wt  F  F rˆ  F (cos  ˆi  sin  ˆj)

F

m w

 F [cos  ˆi  sin  ˆj]   a  m  V

 0

 F dv  m

t

 0

F (cos wt) dt ˆi  m

 speed = | V | 

(b)

= v dv/ ds

F mw

r 

0 t

 (sin wt) dt 0

  V 

F [sin wt ˆi  (1 – cos wt) ˆj]  mw

(sin wt) 2  (1 – cos wt) 2  Speed =

2F sin wt 2 mw

Ans.

s

t





2F ds 2F sin wt 2 dt  V sin wt  ds  Distance is calculated by speed. Then 2 mw dt mw 0 0

S = Distance S = –

2F mw

2

 2 cos

wt 2

t

S = 0

4F  wt  1 – cos  2  2  .... (ii) velocity of particle mw

48

Compact ISC Physics (XII)

will be zero : 0 = S=

2F sin wt 2 mw

sin wt



 0  put value of t = 2π w in (ii)

2

8F Distance 8F/mw 2  (1 – cos )  S =  Average speed = Time 2/w mw 2 mw 2 4F

Average speed = 4F  mw

Ans.

1.106: K = tan  = friction coefficient  When = W = net tangent ial

accelerat ion/

Wx

=

Accel erati on

w

mg sin  cos  – k mg cos  m

in x direction. Now W =



s g co Km

  mg sin

x

= g (sin  cos  – K cos ) = g [sin  cos  – sin ]  Acceleration in x dirn: Wx =



v

W = g sin  (cos  – 1)... (i)

mg sin  – k mg cos  cos   Wx = g sin  [1 – cos ] .... (ii) m

From (i) and (ii) : W = –Wx 



d Vτ  – dt



d Vx  V = –Vx + cos  at t = 0 dt

V

= V0 and Vx = 0  V = – Vx + V0 .... (iii) Then const = V0  Also Vx = V cos  .... (iv) From (iii) and (iv) : V =

V0 1  cos 

Ans.

1.107: Tangential force on system is F then  F 

 (dm) g sin  =  ( R d) g sin  = Rg

t

 sin  d 

 = R f  

in = 0 = –  Rg

0



cos 

f 

  f =  R  f =  F = R

R g gR  R   a =  mR g 1 – cos  R  = x  1 – cos  R   a =  1 – cos  R 

Rg 1 – cos 





y



dm=  R d  d  R (dm)g

f

F x

dm  d

49 1.108: At time of break off. normal reaction will be zero. Where mw0 = Psuedo force because observer at sphere.

mV02 = mg cos  R

R w0= acceleration

– m w0 sin   V20 = Rg cos  – R w0 sin .... (i) Energy equation: Wall forces = Kf – Ki  Wpsuedo + Wmg = Kf – Ki m w0 R sin  + mg [R – R cos ] =

1 m V20 – 0  V20 = 2w0 2

 – V02   R sin  + 2 g R [1 – cos ].... (ii) From (i) and (ii) : V = 2R   R  2 0

+ 2 g R  V20 = 2 g R 3 

1.109 :

V0 =

R g  cos   3

mw0 0 R

0

R

V0 mg

  w0

5  g w 0  3  g    w   3 1   0     g    



Ans.

1 K Given F  n   F = n Where K = Constannt. r r A particle is said to steady if we displace particle away from origin, particle want to regain its position and also it we displaced particle toward origin, particle want to regain its original position. Then. At steady state (mean position) : F =

m V02 K m V02 Then n  .... (i) r r r

m V2 > 0  K r–n – mV2 r–1 > 0  Differentiate both side r with respect to r for small increase of r: –n k r–n–1 dr + mV2 r–2 dr > 0 and It we increases r then. F –

– n K  1   mV02   1       > 0 .... (ii)  From (i) and (ii) n < 1 Ans. r n  r   r   r 

v

v0 r 0

F

O r

50

Compact ISC Physics (XII)

1.110: At steady state: mg sin  = m (R sin ) w2 cos   cos  =

g

. Case (i) : If Rw2 > g then cos

g

is defined and Rw R w2 only one equilibrium position will be exist and will be steady. g Case (ii) : If Rw2 < g then only = 0º will be equilibrium R w2 position because tangential fore along arch of ring due to mg 2

Y R N





will be greater than that of pseudo force and object will come at lower position of ring.

90– 

Y

mg

2

m (R sin  ) w