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INTRODUCTION: So far we have treated objects as if they were particles, having mass but no size. In translation motion each point on a body experiences the same displacement as any other point as time goes on, so that the motion of one particle represents the motion of the whole body. But even when a body rotates or vibrates or translates while rotating, there is one point on the body, called the centre of mass, that moves in the same way that a single particle (having same mas) subject to the same external force would move. Figure 6.1 shows the simple parabolic motion of the centre of mass

fig. 6.1

of a hammer thrown from on person to another; no other point on the hammer moves in such a simple way. Note that, if the hammer were moving in pure translation motion, shown in figure 6.2, then every point in it would experience the same displacement

fig. 6.2

as does the centre of mass in figure 6.1. For this reason the motion of the centre of mass of a body is called the translation motion of the body. Centre of Mass

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When the system with which we deal is not a rigid body, a centre of mass, whose motion can also be described in a relatively easier way, can be assigned, even though the particles that make up the system may be changing their positions with respect to each other in a relatively complicated way as the motion proceeds. First we will define the centre of mass and show how to calculate its position, then we will discuss the properties that make it useful in describing the motion of extended objects or system of particles. DEFINITION OF CENTRE OF MASS Two particle system: Consider first the simple case of a system of two particles m1 and m2 at distances x1 and x2 respectively, from some origin O as shown in figure 6.3(a) . We define a point C, the centre of mass of the system, at a distance xC .M . from the orgin O, as shown in figure 6.3(b), where xC .M . is defined by

xC.M. =

m1 x1 + m2 x2 m1 + m2

...(6.1)

x2 x1 O

x2 x1 m1

m2

X

O

C m1

m2

X

xcm (a)

(b) fig. 6.3

This point has the property that the product of the total mass of the system (= m1 + m2 ) times the distance of this point from the origin is equal to the sum of the products of the mass of each particle by its distance from the origin; that is, ( m1 + m2 ) xC.M. = m1 x1 + m2 x2 In equation 6.1, xC .M . can be regarded as the mass-weighted mean of x1 and x2 . An analogy might help to fix this idea. Suppose, for example, that we are given two boxes of nails. In one box we have n1 nails all having the same lenght !1; in the other box we have n2 nails all having the same length ! 2 . We are asked to get the mean length of the nails. If n1 = n2 , the mean length is simply ( !1 + ! 2 ) 2. But if n1 ≠ n2 , we must

allow for the fact that there are more nails of one length than another by a “weighting” factor of each length. For !1 n2 n1 this factor is n + n and for ! 2 this factor is n + n , the fraction of the total number of nails in each box. Then the 1 2 1 2 weighted-mean length is

 n   n2  ! =  1  !1 +   !2  n1 + n2   n1 + n2  or

!=

n1! 1 + n2! 2 n1 + n2

[Note that when we put n1 = n2 in the above euqation, we get l = (l1 + l2 ) 2 ] The centre of mass, defined in equation 6.1, is then a weighted-mean displacement where the “weighting” factor for Centre of Mass

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each particle is the fraction of the total mass that each particle has. Suppose, for an example, that m2 = 0. Then there is only one particle, of mass m1 , and the centre of mass must lie at the position of that particle. Same arguement can be given for the case when m1 = 0. If m1 = m2 then centre of mass of this two particle system should be at the mid-point of x1 and x2 . Suppose m1 is much greater than m2 . In such a case, centre of mass should lie very close to m1. All these predictions come true if you use equation (6.1) to locate the centre of mass of the system. If we have n particles, m1 , m2 ,........., mn , along a straight line, by definition then centre of mass of these particles relative to some origin is xC.M. =

m1 x1 + m2 x 2 + ....... + mn xn m1 + m 2 + ...... + mn

n

∑m x i

xC.M. =

⇒

i

i =1 n

∑m

...(6.2)

i

i =1

where x1 , x2 ,........., xn are the positions of the masses relative to the same origin. The sum n

∑m i =1

i

=M

is the total mass of the system. We can rewrite equation 6.2 in the form n

MxC.M. = ∑ mi xi

...(6.3)

i =1

If two particles of masses m1 and m2 are separated by a distance d, as shown in figure 6.4(a), find the distance of the centre of mass of particles from each particle. d m1

m2 fig. 6.4(a)

Solution: Let us assume that m1 and m2 lies on the x-axis and their positions are x1 and x2 , respectively, as shown in figure 6.4(b).

Centre of Mass

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d

x2 x1 fig. 6.4(b)

C

O

m1

m2

X

xcm d1

d2

As the particles m1 and m2 lie on the x-axis, their centre of mass must lie on the line segment joining them, i.e., the portion of the x-axis intercepted between m1 and m2 . Hence, we can assume that the centre of mass of m1 and m2 also lies on the x-axis at some point C, as shown in figure 6.4(b). If xC .M . be the position of the point C and d be the distance between m1 and m2 , then we have

d = x2 − x1 and

xC .M . =

...(i)

m1 x1 + m2 x2 m1 + m2

...(ii)

[Using equation 6.1]

If d1 be the distance between C and m1 and d 2 be the distance between C and m2 , then, from figure 6.4(b), we have, d1 = xC .M . − x1 =

m1 x1 + m2 x2 − x1 m1 + m2

=

m2 ( x2 − x1 ) m1 + m2

d m1 + m2

⇒

d1 = m2 ⋅

and

d 2 = x2 − xC .M . = x2 −

=

⇒

m1 x1 + m2 x2 m1 + m2

[Using (ii)]

[Using (i)]

...(6.4)

[Using (ii)]

m1 ( x2 − x1 ) m1 + m2

d 2 = m1 ⋅

d m1 + m2

[Using (i)

...(6.5)

ALTERNATE METHOD: Centre of Mass

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d m1 c

m2

(origin)

d1

X

d2

fig. 6.4(c)

5

Let us assume that the line joining the positions of m1 and m2 is x-axis and choose centre of mass of m1 and m2 as the origin, as shown in figure 6.4(c). If d1 and d 2 be the distances of m1 and m2 from their centre of mass, respectively, then, their position x1 and x2 , respectively, can be written as

x1 = −d1 and x2 = + d 2 .

Using equation 6.1, we have xcm =

m1 x1 + m2 x2 m1 + m2

⇒

m1 x1 + x2 m2 x2 = 0

⇒

−m1d1 + m2 d 2 = 0

⇒

m1 d 1 = m 2 d 2

⇒

d 1 m2 = d 2 m1

⇒

d1 m2 = d1 + d 2 m1 + m2

⇒

d1 = m2

d m1 + m2

d 2 = m1

d m1 + m2

[Since xcm = 0 ]

[∵ d = d1 + d 2 ]

Similarly,

NOTE: *

you should notice that the product of mass of a particle and its distance from the centre of mass is constant. If a particle of mass m is at a distance d from the centre of mass, then, md = constant

or, we can say d ∝

Centre of Mass

...(6.6)

1 i m

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m1

Suppose you have three point masses m1 , m2 and m3 as shown in figure 6.5(a). Suggest some method to predict the approximate location of the centre of mass of these three particles. [In the next section you will learn methods to locate the exact position of the centre of mass of such systems.]

m2

m3

fig. 6.5(a)

Solution: (READ ONLY AFTER GIVING SOME HONEST ATTEMPTS ON YOUR OWN). Let us first consider only m1 and m2 . The centre of mass of the system consisting of m1 and m2 only would be at some position between the positions of m1 and m2 on the line joining m1 and m2 , as shown in figure 6.5(b). Once position of the centre of mass of this system is known, entire mass of the system can be assumed to be concentrated at this position.Hence, we can assume that mass ‘m1 + m2’ is concentrated at the point C’.

d1

m1

d2

You should choose C′ in a way such

C'

that

d1 m2 = d2 m1

m2 fig. 6.5(b)

When we consider m1 and m2 concentrated at C '. the given three particle system reduces to a two particle system, as shown in figure 6.5(c). Now, using the same approach we can find the position of the centre of C′ (m 1 + m 2 )

(m 1 + m 2 )

C′ 1

1

2

m3 = m+m 1 2

C 1

m3

m3 2

fig. 6.5(c)

fig. 6.5(d)

mass of this two point mass system, as shown in figure 6.5(d). Hence, point C is the centre of mass of the given three point mass sytem. DEFINITION OF CENTRE OF MASS FOR A MORE THAN TWO PARTICLE SYSTEM (NON COLLINEAR): suppose now that we have three particles not in a straight line; they will lie in a plane, as shown in figure 6.6. The centre of mass C is defined and located by its coordinates xcm and ycm , where xcm =

m1 x1 + m2 x2 + m3 m3 m1 + m2 + m3

Y y1

m1

C

ycm y3 y2

m3 m2

x2

x1 xcm

x3

X

fig. 6.6 Centre of Mass

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ycm =

7

m1 y1 + m 2 y2 + m 3 y3 m1 + m2 + m3

in which x1 , y1 are coordinates of the particle of mass m1; x2 , y2 are those of m2 ; and x3 , y3 are those of m3 . The coordinates xcm , ycm of the centre of mass are measured from the same arbitrary origin. For a large number of particles lying in a plane, the centre of mass is at xxm , ycm where

and

xcm =

Σm i x i 1 = Σm i xi M Σm i

...(6.7a)

ycm =

Σm i y i 1 = Σm i y i M Σm i

...(6.7b)

where M (= Σmi ) is the total mass of the system. For a large number of particles not necessarily confined to a plane but distributed in space, the centre of mass is at xcm , ycm , zcm , where zcm =

Σm i z i 1 = Σm i z i M Σm i

...(6.7c)

The three scalar equations (6.7a, b, c) can be replaced by a single vector equation $ 1 $ rcm Σmi ri M

...(6.8)

$ in which the sum, Σmi ri , is a vector sum. [Note: In equation (6.8) you should notice that if the origin of our reference frame is at the centre of mass (which $ $ $ $ means that rcm = 0 ), then Σmi ri = 0 for the system.] Equation (6.8) is the most general case for a collection of particles. Previous equations are just special instances of this one. The location of the centre of mass is independent of the reference frame used to locate it (see example 1). The centre of mass of a system of particles depends only on the mass of the particles and the positions of the particles relative to one another.

Locate the centre of mass of three particles of mass m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 3.0 kg at the corners of an equilateral triangle 1.0 m on a side, as shown in figure 6.7(a).

m3 1.0 m

fig. 6.7(a) m1

Centre of Mass

1.0 m

1.0 m

m2

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Solution: Let us choose the x-axis along one side of the triangle and the origin at the positon of one of the particles; as shown in fgure 6.7(b), then, xcm =

and

Y m3

Σmi xi Σmi

! cm r

m1

m1 x1 + m2 x2 + m3 x3 m1 + m2 + m3

=

(1.0kg )(0m) + (2.0 kg )(1.0 m) + (3.0 kg )(1.0 m.cos 60°) (1.0 kg +2.0 kg + 3.0 kg )

=

7 m. 12

=

=

m2

xcm

=

ycm =

ycm x3 X

fig. 6.7(b)

Σmi yi m1 y1 + m2 y2 + m3 y3 = m1 + m2 + m3 Σmi (1.0kg )(0m) + (2.0 kg )(1.0 m) + (3.0 kg )(1.0 m.sin 60°) (1.0 kg +2.0 kg + 3.0 kg )

3 m. 4

NOTE: •

$ Σmi ri $ You could also use the result rcm = to locate C. Σmi

•

You should notice that C is not coinciding with the geometrical centre of the triangle. Why is it not at the geometrical centre of the triangle?

•

If m1 = m2 = m3 , then, xcm =

y + y2 + y3 x1 + x2 + x3 . Therefore, in this case C concides with and ycm = 1 3 3 the geometrical centre of the triangle.

CENTER OF MASS OF EXTENDED BODIES: A rigid body, such as a meter stick, can be thought of as a system of closely packed particles. Hence it also has a centre of mass. The number of particles in the body is so large and their spacing is so small that we can treat the body as it has continuous distribution of mass. To obtain the expression for the centre of mass of a continuous body, let us divide the body into an infinite number of infinitesimal mass elements. Y

Such an element of a body of mass M is shown in figure 6.8. The coordinates of the centre of mass can now be given precisely as

dm

r!

M

xcm =

∫ dm ⋅ x = 1 x ⋅ dm, ∫ ∫ dm M

1

...(6.9)(a)

O

x

y X

fig. 6.8 Centre of Mass

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ycm =

∫ dm ⋅ y = 1 y ⋅ dm, ∫ ∫ dm M

...(6.9)(b)

zcm =

∫ dm ⋅ z = 1 z ⋅ dm, ∫ ∫ dm M

...(6.9)(c)

In these expressions dm is the mass of the element at the point x, y , z and ∫ dm equals M, where M is the mass of the body. The above three scalar equations (6.9 a, b, c) can be reduced to the vector equation $ 1 $ rcm = r dm M∫

...(6.10)

$ $ $ $ Once again we see that if origin of our reference frame is at the centre of mass (that is , if rcm = 0 ), the ∫ r ⋅ dm = 0 for the body.

NOTE: Often we deal with homogeneous objects having a point, a line, or a plane of sysmmetry. Then the centre of mass lie at the point, on the line, or in the plane of symmetry. For example, the centre of mass of a homogenous sphere (which has a point of symmetry) will be at the centre of the sphere, the centre of mass of a homogeneous cone (which has a line of symmetry) will be on the axis of the cone, etc. We can understand that this is so because, from $ symmetry ∫ r ⋅ dm is zero at the centre of a sphere, somewhere on the axis of a cone, etc. It follows from equation $ $ (6.10) that rcm = 0 for such points which means that centre of mass is located at these points of symmetry.

Find the centre of mass of the uniform triangular plate shown in figure 6.9(a).

fig. 6.9(a)

Solution: If a body can be divided into parts such that centre of mass of each part is known, the centre of mass body can usually be found simply. The triangular plate may be divided into narrow strips parallel to one side, as shown in figure 6.9(b). The centre of mass of each strip lies on the mid-point of the strip because mass of each strip is distributed symmetrically about this point. Now each strip can be replaced by a point mass having the same mass as that of the strip and positioned at the mid-point of the strip. centre of mass of all such point masses is basically the centre of mass of the given triangular plate only and obviously the lies somewhere on the line formed by joining the mid-points of all strips, which is shown in figure 6.9(c). Centre of Mass

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As this line joins the mid-point of a side from the vertex opposite to the side, this line is median of the given triangle.

fig. 6.9(c)

fig. 6.9(d)

fig. 6.9(e)

But we can divide the triangle in three different ways, using this process for each of the three sides. Hence, in this way we can get two more medians of the triangle, as shown in figures 6.9(d) and (e). As the centre of mass of the triangle should lie on each of the three medians, it lies at the common intersection point of the three medians, as shown C in figure 6.9(f). fig. 6.9(f)

If a homogeneous thin rod of mass m and l is given, from symmetry, we know that its centre of mass coincides with the geometrical centre (mid-point) of the rod. Prove this by calculation for the centre of mass of the rod.

Solution: Let us choose the x-axis along the length of the rod and origin at the left end of the rod, as shown in figure 6.10. A differential mass element of the rod having mass dm and length dx is also shown in the figure at a distance x from the origin.

dm X

O x

dx fig. 6.10

m  Mass of the element, dm, equals to the linear mass density  mass per unit length =  times the length of the !  element, dx. Therefore, dm =

m ⋅ dx !

If xcm be the coordinate of the centre of mass of the rod, then !

xcm =

∫ x ⋅ dm = ∫ dm

m

∫ x ⋅  !  ⋅ dx 0

m

m   ∵ dm = ! ⋅ dx and ∫ dm = m 

!

1 1 !2 = ∫ x ⋅ dx = × !0 ! 2

= ! 2. Hence, the centre of mass of the rod lies at the mid-point of the rod. Centre of Mass

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A nonuniform thin rod of length ! , placed along the x-axis, as shown in figure 6.11(a) has a linear mass density, λ ( x), given as

λ ( x) = λ0 + kx

X

O

where λ0 and k are positive constants.

fig. 6.11(a)

Find the position of the centre of mass of the rod. Solution: First of all you should notice that in this case distribution of mass is not uniform, hence, the centre mass will not coincide with the point of symmetry of the body and second thing which should be noticed is that mass of the body is not given. To calculate the position of the centre of mass, we have selected an element of the rod having mass dm, length dx and situated at a distance x from the origin, as shown in figure 6.11(b). Mass of the element,

dm X

O x

dx fig. 6.11(b)

dm = (linear mass density at the position of dm ) × (length of dm ) = λ ( x ) ⋅ dx

= (λ0 + kx) ⋅ dx If xcm be the coordinate of the centre of mass, then

xcm =

∫ x ⋅ dm ∫ dm !

∫ x (λ

+ kx) ⋅ dx

∫ (λ

+ kx) ⋅ dx

0

=

0 !

0

0

!

=

!

λ0 ∫ x ⋅ dx + k ∫ x 2 ⋅ dx 0

0

!

!

0

0

λ0 ∫ dx + k ∫ x ⋅ dx ! 2 k !3 + 2 2 3 = 3λ0 ! + 2k ! = k!2 3(2λ0 + k !) λ0! + 2

λ0

Centre of Mass

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NOTE: In the previous example result obtained can be rearranged to obtain    ! 3λ + 2k !  xcm =  0  2  3λ + 3 k !  0  2  ! In the above expression numerator of the term within square bracket is greater than the denominator. Hence, xcm > . 2

Locate the centre of mass of an uniform semicircular thin wire of radius r and mass m.

Y dm r

Solution: Let us choose the reference frame such that the origin is at the centre of the semicircle and x-axis passes through both ends of the wire, as shown in figure 6.12. If an element of the wire, subtending an angle dθ, is chosen at an angular position θ with respect to the x-axis, then, mass of the element,

dθ

y θ

O

x

X

fig. 6.12

 mass of the wire per unit angle   angle substneded  dm =  ×    subtended by its length on its centre   by the element  =

m × dθ π

If the chosen element is considered to be a point mass, then position of the element can be given by

x = r cos θ and

y = r sin θ

If xcm and ycm be the coordinates of the centre of mass of the wire, then, xcm =

∫ x ⋅ dm ∫ dm π

=

m



∫ (r cos θ )  π ⋅ dθ  0

m π

r = ∫ cos θ ⋅ dθ π 0 =

Centre of Mass

r r (sin π − sin 0 ) = (0 − 0 ) = 0 π π

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ycm =

∫ y ⋅ dm ∫ dm π

=

13

m



∫ (r sin θ )  π ⋅ dθ  0

m

π

r = ∫ sin θ ⋅ dθ π 0

=

1 r r r π − cos θ 0 = ( − cos π ) − ( − cos 0 ) = [1 + 1] π π π 2

=

2r π

NOTE: From figure 6.12 it is obvious that the centre of mass should lie on the y-axis because about this line distribution of mass is symmetric Hence, xcm = 0 could be used without any calculation •

If you perform these integrals for an uniform circular wire (with its centre coinciding with the origin) then you would get xcm = 0 and

ycm = 0

In this case too, you could predict the result from the symmetry property of the circle about its center.

Locate the centre of mass of a homogeneous semicircular disc of radius R and mass M. Solution: A semicircular disc can be subdivided into large number of semicircular ring elements, as shown in figure 6.13(a). According to result obtained in the last example, all these elemental rings have their centre of mass on the Y M dm

Ce X fig. 6.13(a)

O

dr r

R

x

X

fig. 6.13(b)

symmetrical axis of the disc (i.e., on the y-axis for chosen reference frame). Now, if we replace all these elemental rings by points masses at their centre of mass position, then, given disc reduces to a system of large number of particles distributed on the y-axis, therefore, to locate the centre of mass of the disc, centre of mass of these particles can be found and used equivalently. If we choose an elemental semicircular ring of radius r and thickness dr, as shown in figure 6.13 (b), then, mass of the chosen element, Centre of Mass

dm = mass per unit area × area of the element

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=

M × (π r ⋅ dr ) (π R 2 2)

=

2M ⋅ r ⋅ dr R2

14

...(i)

If the centre of mass of the chosen element is at the point Ce, as shown in the same figure, then, coordinates of Ce can be given as x = 0; y=

...(ii)

2r π

...(iii)

Now, the chosen element can be treated as a point mass at the point Ce. If we consider all elemental rings shown in figure 6.13 (a) in a similar way, then, the coordinates of the centre of mass of the given semicircular disc, X and Y, can be obtained by using equation 6.7. Therefore, X=

∫ x ⋅ dm ∫ dm

=0

and

Y=

[using equation (ii)]

∫ y ⋅ dm ∫ dm 2r

∫ π ⋅ dm = ∫ dm 2 =π

Centre of Mass

R 0

∫r⋅

2M ⋅ r ⋅ dr R2 M

=

4 π R2

=

4 R3 ⋅ π R2 3

=

4R 3π

R 0

[using equation (iii)]

∫r

2

[using equation (i)]

⋅ dr

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Hence, the centre of mass of the given semicircular disc lies on the symmetrical axis of the disc at a distance of

4R from its center. 3π

NOTE : •

Center of mass of a complete circular disc (uniform) lies at the centre of the disc.

•

While choosing elements, you must take care the following : (a)

the centre of mass of the element itself must be known .(If your chosen element is a point mass then its centre of mass coincides with its position.)

(b)

by using simple integration techniques your element must be able to cover the entire body.

A homogeneous circular plate of radius r has a circular hole cut out with radius r/2, as shown in figure 6.14 (a). Find the centre of mass of the plate.

r/2 r fig. 6.14(a)

Solution: Let us choose the reference frame as shown in figure

Y

6.14 (b) and consider the following three bodies: (a)

complete disc of radius r and mass m,

(b)

removed disc of radius r/2 and mass mremoved,

(c)

remaining disc (with a hole of radius r/2) of radius r and mass mremaining.

C2 C O

C1

X

fig. 6.14(b)

It is obvious from the figure 6.14 (b) that as the distribution of mass in all the three bodies is symmetrical about the x-axis, their centre of mass lie on it only. If C, C1 and C2 be the centre of mass of the original disc (before making the hole), removed disc and remaining disc, respectively, then we have,

C1 ≡ (r / 2, 0) C ≡ (0, 0)

C2 ≡ ( x0 , 0) where x0 is the x-coordinate of the centre of mass of the remaining disc. Now, we have to solve for x0. Mass of the removed disc mremoved = Centre of Mass

m 2 × (π ( r 2) ) π r2 Web: http://www.locuseducation.org

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=m

16

4

Mass of the remaining disc mremaining = m − mremoved m 4

= m− =

3 m 4

Now, if we combine the remaining and removed discs according to their initial relative positioning, we get the complete initial disc. Therefore, we have, X initial =

⇒

mremaining × xremaining + mremoved × xremoved mremaining + mremoved

3 m r m ⋅ x0 + ⋅ 4 2 0= 4 3 m m+ 4 4

⇒

3 x0 +

r =0 2

⇒

x0 = −

r 6

 We have replaced the bodies by the   point masses at their center of mass 

ALTERNATE METHOD : The remaining body can be represented by two discs superimposed, one of mass ‘m’ and radius ‘r’ and the other m of mass ‘ − ’ and radius ‘r/2’ you are urged to solve the given problem using this method also. 4

Centre of Mass

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1.

Must there be mass at the centre of mass of a system? Explain.

2.

Must the centre of mass of a solid body be in the interior of the body? If not, give examples.

3.

A system consists of two point masses M and m(< M). The centre of mass of the system is : (a) At the middle of m and M (b) Nearer to M (c) Nearer to m (d) At the position of large mass.

4.

Find the center-of-mass coordinates xCM and yCM for the object in figure, assuming that distribution of mass is uniform. y

10 m

0

5.

10 m

x

Find x and y coordinates of the centre of mass of the plate shown in figure from which a square of side 2 m is cut out. Assume that the distribution of mass is uniform. Y 6m

2m 6m

6.

0

2m

X

Three identical spheres each of radius R are placed touching each other on a horizontal table as shown in figure. The x and y coordinates of the centre of mass are: y

(a) (R, R)

(b) (0, 0)

R R (c)  ,  2 2

R   (d)  R, . 3 

Centre of Mass

x

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Y

7.

l

l

Three laminar objects of same density a square, a disc and an equilateral triangle are placed as shown in figure. Find the coordinates of the centre of mass of the system of these three bodies.

l

l O

8.

Four particles of masses m1 = 2 kg, m2 = 4 kg, m3 = 1 kg and m4 are placed at four corners of a square as shown in figure. Can mass of m4 be adjusted in such a way that the centre of mass of system will be at the centre of the square, C.

X m1

m2 C

m4

m3

9.

From a uniform disc of radius R, a circular hole of radius R/2 is cut. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of mass of the resulting flat body.

10.

From a uniform disc of radius R, two circular sections each of radius R/4 have been removed as shown in figure. For the reference axes shown the co-ordinates of centre of mass of remaining body are y −3 R −3 R −3 R −3 R , , (a) (b) 112 112 56 56 (c)

−3 R −3 R , 128 128

(d)

−3 R −3 R , . 64 64

x y

11.

Figure shows a thin uniform disc of radius R, from which a hole of radius R/2 has been cut out from left of the centre and is placed on right of the centre of disc. Find the C.M. of the resulting system.

(-R/2, 0) (+R/2, 0)

x

O R

12.

A square hole is punched out from a circular lamina, the diagonal of the square being a radius of the circle. Show that the centre of mass of the remaining body is at a distance R/(4π – 2) from the centre of the circle, where R is the radius of the circular lamina.

13.

A nonuniform thin rod of length L lies along the x axis with one end at the origin. It has a linear mass density

λ kg/m, given by λ = λ0 (1 + x/L). The density is thus twice as great at one end as at the other. (a) Use M = ∫ dm to find the total mass. (b) Find the centre of mass of the rod. 14.

AB is a uniformly shaped thin rod of length L, but its linear mass density varies with distance from one end A of the rod as λ = px 2 + c, where p and c are positive constants. Find out the distance of the centre of mass of this rod from the end A. y

15.

Use integration to find the centre of mass of the right isosceles triangle shown in figure. 10 m

0

Centre of Mass

10 m

x

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MOTION OF THE CENTRE OF MASS : Now we can discuss the physical importance of the centre of mass concept. Consider the motion of a group of particles whose masses are m1, m2,...mn and whose total mass is M. For the time being we will assume that mass neither enters nor leaves the system so that the total mass M of the system remains constant with time. In a later section we shall consider systems in which M is not constant; a familiar example is a rocket, which expels hot gases as its fuel burns, thus reducing its mass. From equation (6.8) we have, for our fixed system of particles, $ $ $ $ M rcm = m1 r1 + m2 r2 + ... + mn rn , $ where rcm is the position vector identifying the centre of mass in a particular reference frame. Differentiating this equation with respect to time, we obtain $ $ $ $ d rcm d r1 d r2 d rn M = m1 + m2 + ... + mn dt dt dt dt

or

$ $ $ $ M vcm = m1v1 + m2 v2 + ... + mn vn

...(6.11)

$ $ $ $ where v1 (= d r1 dt ) is the velocity of the first particle, etc. and d rcm dt (= vcm ) is the velocity of the centre of mass. Differentiating equation (6.11) with respect to time, we obtain $ $ $ $ d vcm d v1 d v2 d vn M = m1 + m2 + ... + mn dt dt dt dt

⇒

$ $ $ $ Macm = m1a1 + m2 a2 + ... + mn an

...(6.12)

$ $ $ where a1 is the acceleration of the first particle, etc., and d vcm dt (= acm ) is the acceleration of the centre of mass $ $ $ of the system. Now, from Newton’s second law, the force F1 acting on the first particle is given by F1 = m1a1 . Like $ $ wise, F2 = m2 a2 , etc. We can the write equation (6.12) as

$ $ $ $ Macm = F1 + F2 + ... + Fn

...(6.13)

Hence, the total mass of the group of particles times the acceleration of its centre of mass is equal to the vector sum of all the forces acting on the group of particles. Among all these forces will be internal forces exerted by the particles on each other. However, from Newton’s third laws, these internal forces will occur in equal and opposite pairs, so that they contribute nothing to the sum on the right hand side of the equation (6.13). Therefore, the right hand sum in equation (6.13) represents the sum of only the external forces acting on all the particles. We can rewrite equation (6.13) as simply

$ $ M acm = Fext

or

$ $ Fext acm = M

...(6.14)(a)

...(6.14)(b)

This states that the centre of mass of a system of particles moves as though all the mass of the system were concentrated at the centre of mass and all the external forces were applied at that point. Centre of Mass

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You should notice that we obtained this simple result without specifying the nature of the system of particles. The system can be a rigid body in which the particles are in fixed positions with respect to one another, or it can be a collection of particles in which there may be all kinds of internal motion. Whatever the system is, and however its individual parts may be moving, its centre of mass moves according to equation (6.14). Only external forces govren the motion of the centre of mass of the system. $ $ $ $ You should also notice that if Fext = 0 , then acm = 0 . In this case there may be internal forces acting on the different parts of the system and hence those parts may be moving with some accelerations.

$ $ Two 3-kg masses have velocities v1 = 2iˆ + 3 ˆj m / s and v2 = 4iˆ − 6 ˆj m / s . Find the velocity of the centre of $ mass. If a constant force F = 24iˆ N is applied on the system for 5 seconds, find the velocity of the centre of mass after the action of force.

Solution: From equation (6.11), we have $ $ m1v1 + m2 v2 $ vcm = m1 + m2

=

3(2iˆ + 3 ˆj ) + 3(4iˆ − 6 ˆj ) m/ s 3+3

= 3iˆ − 1.5 ˆj m / s

During the action of the external force, from equation (6.14), we have $ acm =

=

$ Fext m1 + m2 24 ˆ i m / s2 6

= 4iˆ m / s 2

As the centre of mass of the system has a constant acceleration for the period in which the external force is applied, the velocity of the centre of mass after the action of the force can be given as

$ $ $ vcm final = vcm initial + acm .t

$ $ $ [Using, v = u + at ]

= [(3iˆ − 1.5 ˆj ) + (4iˆ ⋅ 5)] m/s = 23iˆ − 1.5 ˆj m/s

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Two particles of$equal mass m are connected to a spring (massless) and rest on a frictionless horizontal table. If an external force F is applied to one of them, as shown in figure 6.15 (a), describe the motion of the system.

! F

fig. 6.15(a)

Solution: Our two-body system is indicated by the shading in figure 6.15(b). The internal forces are those exerted by the spring and the gravitational attraction between the bodies, which is so small that we can neglect it. m

m fig. 6.15(b)

The external forces acting on this system are the force of gravity exerted on each body, the normal contact force $ exerted by the surface of the table on the particles (which balances the force of gravity) and the applied force F . When we sum over all the forces acting on the system, the internal forces cancel (the force exerted by the spring on a particle is equal and opposite to the force applied by that particle on the spring, as shown in figure 6.15 (c)). Hence, the resultant force on the system $ is sum of all external forces which is equal to the applied force F , because gravity is balanced by the normal contact force from the surface. Therefore the acceleration of the centre of mass of the system is given by $ acm =

! –F1

! F1 m

fig. 6.15(c)

$ F m+m

$ F = 2m Since the masses are equal, the centre of mass lies halfway between the particles. We should note that the net force on the particle on the left side is the spring force exerted $ on it and the net force on the particle on the right side is the sum of the spring force on it and the applied force F . As the spring constant is not known and the spring force on each particle varies with the distance between the particles it is quite complicated to analyze the motion of the individual particles. However, the description of the motion of the centre of mass is simple. Centre of Mass

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$ NOTE: Consider the same example with no spring, i.e., just two equal masses on a smooth table with a force F exerted on one of them. In this case the motion of each particle would be different from the last case, because there is no spring force on any of the particles. However, the centre of mass motion is the same as in the previous case.

: A projectile explodes into two equal pieces, each of mass m, at the top of its flight. One of the pieces drops straight down from rest after the explosion while the other moves off horizontally, so that they land simultaneously. Where does the second piece land ? Solution: Considering the projectile to be the system (whether it is in one piece or two), the only external force exerted on the system before the pieces hit the ground is that due to gravity. The forces exerted during the explosion are interval forces, which do not affect the motion of the centre of mass. After the explosion the centre of mass moves on the rest of the parabola just as if there had been no explosion, as shown in figure 6.16 mm CM

2m m

CM

m

fig. 6.16

Since the centre of mass is halfway between equal masses and we know that one mass drops straight down, the other must land at an equal distance from the centre of mass, as shown in above figure. $ LINEAR MOMENTUM OF A PARTICLE: The momentum of a single particle is a vector p defined as the product of its mass m and its velocity v$ . That is,

$ $ p = mv

...(6.15)

$ Because it is proportional to v$ , the momentum p of a particle depends on the reference frame of the observer.

Newton, in his famous princpia, expressed the second law of motion in terms of momentum (which he called “quantity of motion”). Expressed in modern technology Newton’s second law reads: The rate of change of momentum of a body is proportional to the resultant force acting on the body and is in the direction of that force. In symbolic form this becomes $ d p$ F= dt

...(6.16)

For a single particle of constant mass m, we have $ d p$ F= dt =

Centre of Mass

$ d (mv) dt

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$ d v $ dm =m +v dt dt $ dv =m dt $ = ma

$ d p$ $ $ which is exactly what we have used up to now. The relations F = ma and F = for single particles are dt completely equivalent in classical mechanics.

LINEAR MOMENTUM OF A SYSTEM OF PARTICLES: Suppose that instead of a single particle we have a system of n particles, with masses m1, m2 etc. We shall continue to assume that no mass enters or leaves the system, so that the mass M (= ∑ mi ) of the system remains constant with time. The particle s may interact with each other and external forces may act on them as well. Each particle will $ $ $ have a velocity and a momentum. Particle 1 of mass m1 and velocity v1 will have a momentum p1 = m1v1 , for $ example. The system as a whole will have a total momentum P in a particular reference frame, which is defined to be simply the vector sum of the moment a of the individual particles in that frame, or $ $ $ $ P = p1 + p2 + ...+ pn $ $ $ = m1v1 + m 2 v 2 + ...+ mn vn

.

..(6.17)

If we compare this relation with equation (6.11) we see at once that

$ $ P = M vcm

...(6.18)

which is an equivalent definition for momentum of a system of particles. In words equation (6.18) states : The total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass. Differentiating equation (6.18) with respect to time we obtain $ $ dP d ( M vcm ) = dt dt =M

$ d vcm dt

$ = M acm Comparing above relation with equation (6.14) we see that $ dP $ = Fext dt

...(6.19)

$ $ This equation is the generalization of the single particle equation F = d p dt to a system of particles, no mass entering or leaving the system. In words equation (6.19) states : The rate of change of linear momentum of a system of particles is equal to the net external force acting on the system. Here you should notice that internal forces can not change the momentum of the system as a whole although they can change the momenta of the individual particles of the system on which they are acting. Centre of Mass

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CONSERVATION OF LINEAR MOMENTUM: Suppose that the sum of the external forces acting on a system is zero or there is no external force acting on the system. Then, form equation (6.19), $ $ dP $ = Fnet = 0 dt $ P = constant or ...(6.20) when the resultant external force acting on a system is zero, the total vector momentum of the system remains constant. This simple but quite general result is called the principle of the conservation of linear momentum. The conservation of linear momentum principle is the second of the greatest conservation principles that we have met so far, the first being the conservation of energy principle. For the conservation of linear momentum observers $ in different inertial reference frames would assign different values of p to the linear momentum of the system, but $ $ $ each would agree (assuming Fent = 0 ) that his own value of P remained unchanged as the particles that make up the system move about. The total momentum of a system can only be changed by external forces acting on the system. The internal forces, being equal and opposite, produce equal and opposite changes in momentum which cancel one another. For a system of particles the moment of individual particles may change (due to internal forces), but their sum remains constant if there is no external net force. Momentum is a vector quantity. Equation (6.20), therefore, is equivalent to three scalar equations, one for each coordinate direction. Therefore, we can write if Fx (x - component of the net external force ) = 0

...(6.21)(a)

then Px (x - component of the total momentum of the system) = constant. Similar, if Fy = 0, the Py = constant

...(6.21)(b)

and, if Fz = 0, then Pz = constant.

...(6.21)(c)

NOTE: For a system of particles of constant mass, equation (6.20) could also be obtained in the following way : $ $ if Fext = 0 , then acm = 0 ⇒ ⇒ ⇒

$ Vcm = constant $ M Vcm = constant $ $ P = MVcm = constant

In words this can be stated as : If the resultant external force on a system is zero, the velocity of the centre of mass of the system is constant and the total momentum of the system is constant (i.e., conserved).

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Let us consider the projectile in example 12 once again. Let us imagine that our projectile (shell) explodes into many tiny fragments while in flight, as shown in figure 6.17. We assume that the air resistance is

fig 6.17

negligible. The system is the shell, the earth is our reference frame, and the external force is that of gravity. The shell explodes at x = x1 and shell fragments are blown in all directions. What can we say about the motion of this system thereafter? Solution: The forces of the explosion are all internal forces; they are forces exerted by part of the system on other parts of the system. These forces may change the momenta of all the individual fragments, but they can not change the total momentum of the system. Only an external force can change the overall momentum of the system. Therefore, the centre of mass of the system would move under the action of gravity as there was no explosion and hence the centre of mass of the fragments will continue to move in the parabolic path that the unexploded shell would have followed as shown in figure 6.18.

fig 6.18

Here you should note that the change in total momentum of the system is the same whether the shell explodes or not and this change is attributed to gravity only.

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A man of mass 70 kg and a boy of mass 35 kg are standing together on a smooth ice surface for which friction is negligible. If after they push each other apart, the man moves with a speed 0.3 m/s relative to the ice, how far apart are they after 5.0 sec.? Solution: We take the man and the boy together as the system. The force of gravity on each is balanced by a corresponding normal force of the ice. Since there is no friction, the resultant force on the system is zero. the force exerted by the man on the boy is equal and opposite to that exerted by the boy on the man. Therefore, total momentum of the system is conserved. As the initial momentum of the system is zero (since the boy and the man are standing at rest), it will remain zero even after the moment the boy and the man push each other apart. Therefore, the boy and the man must have equal and opposite momentum. Since, the man has twice the mass of the boy, the boy must have twice the speed of the man. Since the man moves in one direction with sped 0.3 m/s, the boy moves in the opposite direction with speed 0.6 m/s. After 5 sec. The man have moved 1.5 m and the boy 3 m and they are 4.5 m apart. NOTE: •

As the total momentum of the system is always zero, the centre of mass of the system is always at rest in its original position, the position where they initially stood, as shown in figure 6.19.

70 kg

35 kg

Fm

Fb

70 kg

35 kg Vb

Vm

70 kg

35 kg Vb

Vm

70 kg

35 kg Vb

Vm

fig. 6.19

•

As the centre of mass of the system has to be at rest, at some time, if d be the distance between the boy and the man and d1 and d 2 be their distances from the centre of mass, then using equation (6.4), we have,

Centre of Mass

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d1 = m2 ⋅ d = d1

⇒ ∴

27

d m1 + m2

m1 + m2 m2

at t = 5 sec., distance between the man and the boy  mass of the     distance travelled   system  = ×   mass of   by the man     the boy  = (0 ⋅ 3 × 5 ) ×

70 + 35 35

= 0 ⋅ 3× 5× 3 m = 4 ⋅5 m

A bullet of mass 10 g moves horizontally with speed 400 m/s and embeds itself in a block of mass 390 g initially at rest on a frictionless table. What is the final velocity of the bullet and block? Solution: Since there are no horizontal forces on the bullet-block system, the horizontal component of the total momentum of the system is conserved. (There is a small vertical resultant force on the system, the weight of the bullet before it strikes the block. The bullet accelerates toward the earth before it strikes the block. We shall ignore this slight vertical motion.)

Bullet v1x m1

Wooden block

m1 + m2 Frictionless surface

vx

M

M

Before impact: Pix= m1v1x

After impact: Pfx= (m1+m2)v2 fig. 6.20

The total initial horizontal momentum Pix before the bullet strikes the block is just that of the bullet: Pix = m1v1x = (10 g )(400 m/s) = 400 g ⋅ m/s = 4kg ⋅ m/s Afterward the bullet and block move together with a common velocity vx . The total final momentum Pfx is Pfx = (m1 + m2 )vx = (10 g + 390 g )vx = (0.4 kg)vx

Since the total momentum is conserved, the final momentum equals the initial momentum: (0.4 kg)vx = 4 kg ⋅ m/s vx = 10 m/s Centre of Mass

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Since the bullet and block both move together with this velocity, the centre of mass must move with this velocity. We could have found the velocity of the centre of mass (which is constant) before the collision from Equation 8-4: MvCM , x = Σmi vix + m2 (0) vCM, x =

m1v1 x 4 kg ⋅ m/s = = 10 m/s m1 + m2 0.4 kg

A simple computation of the initial and final energies shows that again mechanical energy is not conserved. The original kinetic energy is

1 2

m1vix2 = 12 (0.01 kg) × (400 m/s) 2 = 800 J, and the final energy is

(0.04 kg)(10 m/s) 2 = 20 J. In this case most of the original kinetic energy (780 J out of 800 J) is lost because large nonconservative forces between the bullet and the block deform the bodies. A bullet embedding itself in a block is an example of an inelastic collision. We shall study such collisions in more detail later in this chapter. 1 2

Consider two blocks A and B, of masses m1 and m2 , coupled by a spring and resting on a horizontal frictionless table, as shown in figure 6.21(a).

A

B

m1

m1 fig. 6.21(a)

Let us push the blocks closer and compress the spring and then release the blocks. Describe the subsequent motion. Solution: When we release the system of the given two blocks from the rest, net external force on the system is zero at that moment and thereafter (because three is no external horizontal force acting on the system and the weight of the system is balanced by the normal contact forces on the blocks from the horizontal surface). Therefore, total linear momentum of the system is conserved. If system is released from rest at t = 0, then, momentum of the system at any time t is $ $ P(t ) = M sys vcm (t ) $ = Psys (at t = 0) $ =0

∴

$ $ vcm = 0

Or we can say that the centre of mass of the system is always at rest. You should note that above obtained result could be obtained in the following way also: we have,

⇒

$ $ Fext acm = M system $ $ acm = 0

Therefore,

$ vcm = constant

(∵

$ $ Fext = 0 )

$ = vcm,initial $ =0 Centre of Mass

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Hence, the centre of mass of the system is always at rest. Due to internal forces (spring forces on the two blocks and their corresponding reaction forces on the spring) the two blocks would move on the horizontal surface. While the two blocks move horizontally, following predication can be made: •

As the speed of the centre of mass is always zero, it maintains its initial position. Therefore, at any time, if m1 has moved by a distance d1 away from the centre of mass, m2 must have moved a distance d 2 = away from it in the opposite direction. This is exactly what we discussed in example 14.

•

m1d1 m2

As the net momentum of the system is always zero, at any moment, we can write $ $ Psys = 0 $ $ $ ⇒ m A v A + mB v B = 0 $ $ ⇒ ...(i) mAv A = −mB vB $ m $ v A = − B vB mA v A mB = vB m A

⇒ ⇒

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

Therefore, according to equations (i) and (ii), at any moment, magnitudes of momenta of the blocks A and B must be equal, A and B must be moving in opposite directions. At a certain moment if one of the blocks comes to rest, other must come to rest simultaneously. •

We have, k A 12 mAvA2 = k B 12 mB vB2

vA ( 12 mAvA ) vB ( 12 mB vB ) v = A vB k A mB = k B mA =

⇒

[Using (i)] [Using (ii)]

Therefore, at any time, kinetic energy of a block is inversely proportional to its mass. At this point you should also not that mechanical energy of the system is conserved. •

Figure 6.21(b) shows the system at some instant when the block A and the block B have moved by distances

d1 and d 2 , respectively, away from their initial positions and have speeds v A and vB . vA

vB

m1

 Psys = 0 •

⇒

m2 d1

d2

m1v A = m2 v A ; Displacement of the CM is zero ⇒ m1d1 = m2 d 2 . fig. 6.21(b)

When we release the system from the rest, each block accelerates in the outward direction due to spring force acting on it and in this way compression in the spring decreases and hence the outward spring force on each block also decreases. But as long as there is any compression left in the spring, the two blocks will continue to accelerate in the outward direction, as shown in figure 6.22(a) and (b).

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At a certain moment when the spring acquires its natural length, spring force on each block becomes zero and hence their acceleration also vanishes at that time, as shown in figure 6.22(c). At this moment each block has its maximum speed and hence it continues to move in the same direction which causes elongation in the spring. Now, spring exerts inward force on each block which increases with the distance between the two blocks, as shown in figure 6.22(d). natural length

v1

(b)

(e)

v1 (decreasing)

Fsp

Fsp

m1

m2

REST

m1

m2

Fsp

(i)

v2 (decreasing)

Fsp

m1

m2

v1 (increasing)

REST

Fsp v2(increasing)

m1

m2

v1 , max

Fsp = 0

v2 , max

m1

m2

v1 (decreasing)

(h)

v2 , max

m2

Fsp

Fsp = 0

(g)

v2 (increasing)

Fsp = 0

Fsp

Fsp

(f)

B REST m2 Fsp

Fsp = 0 v1 , max m 1

(c)

(d)

A REST m1

Fsp

(a)

v1 (decreasing)

Fsp m 1

Fsp

m2

REST

REST

m1

m2

Fsp

Fsp

fig. 6.22 Centre of Mass

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Due to inward spring force each block suffers continuous decrease in its speed and eventually comes to rest, as shown in figure 6.22(d) and (e). when blocks come to rest, elongation in the spring is maximum and hence inward spring force on each block has its maximum value. Obviously, blocks can not remain in rest with spring having maximum elongation. Blocks now move with increasing speed in the inward direction and in this way elongation in the spring decreases and hence the force on each block, as shown in figure 6.22(f). When the spring acquires its natural length once again, as shown in figure 6.22(g), spring force on them becomes zero and hence they acquire maximum inward speed. As the mechanical energy of the system is conserved (since work done by nonconservative forces is zero in this case), when system repeats its configuration, kinetic energy of the system must be repeated. Hence, maximum outward speed and maximum inward speed of each block are same, because they occur when the spring has its natural length. Blocks A and B would continue their inward motion due to their inward velocity but as soon as they pass the natural length position of the spring, spring becomes compressed and spring force retards the inward motion of the two blocks, as shown n figure 6.22(h). When the blocks stop, as shown in figure 6.22(i), compression in the spring is equal to the initial compression, which can be proved in the following way: E f = Ei

[∵ wnoncon = 0]

⇒

k f + U f = ki + U i

⇒

U f = Ui

⇒

final compression = initial compression

[∵ ki = 0 and k f = 0]

Now, the system has regained its initial state and thereafter it will go on repeating this cycle Positions of the blocks when the spring has its natural length are defined as their equilibrium positions because net force on them is zero. Therefore, here we say that the centre of mass of the system remains at its initial position and the blocks and B oscillate about their equilibrium positions. If friction were present, the motion will die out as the energy is dissipated. What can be said about the linear momentum of the system in this case? NOTE: Let us consider a very interesting and frequently observed case separately. If the centre of mass of a system of constant mass is initially at rest and net external force on the system is zero, then the results obtained in the last example can be used in all such cases. In such cases the centre of mass always remains at its inition position. Therefore,

$ ∆rcm = 0

⇒

⇒

$ $ m1∆r1 + m2 ∆r2 =0 m1 + m2

$ $ $ m1∆r1 + m2 ∆r2 = 0

...6.22(a)

[for a two particle system]

...6.22(b)

Above vector equation can also be written in the following one dimensional form: m1∆x1 + m 2 ∆x2 = 0

Centre of Mass

...6.22(c)

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From equation 6.22(c), we have, ∆x1 = −

m2 ∆x 2 m1

which says that the two parts of the system always move in opposite directions and magnitudes of their displacements are inversely proportional to their masses.

l

A man of mass m1 is located on a narrow raft of mass m1 a float on the surface of a lake. The man travels through a distance l with respect to the raft and then stops, as shown in figure 6.23(a). The resistance of the water is negligible. Find the corresponding displacement d of the raft relative to the water.

fig. 6.23(a)

Solution: Let us consider the man and the raft as a single system. As the centre of mass fo the external force on the system, we can apply equation 6.22(c) in this case. Let us define the horizontal direction as x-direction and assume that the raft has moved by a distance d with respect to the water, as shown in figure 6.23(b). Applying equation 6.22(c), we get,

l (ωrt raft)

+ve X

m1 m2 fig. 6.23(b)

d

m1∆x1 + m2 ∆x2 = 0 m1 ∆x1 m2

⇒

∆x2 = −

⇒

(− d ) = −

⇒

m1d + m2 d = m1l

⇒

m1 (l − d ) m2

d=

∵ ∆x2 = displacement of the raft = −d and   ∆x = displacement of the raft = (l − d )    1

m1 l m1 + m2

$ In this last example if the man while walking on the raft moved with a velocity v1 (t ) with respect to the raft, find: (a) (b)

velocity expression of the raft with respect to the water for the same time interval the horizontal component of the force with which the man acted on the raft during the motion.

Solution: In the last example, obtained result is d=

m1 l m1 + m2

where d is the distance traveled by the raft with respect to the water and l is the distance traveled with respect to the Centre of Mass

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raft. As we know that the man and the raft move in opposite directions, in vector form above equation can be written as $ d =−

m1 $ l m1 + m2

If we assume that these distances are some intermediate distances while the movement of the man and the raft, then differentiating the above expression with respect to the time, we get, $ $ d (d ) m1 d (l ) =− dt m1 + m2 dt

⇒

$ v2 (t ) = −

m1 $ v1 (t ) m1 + m2

$ $ $ dl $ d (d ) where v1 (t ) = is the velocity of the man with respect to the raft and v2 (t ) = is the velocity of the raft with dt dt respect to the water. ALTERNATE METHOD:

$ When the man moves on the raft with velocity v1 (t ) with respect to it, let us assume that the raft moves with velocity $ v2 (t ) with respect to the water, as shown in figure 6.22(c). As the external horizontal force on the system “man + raft” is zero, linear momentum of the system in the horizontal direction must be conserved. Therefore, at some time t, the linear momentum of the system $ $ v1(t) (ωrt raft) Psys (t ) = Psys, initially =0 ⇒

$ $ $ $ m1 [v1 (t ) + v2 (t ) ] + m2 v2 (t ) = 0

⇒

$ $ $ m1v2 (t ) + m2v2 (t ) = −m1v1 (t )

⇒

$ v2 (t ) = −

! v2(t) (ωrt water)

m1 $ v1 (t ) m1 + m2

fig. 6.23(c)

$ $ where ‘ v1 (t ) + v2 (t ) ’ is the velocity of the man with respect to the water.

A smooth wedge of mass M with a small block of mass m at the highest point of its inclined surface is released from rest on a fixed smooth horizontal surface, as shown in figure 6.24(a). Initially block is at rest with respect to the wedge surface and the angle of inclination of the inclined surface with respect to the horizontal is θ. Find: (a) distance traveled by the wedge with respect to the horizontal surface when the block reaches the lowermost point of the inclined part of the wedge (b) speed of the wedge with respect to the horizontal surface at the same moment. Centre of Mass

m

oo sm

H

th M

θ

smooth fig. 6.24(a)

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Solution: (a) If we consider the wedge and the block as a single system, then, there is no external force acting on the system in the horizontal direction. Therefore, the centre of mass of the system suffers zero horizontal displacement because its initial velocity in the horizontal direction was also zero. If we define horizontal direction as x-direction, as shown in figure 6.24(b), and assume that the wedge has moved by a distance x along the +ve x-direction, then, for the system, we can write,

+ve x-direction

[∵ ∆xcm = 0]

m1∆x1 + m2 ∆x2 = 0 ⇒

m∆x1 + M ∆x2 = 0

⇒

m ( x − Hcotθ ) + Mx = 0

⇒

mx + Mx = mHcotθ

⇒

m x= Hcotθ m+M

M θ

m x Hcot θ

fig. 6.24(b)

Here ∆x1 is the displacement of the block along +ve x-direction (or we can say change in horizontal position of the block). I have used ∆x1 = x − Hcosecθ . How did I got this relation? Try to find it out from the figure 6.24(b) only. (b)

when the block reaches the lowermost point of the inclined surface of the wedge, let its speed with respect to the wedge be u and the wedge has a speed v with respect to the horizontal surface, as shown in figure 6.24(c).

N

v M

u

mg

N

N'

m

Mg fig. 6.24(c)

fig. 6.24(d)

When the block slides over the wedge, free body diagrams of the block and the wedge are shown in figure 6.24(d). When we consider the block and the wedge as a single system, then, normal contact forces applied by the block and the wedge on each other do not contribute to the net force on the system. Remaining forces acting on the system are weight of the block, weight of the wedge and the normal contact force on the wedge from the horizontal surface. All these three forces act in vertical direction and hence, the net force acting on the system has no horizontal component, therefore, linear momentum of the system is be conserved in the horizontal direction. Applying conservation of linear momentum in the horizontal direction between the moments when the system was released from rest and when the block just reaches the lowermost point of the inclined surface of the wedge, we get Psys, x, fin = Psys, x, initially

⇒

Mv + m(v − u cos θ ) = 0

...(i)

Here, reference +ve direction and final momentum of the two bodies used are shown in figure 6.24(e). Centre of Mass

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LOCUS

m θ

v

M

35

u cosθ m

v

u

v

u sin θ

+ve direction

M

v

+ve direction

fig. 6.24(e)

In equation (i) we have two known (u and v), therefore, to solve for v we need one more equation. In this case work done by nonconservative forces (normal contact forces) on the system is zero, therefore, mechanical energy of the system is conserved. Therefore, gain in K.E. of the system = loss in P.E. of the system ⇒

1 1 Mv 2 + m  (v − u cos θ ) 2 + (u sin θ ) 2  = mgH 2 2

...(ii)

Solving equations (i) and (ii), we can get the expression for v.

Consider now the apparently simple example of a ball thrown up from the earth by a person and then caught by him on its return. To simplify matters we can consider the person to be part of the earth since he does not loose contact with it. We also assume that air resistance is negligible. Now, we shall go through a comprehensive analysis of this situation from the viewpoint of this topic. Solution: The system being considered consists of the earth and the ball. The gravitational forces between the parts of the system are now internal forces. Let us choose a reference frame in which the system (earth + ball) is at rest. When the ball is thrown up, the earth must recoil as seen by an observer in this reference frame. The momentum of the system (earth + ball) is zero initially and no external forces act. Therefore, momentum is conserved and the total momentum remains zero throughout the motion. The upward momentum acquired by the ball is balanced by an equal and opposite downward momentum of the earth. We have Initial momentum = final momentum,

$ $ 0 = mB vB + mE vE , $ $ mB vB = −mE vE .

$ $ Here mB and mE are the masses of ball and earth respectively and vB and vE are the velocities of the ball and the earth in our selected reference frame. Owing to the enormous mass of the earth in comparison with the ball, the recoil speed of the earth is negligibly small. As the ball and earth separate, the internal force of gravitational attraction pulls them together until they cease separating and begin to approach one another. As the ball falls toward the earth, the earth falls toward the ball with an equal but oppositely directed momentum. As the ball is caught, its momentum is neutralized by (and it neutralizes) the momentum of the earth. Both objects lose their relative motion, the total momentum is still zero, and the original situation before throwing is restored. You will recall that when we discussed the conservation of energy in the presence of gravitational potential, we neglected to consider the motion of the earth itself. We took the surface of the earth as our zero level of gravitational potential energy. The reference position did not matter, since we were concerned only with changes in potential energy. However, in computing changes in kinetic energy, we assumed that the earth remained stationary, Centre of Mass

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as in the case of the ball thrown up from the earth. In principle, we cannot ignore the change in the kinetic energy of the earth itself. For example, when the ball falls toward the earth, the earth is slightly accelerated toward the ball. We neglected this fact before because we assumed that the change in kinetic energy of the earth is negligible. This result is not obvious, because although the earth’s speed will certainly be small, its mass is enormous and the kinetic energy acquired may be significant. To settle the point we compute the ratio of the kinetic energy of the earth to that of the ball. Using mE vE = mBvB from momentum conservation, we have K E 12 mE vE2 12 (mE vE ) 2 mB mB = = ⋅ = . K B 12 mB vB2 12 (mB vB ) 2 mE m E Since the mass of the ball mB is negligibly small compared to the mass of the earth mE , the kinetic energy acquired by the earth, K E , is negligibly small comapred to that of the ball, K B . For example, if mB = 6 kg (a rather massive ball), then, since mE = 6 × 1024 kg, K E K B = 10−24 !

A ball of mass m, moving with a velocity u along x-axis, strikes another ball of mass 2m kept at rest. the first ball comes to rest after collision and the other breaks into two equal pieces. One of the pieces starts moving along Y-axis with a speed v1. What will be the velocity of the other piece after the collision? Assume that the only forces present are the forces applied by the balls on each other during the collision. Solution: If we consider the two balls as a single system, then, the overall linear momentum of the system must be conserved because there is no external force acting on the system. Just before the collision system is shown in figure 6.25(a) and just after the collision it is shown in figure 6.25(b). y

y v1 m

u m

REST 2m

x

x

θ

m

m v2 fig. 6.25(a)

fig. 6.25(b)

! P1

As the initial momentum of the system is along +ve direction of x-axis, its final momentum must also be along the same direction. Therefore, velocity of the second part of the ball of the larger mass is chosen in such a way that when its momentum is added with the momentum of the part moving along +ve Y axis, we may get net momentum along +ve x-axis, as suggested in figure 6.25(c).

θ

! Pnet ! P2

fig. 6.25(c)

Applying conservation of linear momentum on the system, from figures 6.25(a) and (b), we get, ALONG y-axis: Py , finally = Py , initially ⇒ Centre of Mass

mv1 − mv2 sin θ = 0 Web: http://www.locuseducation.org

PHYSICS

⇒

LOCUS

v2 sin θ = v1

37

...(i)

ALONG x-axis: Px , finally = Px , initially ⇒

mv2 cos θ = mu

⇒

v2 cos θ = u

...(ii)

solving equations (i) and (ii), we get, v2 = v12 + u 2

and

Centre of Mass

θ = tan −1 ( v1 u ).

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Two bodies of masses 1 kg and 2 kg are moving in two perpendicular direction with velocities 1 m/s and 2 m/s as shown in figure. The velocity of the centre of mass (in magnitude) of the system will be; 1 m/s

1 kg

(a) 3 m/s (b) 1.67 m/s

2 m/s

(c) 1.5 m/s (d) 1/37 m/s.

2 kg

2.

Two particles A and B initially at rest, move towards each other mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, the speed of the centre of mass of the system is: (a) 3 V (b) V (c) 1.5 V (d) zero

3.

A particle of mass 4 m which is at rest explodes into three fragments. Two of fragments, each of mass m are found to move with a speed v each in mutually perpendicular directions. The total energy released in the process is;

4.

(a)

1 mv² 2

(b) mv²

(c)

3 mv² 2

(d)

$ $ $ $ $ $ Two particles having position vectors r1 = (3i + 5 j ) meter and r2 = (−5i − 3 j ) meter are moving with $ $ $ $ $ $ velocities v1 = (4i + 3 j ) m/s and v2 = (ai + 7 j ) m/s. If they collide after 2 second, the value of a; (a) 2 (c) 6

5.

the momentum of all the fragments is zero the momentum of all the fragments increases the K. E. of all the fragment remain zero the K. E. of all the fragment is more than zero.

A shell is fired from a cannon with a velocity v(m/s) at an angle θ with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed (m/s ) of the other piece immediately after the explosion is : (a) 3 v cos θ (c)

7.

(b) 4 (d) 8.

A bomb at rest explodes into large number of tiny fragments. Then: (a) (b) (c) (d)

6.

5 mv². 2

3 v cos θ 2

(b) 2 v cos θ (d)

3 v cos θ 2

A body A of mass M while falling vertically downward under gravity breaks into two parts; a body B of 2M M mass and a body C of mass . The centre of mass of bodies B and C taken together shifts 3 3 compared to that of body a towards: (a) Body (b) Body B (c) Depends on height of breaking (d) Does not shift.

Centre of Mass

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8.

The center of mass of a two-particle system is alwasy closer to the more massive particle.

9.

The total momentum of a system equals the product of the total mass and the velocity of center of mass.

10.

The momentum of a system can be conserved even if the mechanical energy is not.

11. 12.

Internal forces do not affect the motion of the center of mass of the system. 1 m/s A 4-kg fish is swimming at 1 m/s to the right. He swallows 3 m/s 1 -kg a8 fish swiming toward him at 3 m/s to the left (figure. Waht is the veocity of the larger fish immediately after his 1 4 kg 8 kg lunch? The variation of momentum with time of one of the body in a two body collision is shown in Fig. The instantaneous force is maximum corresponding to point: • (a) P •R S (b) Q (c) R p •P •Q (d) S. An isolated particle of mass m is moving in horizontal plane (x-y), along the x-axis, at a certain height above the ground. It suddenly explodes into two fragment of masses m/4 and 3m/4. An instant later, the smaller fragment is an y = +15 cm. the larger fragment at this instant is at: (a) y = –5 cm (b) y = +20 cm (c) y = +5cm (d) y = –20 cm A child is sitting at one end of a long trolley moving with a uniform speed v on a smooth horizontal track. If the child starts running towards the other end of the trolley with a speed u, the speed of the centre of mass of the system will: (a) u + v (b) v – u (c) v (d) none.

14.

15.

16.

17.

A pulley fixed to the ceiling carries a thread with bodies of masses m1 and m2 attached to its ends. The mass of the pulley and the thread are negligible, friction is absent. Find the acceleration of the centre of inertia of this system.

18.

A bomb explodes in air when it has horizontal speed of v. It breaks into two identical pieces of equal mass. If one goes vertically up at a speed of 4v, find the velocity of there immediately after the explosion. 2R

19.

20.

A ball of mass M and radius R is placed inside a spherical shell of same mass M and the inner radius 2R. The combination is at rest on a table top is the position shown in figure. The ball is released, rolls back and fourth inside, and finally comes at rest at the bottom of the shell. Find the maximum displacement of the shell during the process.

R

Smooth surface

A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments each of mass m are found to move with a speed v each in mutually perpendicular directions. The total energy released in the process of explosion is......

Centre of Mass

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PHYSICS

21.

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40

A pendulum consists of a wooden bob of mass m and length l. A bullet of mass m1 is fired towards the v1 pendulum with a velocity v1. The bullet comes out of the bob with speed and the bob just completes 3 motion along a vertical circle. The velocity v1 is : (a)

3 m    5 gl 2  m1 

m (b) m 5 gl 1

(c)

3 m    gl 2  m1 

(d)

3  m1    5 gl . 2 m 

22.

A particle of mass m, moving in a circular path of radius R with a constant speed v2 is located at point (2R, 0) at time t = 0 and a man starts moving with a velocity v1 along the positive y-axis from origin at time t = 0. Calculate the linear momentum of the particle w.r.t. man as a function of time.

23.

Ball B, of mass mB, is suspended from a string of length l attached to cart A, of mass mA, which may roll freely on a frictionless horizontal surface. If the ball is given an initial horizontal velocity v0 while the cart is at rest, determine (a) the velocity of B as it reaches its maximum height, (b) the maximum height h through which B will rise.

mA Cart A

l v0

A mB

24.

The 60 kg bullet is fired at the two blocks resting on a surface where the coefficient of kinetic friction is 0.50. The bullet passes through the 8 kg block and lodges in the 6 kg block. the blocks slide the distances shown. Compute the initial velocity v of the bullet. 0.8 m v 8 kg

6 kg

60 g 1.2 m

25.

Two blocks of masses m1 and m2 are connected by a spring of force constant k. Block of mass m1 is pulled by a constant force F1 and other block is pulled by a constant force F2. Find the maximum elongation that the spring with suffer.

26.

An astronaut outside a spaceship hammers a loose rivet back in place. What happens to the astronaut as he swings the hammer ? (a) Nothing. The space ship takes up the momentum of the hammer. (b) He moves away from the space ship. (c) He moves towards the space ship. (d) He moves towards the space ship as he pulls the hammer back and moves away from it as he swings the hammer forward. (e) He moves away from the space ship as he pulls the hammer back and moves toward it as he swings the hammer forward.

Centre of Mass

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IMPULSE: Let us consider equation (6.16) once again. We have, $ dp $ =F dt $ $ Therefore, d p = F ⋅ dt $ $ Here d p is the change in momentum caused by the force F in the time interval dt. Change in momentum caused by $ the force F for a time interval [t1 , t2 ] can be found by the following equation t

$ 2 $ ∆p = ∫ dp t1

⇒

t $ 2 $ ∆p = ∫ F ⋅ dt

...(6.23)

t1

$ The change in momentum caused by a force, defined by equation (6.23), is called as impulse, J , of that force. Impulse is a vector quantity and has same units and dimensions as that of momentum. $ If force F is the only force acting on the system, then, change in momentum contributed by it is the net change in momentum of the system. Therefore, for such a case, equation (6.23) reduces to the following form: t $ 2 $ ∆p = ∫ F ⋅ dt t1

⇒

t $ $ 2 $ p f − pi = ∫ F ⋅ dt t1

⇒

t2 $ $ $ p f = pi = ∫ F ⋅ dt

...(6.24)

t1

$ For a constant force, F , equation (6.23) reduces to the following form: $ $ ∆p = F ⋅ ∆t

...(6.25)

where ∆t is the length of the time interval for which the impulse of the force is being calculated. NOTE: •

•

F

It is obvious from equation (6.23) that for a force constant in its direction, its impulse is area under force-time graph and the direction of the impulse is along the direction of the force only, as suggested by equation (6.25). Now, we have one more alternate way to explain the conservation of linear momentum. If net force on a system is zero, thus, impulse obtained by the system is zero and hence, it momentum remains unchanged .

Centre of Mass

t2

∫

area= F⋅dt t1

= impulse, J

O

t1

t2

t

fig. 6.26

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•

•

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42

From this point of view also it is clear that internal forces can not change the overall momentum of a system. Impulse contributed by any force for any time interval is negative of that contributed by its reaction force for the same time interval. Therefore, for any time interval, combined impulse due to of the action-reaction pair is zero. $ Using the concept of impulse we can establish an expression for the time-average of a force. If a force F $ $ acts for the time interval [t1 , t2 ], then, for this interval, average value of the force F , Fav , would impart the $ same impulse as the force F itself. Therefore, ⇒

t2 $ $ Fav. × ( length of the time interval ) = ∫ F ⋅ dt t1

$ F ∫ ⋅ dt

t2

⇒

•

$ Fav =

t1

...(6.26)

t 2 − t1

From above discussion it is obvious that if more than one force act on a system, then, for a time interval ∆t, average value of the net force can be given as $ $ ∆psys Fnet, av = t2 – t1

...(6.27)

It could also obtained using equation (6.19).

$ A body of mass m is thrown at an angle to the horizontal with the initial velocity u. Assuming the air drag to be negligible, find: $ (a) the change in momentum, ∆p, that the body acquires over the first t seconds of motion $ (b) the modulus of the momentum change, ∆p, during the total time of flight, T. SOLUTION: At some time ‘t’ during its flight the body is shown in figure 6.27. During the entire flight of the body, the body is $ experiencing only the gravitational force, mg , which is constant. Hence, for the first ‘t’ seconds, the change in momentum of the body, $ ∆p = impulse imparted by gravity $ = mg ⋅ t

! v ! u

! mg

fig. 6.27

For the total time of flight, the change in momentum of the body is $ $ ∆p = mg ⋅ T and its magnitude is mgT.

Centre of Mass

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At some time a particle of mass 2 kg has a velocity (2iˆ + 2 ˆj ) m/s. After 10 seconds its velocity becomes (4iˆ + 6 ˆj ) m/s. Find average net force on the particle for these 10 seconds. Solution: Average net force on the particle, $ $ change in momentum of the particle, ∆p Fav = length of the time interval, ∆t =

=

$ $ p f − pi ∆t

=

$ $ mv f − mvi ∆t

$ $ m(v f − vi ∆t

= (2 kg) ⋅

(2iˆ + 2 ˆj ) − (4iˆ + 6 ˆj ) m/s 10 sec

= −(4iˆ + 8 ˆj ) kg ⋅ m/s 2

A small ball of mass m hits a rigid wall perpendicularly with a speed u. If the ball comes to rest immediately after colliding with the wall and duration of collision be ∆t, find the magnitude of the average force exerted by the wall on the ball. SOLUTION: If Fav be the magnitude of the average force exerted by the wall on the ball for the duration of collision, then, we have, $ | ∆p | Fav = ∆t

=

$ $ | p f − pi | ∆t

$ | − pi | = ∆t

Centre of Mass

=

pi ∆t

=

mu ∆t

[∵

$ $ p f = 0]

 pi is the magnitude of the momentum  of the ball just before hitting the wall   

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A cannon of mass M starts sliding freely down a smooth inclined plane at an angle α to the horizontal. After the cannon covered the distance, l, a shot was fired, the shell leaving the cannon in the horizontal direction with a $ momentum p. As a consequence, the cannon stopped. Assuming the mass of the shell to be negligible, as compared to that of the cannon, determine the duration of the shot. SOLUTION: When the cannon slides freely on the inclined plane with angle of inclination α, its acceleration is g sin α down the incline. After covering a distance l on the inclined surface its speed becomes v = 2 ⋅ g sin α ⋅ l . Situations just before the shot, during the shot and just after the shot are shown in figures 6.28(a), (b) and (c), respectively. y

y

y

O

O N

x

x

! F in α

v = 2 gl sin α

just before the shot

Mg cosα

α

α

during the shot

just after the shot.

fig. 6.28(b)

$ In figure 6.28(b) F is the force applied$ by the cannon on the shell at some instant during the shot and − F is reaction applied by the shell on the cannon. N is the normal contact force acting on the cannon from the inclined surface at the same instant and mg sin α any mg cos α are components of weight of the cannon along the inclined surface and perpendicular to the inclined surface, respectively.

If we consider the shell and the cannon as a single system, thenduring the shot, external forces acting on the system are weight of the cannon and the normal contact force on the cannon from the inclined surface, as shown in figure 6.28(d). It is obvious that only these forces are responsible for the change in the overall momentum of the system.

fig. 6.28(c)

y O N

x system

in α

fig. 6.28(a)

REST

! P

M gs

α

x ! –F

M gs

v

O

Mg cosα

α

fig. 6.28(d)

If the duration of the shot be ∆t, then, change in x-component of the momentum of the system is ∆Px = Px,fin − Px , in

$ ⇒ ∆Px =| p | cos α − Mv ...(i) [Using figures 6.28(a) and (c)] $ where | p | is the magnitude of the momentum of the shell just after the shot. For the duration of shot, as mg sin α is the only external force along the x-direction, we can write ∆Px = Mg sin α ⋅ ∆t Centre of Mass

...(ii) Web: http://www.locuseducation.org

PHYSICS

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45

From equations (i) and (ii), we have, $ mg sin α ⋅ ∆t =| p | cos α − Mv ⇒

$ | p | cos α − Mv ∆t = Mg sin α

where v = 2 gl sin α , NOTE: Here you should try to notice that during the shot normal contact force on the cannon, N, from the inclined surface has a much greater magnitude than mg cos α , although its magnitude before and after the shot is Mg cos α .

In the last example find the average value of N for the duration of shot. Solution: Using figure 6.28(d), we have change in momentum of the system along Y direction = average force along Y-direction ×∆t ⇒

∆Py = ( N av − Mg cos α ) ⋅ ∆t

⇒

N av =

=

∆Py ∆t

+ Mg cos α

Py ,fin − Py ,in ∆t

+ Mg cos α

$ | p | sin α − 0 = $ + Mg cos α (| p | cos α − Mv) Mg sin α

 Using result of last example for ∆t and  figures 6.28(a) and (c) for P and P  y ,in y ,fin  

$ | p | ⋅Mg ⋅ sin 2 α = $ + Mg cos α | p | cos α − Mv where,

v = 2 gl sin α .

ALTERNATIVE METHOD:

$ $ For the duration of shot the only external forces on the system are gravity, Mg , and normal contact force, N . Therefore, we have, $ $ $ $ p f − pi = Mg ⋅ ∆t + N ⋅ ∆t ⇒

Centre of Mass

$ $ $ $ Mv − p N = Mg − ∆t

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46

The only force acting on a 2.0-kg object moving along the x axis is shown. If the velocity vx is –2.0 m/s at t = 0, what is the velocity at t = 4.0 s? Fx(N) 4

0

1

2

3

4

t(s)

-4

-8

(a) (c) (e) 2.

3.

–2.0 m/s –3.0 m/s +5.0 m/s

(b) (d)

–4.0 m/s +1.0 m/s

A 3.0 kg ball with an initial velocity of (4i + 3j) m/s collides with a wall and rebounds with a velocity of (–4i + 3j) m/s. What is the impulse exerted on the ball by the wall ? (a) +24 i Ns (b) –24 i N s (c) +18 j N s (d) –18 j Ns (e) +8.0 i N s A 1.2-kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and emerges with a speed of 6.0 m/s in the opposite direction. If the object is in contact with the wall for 2.0 m/s, what is the magnitude of the average force on the object by the wall ? (a) 9.8 kN (b) 8.4 kN (c) 7.7 kN (d) 9.1 kN (e) 1.2 kN

4.

An astronaut outside a spaceship hammers a loose rivet back in place. What happens to the astronaut as he swings the hammer ? (a) Nothing. The space ship takes up the momentum of the hammer. (b) He moves away from the space ship. (c) He moves towards the space ship. (d) He moves towards the space ship as he pulls the hammer back and moves away from it as he swings the hammer forward. (e) He moves away from the space ship as he pulls the hammer back and moves toward it as he swings the hammer forward.

5.

A 2000-kg truck traveling at a speed of 6.0 m/s makes a 90° turn in a time of 4.0s and emerges from this turn with a speed of 4.0 m/s. What is the magnitude of the average resultant force on the truck during this turn? (a) 4.0 kN (b) 5.0 kN (c) 3.6 kN (d) 6.4 kN (e) 0.67 kN

Centre of Mass

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6.

LOCUS

The magnitude of the force (in newtons) acting on a body varies with time t (in microseconds) as shown in the fig. AB, BC and CD are straight line segments. The magnitude of the total impulse of the force on the body from t = 4 µs to t = µs is ........N-s. C

Force (N)

800 600 400 A

200 0

7.

47

B 2

E F D 4 6 8 10 12 14 16 Time (µs)

A body of mass 3 kg is acted on by a force which varies as shown in the graph. The momentum acquired is given by: (a) Zero 10

8. 9.

Force/N

(b) 5 N-s (c) 30 N-s (d) 50 N-s.

0

2

4

6

A 0.4-kg particle initially moving at 20 m/s is stopped by a constant force of 50 N, which lasts for a short tiem ∆t. (a) What is the impulse of this force? (b) Find the time interval ∆t. A girl can exert an average force of 200 N on her machine gun. Her gun fires 20-g bullets at 1000 m/s . How many bullets can she fire per minute? $

10.

ˆ is applied at t = 0 to the two-particle system of Exercise 5. (a) Find the velocity A constant force F = 24iN of the center of mass at t = 5 s. (b) If the center of mass is at the origin at t = 0, where is it at t = 5 s?

11.

Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on frictionless horizontal surface. An impulse gives velocity of 14m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is: (a) 30 m/s (b) 20 m/s (c) 10 m/s (d) 5 m/s

12.

A stream of glass beads comes out of a horizontal tube at 100 per second and strikes a balanced pan, as shown in figure. They fall a distance of 0.5 m to the balance and bounce back to the same height. Each bead has mass 0.5 g. How much mass m must be placed in the other pan of the balance to keep the pointer reading zero?

Centre of Mass

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48

CENTER OF MASS FRAME: In many cases when we examine only the relative motion of particles within a system, but not the motion of this system as a whole, it is most advisable to chose the reference frame in which the centre of mass is at rest. Then we can significantly simplify both the analysis of the phenomena and the calculations. The reference frame rigidly fixed to the centre of mass of a given system of particles and translating with respect to inertial frames is referred to as the frame of the centre of mass, or, C frame. In C frame the centre of mass is always at rest. Therefore, the change in the position of the centre of mass for any time interval must be zero hence. $ $ ∆rcm = 0

...(6.28(a))

or, for a one dimensional analysis ∆xcm = 0

...(6.28(b))

As the velocity of the centre of mass in the C frame is always zero, the overall linear momentum of the system in this frame is $ $ Psys = M sys vcm ⇒

$ $ Psys = 0

$ $ [∵ vcm = 0]

...(6.29)

Hence, the overall momentum of a system of particles in its C frame is always zero. As the acceleration of the centre of mass of a system of particles in its C frame is always zero, net external force on the system, $ $ Fext = M sys acm ⇒

$ $ Fext = 0

$ $ [∵ acm = 0]

...(6.30)

Hence, Net external force on a system of particles in its C frame is always zero. Let us find the relationship between the values of the mechanical energy of a system in some frame, let say k frame, and C frame. Let us begin with kinetic energy, K.E., of the system in k frame. The velocity of the ith particle in k frame may be represented as $ $ $ vi = ui + vc $ $ where ui is the velocity of that particle in the C frame and vc is the velocity of the C frame with respect to the k frame. Now, we can write, kinetic energy of the system in k frame, K .E. = Σmi vi2 / 2 $ $ $ $ (ui + vc ) ⋅ (ui + vc ) = Σmi 2 Centre of Mass

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49

$ $ = Σmi ui2 2 + vc ⋅ Σmi ui + Σmi vc2 2

$ $ Since, in the C frame Σmi ui (net momentum of the system) = 0 , the previous expression reduces to the K .E . = Σmi ui2 2 + Σmi uc2 2

⇒

K . E . = K .E ′. +

mVc2 2

...(6.31)

where K .E.′ is the total kinetic energy of the particles in C frame and m is the total mass of the system. Equation (6.31) can be rearranged to botain,

K .E. = K .E.′ +

P2 2m

...(6.32)

where P is the magnitude of the total momentum of the system in the K frame. Thus, the kinetic energy of a system of particles comprises the total kinetic energy in the C frame and the kinetic energy associated with the motion of the system of particles as a whole. This important conclusion will be repearedly utilized hereafter (specifically, in the next topic, rotational motion, which basically deals with the motion of solid bodies). It follows from equation (6.31) that the kinetic energy of a system of particles is minimum in the C frame, another $ $ distinctive feature of that frame. In C frame vc = 0, and equation (6.31) gives K .E. = K .E.′ . Let us consider the following example: Two blocks A and B of masses m1 and m2 , respectively, connected by a massless spring of spring constant k, are placed on a smooth horizontal surface, as shown in figure 6.29. Block A is given a horizontal speed u towards the

u

A m1 SMOOTH

K

B m2

Natural length

fig. 6.29

block B when the spring has its natural length. Now, we have to analyze the subsequent motion of the two blocks. In example 15, we have already considered a case very similar to this one in much detail. But even then for the sake of clarity we will discuss this case first in the frame of the horizontal surface on which the blocks are placed and then only we would analyze its motion from its C frame. When the block A is given a speed towards right, the spring starts getting compressed due to which at the same time the spring also starts exerting a force on the block B towards right and a force on the block A towards left. Initially this force accelerates the block B and retards the motion of the block A, as shown in figure 6.30. The magnitude of the spring force on each block increases with the compression in the spring. After same time speeds of the two blocks become equal and hence compression in the spring becomes maximum. Thereafter, due to spring force, speed of the block B becomes greater than the speed of the block A and hence spring starts getting relaxed. But when we consider the two blocks and the spring as a single system, the mechanical energy of the system and the linear momentum of the system must be conserved because there is no net external force is acting upon the system. Hence, the centre of mass of the system moves with a constant m1u   m1u + m2 (0) velocity  = m + m = m + m  towards right and the vB (increasing) 1 2 1 2   vA (decreasing) potential energy stored in the spring and the kinetic energy of kx kx the system keeps varying as the system moves but their sum remains constant. Here potential energy in the spring increases with the compression in the spring but at the cost of kinetic energy of the system. When the spring relaxes, the potential is fig. 6.30 converted back into the kinetic energy. Centre of Mass

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50

Here we see that when we analyze the two blocks from the frame of the horizontal surface, their individual motion is quite complicated although their centre of mass moves with uniform velocity. Now, let us analyze the motion of the two blocks from the C frame of the system. When we move the C frame we subtract the velocity of the centre of mass of the system from the velocities of the each block to get their m1 u vc = m1 + m2

u A m1

m2 u = m+m 1 2

m1 u vc = m1 + m2

K

m2

A

B

K

m1

m2

p1 = m1vA m1m2 = m+m u 1 2

fig. 6.31

m1 u m1 + m2

vB =

B

p2 = m2vB m1m2 = m +m u 1 2

velocity in the new frame, as shown in figure 6.31. In this case it is suggested that we should change the frame at the initial moment, when the block A is set in motion, only. In this way it is relatively easier to carry out the formalities of frame change due to following two reasons: (a) (b)

speed of each block is known; the block B is at rest.

In cases when the centre of mass has an acceleration, we will have to also apply pseudo forces when we move to the C frame, which was not required here. You should notice that equations 6.28, 6.29 and 6.30 hold true in this frame and it is relatively easier to predict the subsequent motion of the blocks. Each block initially compresses the spring and loses its own speed. When one block comes to rest, second one also comes to rest because net momentum of the system has to be zero in this frame. Thereafter, due to the spring force the blocks move away from each other. Eventually, in C frame the two blocks oscillate about their mean positions. Detailed analysis is very similar to what we did in the example 15. But we should not forget that the C frame itself is moving towards right with a constant speed in the frame attached with the horizontal surface.

Find the maximum compression in the spring during the subsequent motion of the system given in figure 6.29. Assume l0 as the natural length of the spring. Solution: Let the block A was set in motion was set in motion at t = 0, then at some time ‘t’, if x be the compression in the spring, vA and vB be the speeds of the two blocks, as shown in figure 6.32, we have, rate of change of distance between the blocks = speed of the blocks B–speed of the block A ⇒

dl = vB − v A dt

⇒

d (l0 − x) = vB − v A dt

⇒

dx = v A − vB dt

When compression in the spring, x, is maximum, Centre of Mass

REST

u A

B l0

At some time ‘t’

vA

vB

A

B l0 – x =l fig. 6.32 Web: http://www.locuseducation.org

PHYSICS

LOCUS

dx =0 dt

⇒

v A − vB = 0

⇒

vA = vB

51

...(i)

i.e., the two blocks have common speed or we can say their relative speed is zero. Again, using principle of conservation of linear momentum, we have Pin = Pfin

⇒

m1u = m1vA + m2vB

⇒

m1u = m1vA + m2vA

⇒

vA =

m1u m1 + m2

[Using ...(i)] ...(ii)

Applying, conservation of mechanical energy, we have, Ein = E fin

⇒

K in + U in = K fin + U fin

⇒

1 1 1 m1u 2 + 0 = ( m1 + m2 )v A2 + kx02 2 2 2

...(iii)

where, x0 i the required maximum compression in the spring. Solving equation (ii) and (iii), we can get expression for x0 Alternate Method-I: When the compression in the spring is maximum, relative speed of the two blocks is zero can also be proved using 1 equation 6.31, which is K .E. = K .E.′ + msys vc2 . 2 When compression in the spring is maximum, the potential energy stored in the spring is also maximum, hence, at that moment kinetic energy of the system must be minimum because mechanical energy of the system is conserved. In the frame attached with horizontal surface kinetic energy is minimum when K .E.′ is zero because in the above m1u expression vc is equal to m + m and is always there. When kinetic energy in C frame K .E.′ , is zero, both the 1 2 blocks must be at rest in that frame. Hence, at that instant their velocities in the frame attached with horizontal surface are equal to the velocity of their centre of mass. Therefore, when compression in the spring is maximum, m1u   both the blocks move with the same velocity  of magnitude vc = with respect to the horizontal surface. m1 + m2   Now, using principle of conservation of mechanical energy, as we did in last method, we can solve for maximum compression x0 .

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52

Alternate method - II: I am very sure that till now you must have developed a good understanding of such situations. But still I would like to stretch the discussion. Now, I will analyze the relative motion of the two blocks in the centre of mass frame from the initial moment only. This approach would provide you a smart approach for even more complicated situations. When the block A is given a velocity u, the velocity of the centre of mass at that moment is vc =

m1v1 + m2 v2 m1 + m2

=

m1u + 0 m1 + m2

=

m1 u m1 + m2

as, shown in figure 6.33. u

SYSTEM OF MASS 'm1 +

REST

m2 '

vc =

m 1u m 1+m2

K m1

A

B

m2

Natural length

fig. 6.33

As the net external force on the system is zero, the centre of mass of the system always moves with this velocity only. Hence, while analyzing the motion of the individual blocks from the C frame, we must subtract the velocity of the centre of mass, vc , from the velocities of the blocks with respect to the horizontal surface, as we did in figure (6.31). We do so at the initial moment only, then, velocities of the blocks in C frame at this moment would be as shown in figure 6.34. m2u uA = m + m 1 2

m1u uB = m + m 1 2

m1

m2

kx0 REST m1

REST kx0 m2

Natural lenth

Maximum compression state

fig. 6.34

fig. 6.35

It is clear from the figure 6.33 that as the blocks move along the directions of their initial velocities, spring forces acting on them would retard their motion and after some time they come to rest. When one block comes to rest, other one comes to rest at the same moment, because momentum of the system in this frame must be zero all the time. When the blocks come to rest, compression in the spring becomes maximum. If x0 be the maximum compression then in the maximum compression state each block is being acted upon by an outward force of magnitude due kx0 (as shown in figure 6.35) due to which blocks start moving in the outward direction. Hence, the spring acquires the maximum compression state just for a moment. In this frame, as there is no nonconservative force is doing work on the system, we have Ein = E fin

⇒

Centre of Mass

E (initially) = E (at maximum compression state)

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53

⇒

1 1 1 m1u A2 + m2u B2 = kx02 2 2 2

⇒

1  m2u  1  m1u  1 2 m1 + m2 + kx0 2  m1 + m2  2  m1 + m2  2

⇒

m1m2u 2 1 m + m2 ] = kx02 2 [ 1 2(m1 + m2 ) 2

⇒

 m1m2  x0 =   u k m + m ( ) 1 2  

2

2

12

NOTE: •

Further analysis of the situation would suggest you that in the C frame both the blocks oscillate about their m2 x0 equilibrium position (when the net force on the each block is zero) with amplitudes m + m (for m1 ) and 1 2 m2 x0 (for m2 ) . m1 + m2

•

Maximum elongation in the spring is x0 only. (This can be proved by applying the conservation of energy in the C frame between the moments when the blocks are at their extreme ends, as shown in figure 6.36). Natural length

kx0

Natural length

v=0

v=0 m1

kx0

v=0

m2

m1

m1 x0 x2 = m + m 1 2 maximum compression state x1+ x2= x0

•

v=0 m2

x1

m 2 x0 x1 = m + m 1 2

(a)

kx0

kx0

x2

maximum elongation state x1+ x2= x0 fig. 6.36

Suppose the given system has to be replaced by a spring of same spring constant and a single block of equivalent mass meq , ass shown in figure 6.37. Now, if meq is given the same speed u. then, we must get the compression in the spring. In this case by applying conservation of mechanical energy between the moments when the spring has its natural length and when the compression in the spring is maximum, we get, meq =

⇒ Centre of Mass

m1m2 m1 + m2

Natural lenth

u k

meq fig. 6.37

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54

1 1 1 = + meq m1 m2

⇒

This equivalent mass is smaller than both m1 and m2 , hence, it is defined as reduced mass of the system. Hence, we have 1 mreduced

=

1 1 + m1 m2

...(6.33)

Above result is quite frequently used to simplify the complicated systems before detailed analysis of the system.

Two blocks, of masses m1 and m2 , connected by a weightless spring of spring constant K and natural length l0 rest on a smooth horizontal plane. A constant force F starts acting on one of the blocks as shown in figure 6.38. Find the maximum distance between the blocks during the subsequent motion of the system. K m1

SMOOTH

m2

F

fig. 6.38

Solution: As F is the net external force acting on the system (weight of the system is balanced by the normal contact forces acting on the two blocks from the horizontal surface), the acceleration of the centre of mass of the system with respect to the horizontal surface is along the same direction as that of F, and has the magnitude equal to F (m1 + m2 ), as shown in figure 6.39.

acm = m1

F m1+m2 m2

F

Fig 6.39

Now, let us move to the C frame of the system. As the centre of mass of the system is initially at rest, we need not to subtract anything from the velocities of the two blocks if we change the frame at the initial moment only. Hence, in the C frame too, the blocks are initially at rest. When we move to the C frame, we must apply a pseudo force on each block, as shown in figure 6.40. m1acm

m2acm

m1F/(m1+m2)

F

K REST

m1

m2 l0

m1F/(m1+m2) K

REST

REST

m1

m2

REST

l0

Fig 6.40

Now, from figure 6.40, it is quite easy to predict the subsequent motion of the system in this frame. It is quite obvious that equations (6.28), (6.29) and (6.30) hold true in this case. If m1 stops after covering a length x1 , m2 stops at the same moment, as shown in figure 6.41. m1F m1+m2 REST

K(x1+x2)

K(x1+x2)

m1

m1F m1+m2 m2

x1

l0

REST

x2

Fig 6.41 Centre of Mass

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55

Till now if m2 has covered a length x2 , then maximum elongation in the spring is x1 + x2 . Applying wnoncon = ∆U + ∆k between the moments shown in figure 6.40 and figure 6.41, we get,

m1F m1 F 1  ⋅ x1 + ⋅ x2 =  K ( x1 + x2 ) 2 − 0 + [0 − 0] 2  m1 + m2 m1 + m2 ⇒

1 m1 F ( x1 + x2 ) = K ( x1 + x2 ) 2 2 m1 + m2

⇒

x1 + x2 =

2m1 F K ( m1 + m2 )

=

2m1 F K ( m1 + m2 )

NOTE: •

A little more analysis of the situation, quite similar to what we did in the example 5 of the chapter WORK POWER ENERGY, would lead you to the fact that in the C frame both the blocks oscillate about their m1 F mean positions, which is when the elongation in the spring is K ( m + m ) 1 2

•

The spring never gets compressed during the subsequent motion, i.e., minimum elongation in the spring during the motion of the system is zero. Hence, minimum distance between the blocks is l0 .

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LOCUS

56

A massless spring of force constant 1 kN/m is compressed a distance of 20 cm between masses of 8 and 2 kg. The spring is released on a smooth table. Since the masses are not attached to the spring, they move away with speed v1 and v2 . (a) Show that the kinetic energy of the masses when they leave the spring can be written Ek = 12 p 2 (1/m1 + 1/m2 ), where p is the momentum of either mass. Use conservation of energy to find p. (b) Find the velocity of each mass as it leaves the spring. (c) Find the velocity of each mass if the system is given an initial velocity of 4 m/s perpendicular to the spring, as shown in figure. (d) What is the energy of center-of-mass motion in this case? Wnat is the energy of motion relative to the center of mass? (e) Find the velocity of each mass if the system is given in initial velocity of 4 m/s in the direction along the spring, as shown in figure. What is the energy of center-of-mass motion in this case? What is the energy of motion relative to the center of mass? 2 kg

8 kg

2 kg

4 m/s

8 kg 4 m/s

(a)

2. 3. 4.

(b)

In a reference frame in which the center of mass is at rest, the total momentum of a system is zero. In a reference frame in which the center of mass is at rest, it is possible for all the kinetic energy to be lost in a collision. A 3-kg body moves at 5 m/s to the right. It is chasing a second 3-kg body moving at 1 m/s also to the right. (a) Find the toatl kinetic energy of the two bodies in this frame. (b) Find the velocity of the center of mass. (c) Find the velocities of each body relative to the center of mass. (d) Find the kinetic energy of motion relative to the center of mass. (e) Show that your answer for part (a) is greater than that for part (d) by the 2 , where vCM is the velocity of the center of mass and M is the total mass. amount 12 MvCM

5.

Two heavenly bodies S1 and S2 not far from each other are seen revolving in orbits: (a) Around their common centre of mass (b) Which are arbitrary (c) With S1 fixed and S2 moving round S1 (d) With S2 fixed and S1 moving round S2.

6.

A closed system consists of two particles of masses m1 and m2 which move at right angle to each other with velocities v1 and v2. Find: (a) the momentum of each particle and (b) the total kinetic energy of the two particles in he reference frame fixed to their centre of inertia.

7.

Two point masses m1 and m2 are connected by a spring of natural length l0. The spring is compressed such that the two point masses touch each other and then they are fastened by a string. Then the system is moved with a velocity v0 along positive x-axis. When the system reaches the origin the string breaks (t = 0). The position of the point mass m1 is given by x1 = v0 t − A(1 − cos ω t ) where A and ω are constants. y v1

v2

R (0,0)

m

x

Find the position of the second block as a function of time. Also find the relation between A and l0. Centre of Mass

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8.

9.

10. 11.

LOCUS

57

Two bodies are connected by a spring. Describe how the bodies can move so that (a) the total kinetic energy is just the energy of center-of-mass motion and (b) the kinetic energy is entirely energy of motion relative to the center of mass. Two blocks of masses 5 kg and 2 kg are placed on a frictionless surface and connected by a spring. An external kick gives a velocity 14 m/s to the heavier block in the direction of lighter one. deduce (a) the velocity gained by the centre of mass and (b) the separate velocities of tthe two blocks in the centre of mass frame just after hte kick. Describe how a solid ball can move so that (a) its total kinetic energy is just the energy of center-of-mass motion and (b) it total kinetic energy is energy of motion relative to the center of mass. Two blocks of masses m1 and m2 are connected by a spring of force constant k. Block of mass m1 is pulled by a constant force F1 and other block is pulled by a constant force F2. Find the maximum elongation that the spring with suffer.

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SYSTEM OF VARIABLE MASS: So far we have dealt only with systems in which the total mass of the system, M sys , remained constant with time. Now, we will consider systems in which mass enters or leaves the system while we are observing it. The rate of change of mass of the system, dM dt , is positive when mass enters the system and is negative when mass leaves the system. Figure 6.42(a) shows a system of mass M whose centre of mass is moving with velocity v$ as seen from a particular $ reference frame. An external force Fext acts on the system. Y

Y

At some time 't' M

At time 't +∆t'

M-∆M

∆M cm

cm

O

v + ∆v cm

u

v

X Fig 6.42 (a)

O

X Fig 6.42 (b)

At a time ∆t later the configuration has changed to that shown in figure 6.42(b). A mass ∆M has been ejected from the system and its centre of mass is moving with velocity u$ with respect to the initially chosen reference frame. The mass of the system has reduced to M-∆M and the velocity of the centre of mass of the system is changed to v$ + ∆v$. You may imagine that the system of figure 6.42 represents a rocket. It ejects hot gas from its orifice at a fairly high speed, decreasing its own mass and increasing its own speed. In a rocket the loss of mass is continuous during the $ burning process. The external force Fext is not the thrust of the rocket but is the force of gravity on the rocket and the resisting force of the atmosphere. In such motion fuel is burned in the rocket and exhaust gas is expelled out the back of the rocket. The force exerted by the exhaust gas on the rocket (which is equal and opposite to the force exerted by the rocket on the gas to expel it) propels the rocket forward and is called as thrust force. To analyze the situation let us, for the time being, define the system to be one of the constant mass. This means that in figure 6.42(b), we shall include in our system not only the M-∆M of the body but also the ejected mass ∆M. Therefore, the total mass of the system has remained M as in the figure 6.42(a). Doing so we can apply the results we have derived so far for systems of constant mass. We will see that this approach leads us to the form of Newton’s second law for systems in which the mass is not constant. From equation (6.19), we have $ $ dP Fext = dt For the time interval ∆t, approximately, we can write,

$ $ $ $ ∆P Pf − Pi Fext ≅ = ∆t ∆t $ $ in which Pf is the final momentum, as shown in figure 6.42(b) and Pi is the initial momentum, as shown in figure 6.42(a). Therefore, we have, $ $ $ $ $ ( M − ∆M )(v + ∆v ) + ∆Mu ] − [Mv ] [ Fext ≅ ∆t Centre of Mass

$ $ $ $ ∵ Pf = ( M − ∆M )(v + ∆v ) + ∆Mu $  $  and Pi = Mv

  

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⇒

LOCUS

$ $ ∆v $ $ $ ∆M Fext = M + [u − (v + ∆v ) ] ∆t ∆t

59

...(6.33)

Now, if we let ∆t approach zero, the configuration of figure 6.42(b) approaches that of figure 6.42(a); that is, $ $ ∆v ∆t approaches dv dt , the acceleration of the body in figure 6.42(a). The quantity ∆M is the mass ejected in ∆t ; this leads to a decrease in the mass M of the original body. Since dM dt , the change in mass of the body with

time, is negative in this case, the positive quantity ∆M ∆t is replaced by −dM dt as ∆t approaches zero. Finally, $ ∆v goes to zero as ∆t approaches zero. Making these changes in the last equation, we get $ $ dv $ dM $ dM Fext = M +v −u dt dt dt $ $ dv $ $ dM Fext = M − (u − v ) ⇒ ...(6.34a) dt dt $ $ dv $ dM Fext = M − vrel ⇒ ...(6.44b) dt dt $ The quantity u$ − v$ in equation (6.34a) is just vrel , the relative velocity of the ejected mass with respect to the main body, as written in equation (6.34 b). $ dM , is the rate at which momentum is being transferred into (or out of) the The last term in equation (6.34b), vrel dt system by the mass that the system has ejected (or collected). It can be interpreted as the force exerted on the system by the mass that leaves it (or joins it). For a rocket this term is called the thrust and it is the rocket designer’s aim to make it as large as possible. Equation (6.34) suggests that this requires that the rocket eject as much mass per unit time as possible and that the speed of the ejected mass relative to the rocket be as high as possible. We can rewrite equation (6.34) as $ $ dv $ M = Fext + Fthrust ...(6.35) dt $ $ in which Fthrust ( = vrel dM dt ) is the reaction force exerted on the system by the mass that leaves (or joins) it.

$ $ Here you should note that the thrust force acting on the system vrel dM dt , is along vrel if dM dt is ne gative, i.e., mass of the system is being reduced.

A machine gun is mounted on a car on a horizontal frictionless surface as shown in figure 6.43 (a). The mass of the system (car + gun) at a particular instant is M. At that same instant the gun is firing bullets of mass m whose velocity, $ $ in the reference frame shown, is u . The velocity of the car in this frame is v and the velocity of the bullets with respect to the car is u$ − v$. The number of bullets fired per unit time is n. What is the acceleration of the car? u m m

v

fig 6.43 (a) Centre of Mass

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Solution: Let us select the car and gun as our system. Since mass of the system, M, is variable, we apply Newton’s second law in the form given in equation (6.35), which is $ $ dv $ M = Fext + Fthrust dt $ $ Since no net external force acts on the system, we have Fext = 0 in above equation. Hence, we have $ $ dv M + = Fthrust dt $ dM = vrel dt $ dv dM $ $ $ Now, is a$ , the acceleration of the system; vrel is u − v , pointing to the left in figure 6.43(b), and is –mn. dt dt

Inserting these in above equation, we get u m m

v Freaction= vrel dM/dt = – vrel mn

system boundary fig 6.43 (b)

$ $ v ( mn) $ dv a= = − rel dt M $ $ (u − v )( mn) =− M

NOTE: In figure 6.43(c), the analogous situation for a rocket is shown. Here we will consider the system from the viewpoint of Newton’s third law and the momentum principle. v u

M Freaction= vrel dM/dt

fig 6.43 (c)

Let us choose a fixed-mass system (rocket + gas) and observe if from its C frame. The rocket forces a jet of hot gases from its exhaust; this is the action force. The jet of hot gases exerts a force on the rocket, propelling it forward; this is the reaction force. These two forces make an action-reaction pair and are internal forces of the system under consideration. In the absence of external forces the total momentum of the system is conserved (the centre of mass, initially at rest, remains at rest). The individual parts of the system may change their momentum, however; the hot gases acquire momentum in the backward direction and the rocket acquires an equal magnitude of the momentum in the forward direction. You can analyze the system (bullet + car) in a similar way. Centre of Mass

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Sand drops from a stationary hopper at a rate µ = dM/dt $ onto a conveyor belt moving with velocity v , as shown in figure 6.44(a). What force is required to keep the belt moving at a constant speed? (Assume that sand drops on the belt from the negligible height.)

dM dt

v

Solution: This is a clear-cut example of a force associated with change of mass alone, the velocity being constant. Let us take our system the belt of varying mass, so that we can apply the equation (6.35), which is $ $ dv $ M = Fext + Fthrust dt $ dv $ = 0 in the above equation because we must put dt the velocity of the belt is constant. Hence, we get $ $ $ Fext + Fthrust = 0 ⇒

$ $ Fext = − Fthrust

...(i)

fig 6.44 (a)

dM dt

v

F=v

system fig 6.44 (b)

bound a

dM dt

ry

Hence to keep the belt moving with a constant velocity an external force must be applied on it, which should be $ equal and opposite to the thrust force, vrel dM dt , acting on it. To an observer at rest on the belt, the falling sand (and the hopper) would appear to have a horizontal motion with speed v in a direction opposite to that shown for the belt. $ $ vrel = −v More formally we can write, $ $ $ vrel = u − v $ $ but u = 0, so that

$ $ vrel = −v Using these substitutions in equation (i), we get $ $ dM Fext = −vrel dt $ = −( −v )( µ )

$ = µv

Centre of Mass

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In this case, q dM dt (= µ ) is positive because the system is gaining mass with time,. Hence thrust force on the $ system is acting along the direction of vrel , i.e., in the direction opposite to that of the motion of the belt. Therefore, to keep the belt moving with constant velocity, external force on it must be applied in the direction of its motion. NOTE: From the view point of newton’s second law this situation can be explained as follows. As the sand has negligible speed before it falls on the moving belt, the belt has to apply a force on the falling sand along the direction of its own motion to move the sand with it and hence the falling sand applies a reaction force on the belt in the direction opposite to that of the motion of the belt. Thrust force on the belt is basically this reaction force only. Therefore, to keep the belt moving with constant velocity an external force must be applied on it along the direction of its motion and the magnitude of the external force should be equal to that the thrust force. •

For a one-dimensional case, from next time onwards we will use equation (6.35) in a more simple way. We can rewrite it as Fext ± Fthrust = Ma

...(6.36)

we will judge the direction of the thrust force from the viewpoint of Newton’s third law and its magnitude can be written as vrel .

dM . If the thrust force is along the direction of acceleration, a, then use ‘+’ sign, dt

otherwise ‘–’ sign.

A freight car filed with sand has a hole so that sand leaks out through the bottom at a constant rate of magnitude λ.

λ has an unit of kg/sec. A constant force F acts on the car in the direction of its motion. At t = 0 if v0 be the speed of the car and m0 be its mass, then, find its speed after some time t. Solution: Here, the first thing you should notice is that as sand is just leaking from the bottom, it is separating from the car with zero relative velocity. Hence, if we chose the car as our system, the thrust force on it is zero. At some time t if v be the velocity of the car and a be its acceleration, as shown in figure 6.45, from equation (6.36), we have F

SYSTEM BOUNDARY

v a Petrol

fig. 6.45

Fext ± Fthrust = ma

Centre of Mass

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⇒

F = (m0 − λt ) ⋅

⇒

dv F = dt m0 − λt

⇒

v

t

v0

0

dv dt

63

[∵

Fext = + F and Fthrust = 0]

dt 0 − λt

∫ dv = F ⋅ ∫ m

⇒

v − v0 =

−F F 0 t ln( m0 − λ t ) 0 = ln( m0 − λ t ) t λ λ

⇒

v = v0 +

F  m0  ln λ  m0 − λ t  .

A rocket is fired vertically upward near the surface of the earth where the free-fall acceleration of gravity is g. Show that if the rocket starts from rest, its final velocity is given by v = uex ln( mi m f ) − g ⋅ ∆t , where uex is exhaust speed of the gas with respect to the rocket and ∆t is the time for the fuel to burn. mi is the mass of the rocket at the v a start of the interval and m f is the mass at the end of the interval. At some time 't'

Solution: Let the rocket is fired at t = 0. Situation at some time t is shown in figure 6.46. As the rocket is ejecting gases in the vertically downward direction, thrust force on the rocket acts in the vertically upward direction. Hence, using equation (6.36), we get, mg

Fext ± Fthrust = ma ⇒

− mg + uex

Fthru

dm dv =m dt dt

⇒

− mg − uex ⋅

dm dv =m dt dt

⇒

dv = −uex ⋅

dm − g ⋅ dt m

⇒

⇒

Centre of Mass

v

mf

0

mi

∫ dv = −uex ⋅

∫

SYSTEM BOUNDARY

Fig 6.46

dm dm   dm ∵ dt is negative, dt = − dt   

∆t

dm − g ∫ dt m 0

 mf  v = −uex ⋅ ln   − g ⋅ ∆T  mi 

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m v = uex ⋅ ln  f  mi

64

  − g ⋅ ∆T 

In this case as the external force, that is due to gravity, is acting in the opposite direction of the acceleration it is take to be negative and as the thrust force is acting along the acceleration it is taken to be positive.

A chain hangs on a thread and touches the surface of a table by its lower end AFter the thread has been burned through, find the forces exerted by the table on the chain. Solution: Let L be the total length of the chain and M be its total mass. The part of the chain moving vertically downwards is in free falling state, hence, having fallen by a distance x, the air falls with speed 2 gx , as shown in figure 6.47(a). If we consider the part of the chain fallen on the surface of the table as our system, then, at the moment shown in the figure, mass of the system is increasing and the system is at rest. In the next time interval dt if length dx adds up in our system then increase in mass of the system,

x

+ve direction

L (L-x)

fig. 6.47(a)

N

= (mass per unit length) × length of element

∴

M ⋅ dx L

v = 2 gx

g

system

dm = mass of length dx of the chain

=

initial level

+ve direction system m(x).g

Fth

Rate of change of mass of the system,

fig. 6.47(b)

dm M dx = ⋅ dt L dt =

M ⋅v L

=

M ⋅ 2 gx L

...(i)

It can be explained in various ways that the adding mass exerts thrust force on the system in the vertically downward direction, as shown in figure 6.47(b) Weight of the system and the normal reaction force acting on the system from the surface of the table are also shown in figure 6.47(b). Applying equation (6.36) to this system, we get Fext ± Fthrust = ma ⇒

Centre of Mass

(mg − N ) − Fthrust = 0

∵ a = 0 and downward direcction   is chosen as positive direction   

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⇒

M dm ⋅ x ⋅ g − N + vrel ⋅ =0 L dt

⇒

N= =

65

 mass of the system,   m = mass of length x of chain   

Mgx dm + vrel ⋅ L dt Mgx M + vrel ⋅ . 2 gx L L

 dm  ∵ dt = + 2 gx , using (i) 

As the system is at rest, relative velocity of adding mass is same as its velocity. Therefore, N=

Mgx M + 2 gx ⋅ ⋅ 2 gx L L

=

Mgx M + 2 gx ⋅ ⋅ 2 gx L L

=

3Mgx L

NOTE: •

At the moment under consideration the force exerted by the table on the chain is three times the weight of the chain resting on the table.

•

Thrust force acting on the resting part of the chain is twice as great as the weight of this part.

•

At some time ‘t’ if x be the length of the system, then, mass of the system,  mass per unit   lenth of the  m( x ) =  ×  lenth    system  =

•

M ⋅x L

If mass per unit length is defined as linear mass density and is denoted by λ, then, m( x) = λ ⋅ x

•

Rate of change of mass, dm d (λ ⋅ x ) dx = =λ⋅ dt dt dt

for uniform distribution  of mass λ is constant.   

•

This problem can also be solved by treating the falling part as system with decreasing mass.

•

Let us solve this problem using impulse concept. Consider the situation at the moment shown in figure 6.47(a). In the next time interval dt if dx length of the falling part of the chain hits the surface of the table and comes to rest, then, change in momentum of that element, dp = p f − pi = 0 − ( dm ⋅ v )

 dm is the mass of the  element of length dx   

= − dm ⋅ v Centre of Mass

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Therefore, the average force acted on this element from the part of the chain resting on the table is F=

=

change in momentum length time interval dp − dm ⋅ v = dt dt

= −v ⋅

dm = −λ v 2 dt

dm    Using dt = λ v 

Therefore, force acted on the part of the chain resting on the table by the adding mass or we can also say thrust force acting on that part, Fthrust = − F = λv2 =

M ⋅ 2 gx L

[Using v 2 = 2 gx ]

As the time interval is infinitesimally small, this expression can be regarded as expression for instantaneous value of thrust force.

Centre of Mass

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A girl can exert an average force of 200 N on her machine gun. Her gun fires 20-g bullets at 1000 m/s . How many bullets can she fire per minute? A flat car of mass m0 starts moving to right due to a constant horizontal force F (Fig.) Sand spills on the flatcar from a stationary hopper. The velocity of loading is constant and equal to µ kg/s. Find the time dependence of the velocity and the acceleration of the flatcar in the process of lading. The frictions negligible small.

3.

A rocket ejects a steady jet whose velocity is equal to u relative to the rocket. The gas discharge rate equals µ kg/s. Demonstrate that the rocket motion equation in this case takes the form. mw =F – µu. Where m is the mass of the rocket at a given moment, w is its acceleration, and F is the external force.

4.

A rocket moves in the absence of external forces by ejecting a steady jet with velocity u constant relative to the rocket. Find the velocity v of the rocket at tthe moment when its mass is equal to m, if at the initial moment it possessed the mass m0 and its velocity was equal to zero. Make use of the formula given in the foregoing problem.

5.

Find the law according to which the mass of the rocket varies with time, when the rocket moves with a constant acceleration w, the external forces are absent, the gas escapes with a constant velocity u relative to the rocket, and its mass at the initial moment equals m0.

6.

A cylindrical solid of mass 10–2 kg and cross -sessional area 10–4 m2 is moving parallel to its axis (the xaxis) with a uniform speed of 103 m/s in the positive direction. At t = 0, its front face passes the plane x = 0. The region to the right of this plane is filled with stationary dimention of the cylinder remains practically unchanged and that the dust particle of uniform density 10–3 kg/m3. When a dust particles collides with the face of the cylinder, it sticks to its surface. Assuming that the dimensions of the cylinder remains practically unchanged and that the dust sticks only to the front face of the x-co ordinate of the front of the cylinder find the x-coordinate of the front of the cylinder at t =150 s.

7.

A chain of length L weighing l per unit length begins to fall with constant acceleration through a hole in the ceiling. (a) When the lowest end of the chain has fallen through a distance x determine its velocity v(x). (b) Find the constant acceleration of the falling chain. (c) Find the energy loss when the last link of the chain has left the ceiling.

8.

The end of a chain of length L and mass ρ per unit length that is piled on a platform is lifted vertically with a constant velocity v by a variable force P. find P as a function of the height x of the end above the platform. Also find the energy lost during the lifting of the chain. P

x

(Q.7)

(Q.8) x

Centre of Mass

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COLLISIONS What is a collision? We learn much about atomic, nuclear, and elementary particles experimentally by observing collisions between them. On a larger scale we can better interpret such things as the properties of gases in terms of particle collisions. In this chapter we apply the principles of conservation of energy and conservation of momentum to the collisions of particles. In a collision a relatively large force acts on each colliding particle for a relatively short time. The basic idea of a collision is that the motion of the colliding particles (or of at least one of them) changes rather abruptly and that we can make a relatively clean separation of times that are “before the collision” and those that are “after the collision”. When a bat strikes a baseball for example, the begimning and the end of the collision can be determined fairly precisely. The bat is in contact with the ball for an interval that is quite short in comparison to the time interval for which we are watching the ball. During the collision the bat exerts a large force on the ball, as shown in figure 6.48. This force varies with time in a complex way that we can measure only with difficulty. Both the ball and the bat are deformed during the collision. Forces that act for a time that is short compared to the time of observation of the system but bring noticeable change in the momentum of the part of the system on fig. 6.48

which they act are called impulsive forces. Consider now a collision between two particles, such as those of masses m1 and m2 , as shown in figure 6.49. During the short time of collision these particles exert large forces on one another. At any $ $ instant F1 is the force exerted on particle 1 by particles 2 and F2 is the force exerted on particle 2 by particle 1. By Newton’s third law these forces at any instant are equal in magnitude but are $ $ oppositely directed, i.e., F1 = − F2 . The change in momentum of particle 1 resulting from the collision is $ ∆p1 =

tf

$

m1

m2

F1

F2

fig. 6.49

$

∫ F1 ⋅ dt = F1,av ⋅ ∆t

ti

where the time interval of collision, ∆t = t f − ti . The change in momentum of particle 2 resulting from the collision is

$ ∆p2 =

tf

$

$

∫ F2 ⋅ dt = F2,av ⋅ ∆t ti

$ $ If no other forces act on the particles during collision, the ∆p1 and ∆p2 give the total change in momentum for $ $ $ $ each particle. But have seen that at each instant F1 = − F2 , so that F1,av = − F2,av , and therefore

$ $ ∆p1 = -∆p2 If we consider the two particles as an isolated system, the total momentum of the system is $ $ $ p = p1 + p2 , and the total change in momentum of the system as a result of the collision is zero, that is, Centre of Mass

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$ $ $ $ ∆p = ∆p1 + ∆p2 = 0

Hence, if there are no external forces the total momentum of the system is not changed by the collision. The impulsive forces acting during the collision are internal forces which have no effect on the total momentum of the system. We have defined a collision as an interaction which occurs in a time interval ∆t that is negligible compared to the time during which we are observing the system. We can also characterize a collision as an event in which external forces that may act on the system are negligible compared to the impulsive collision forces. When a bat strikes a ball, a golf club strikes a golf ball, or one billiard ball strikes another, external forces do act on the system. Gravity and friction exert forces on these bodies, for example; these external forces may not be the same on each colliding body nor are F(t) they necessarily cancelled by other external forces. Even then it is Fimp quite safe to neglect these external forces during the collision and to apply conservation of linear momentum provided that impulse of force arising due to the the external forces are negligible compared to the impulsive collision forces of collision. impulse of nonimpulsive As a result the change in momentum of a particle during a collision external forces arising from an external force is negligible compared to the change in momentum of that particle arising from the impulsive collisional force, as shown in figure 6.50.

Fext

For example, when a bat strikes a ball, the collision lasts only a t small fraction of a second. Since the change in momentum is large 0 and the time duration is small, it follows from fig. 6.50 $ $ ∆p = ∆Fav ⋅ ∆t $ that the average impulsive force Fav is relatively large. Compared to this force, the external force of gravity is negligible. During the collision we can safely ignore this external force in determining the change in motion of the ball; the shorter the duration of collision the more likely this is to be true. In practice, therefore, we can apply the principle of momentum conservation during collisions if the time of collision is small enough. We can then say that the momentum of a system of particles just before the particles collide is equal to the momentum of the system just after the particles collide. Later in this chapter we will see that in some cases of collision external forces of relatively high magnitude act on the system during the collision or we can say external forces acting during collision are also impulsive. In such cases conservation of linear momentum can not be applied on the system. SYSTEM BOUNDARY

Consider the example 15 once again. As the horizontal surface, on which the block is placed, is smooth no frictional force act on the system “bullet + block” during the collision. Force applied by the bullet on the block and that applied by the block on the bullet are internal forces.

!

(direction of motion)

!

F1

!

F2 fk

!

F1 = – F2 fk =µm2 g m2

External force on the system along the horizontal direction. fig. 6.51

Therefore, there is no net external force on the system along the horizontal direction. Hence, we used conservation of linear momentum along the horizontal direction and obtained the desired result. Now, suppose that the horizontal surface is made rough. Is conservation of linear momentum applicable now? The system “bullet + block” is shown $ $ in figure 6.51. at an instant during the collision. F2 is the force applied on the block by the bullet due to collision. F1 $ $ is the reaction force of F2 ⋅ f k is the frictional force acting on the block from the horizontal surface. If the bullet Centre of Mass

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strikes the block at t = t1 and the system starts moving with common velocity or we can say collision ends at t = t2 , then, for the time interval ∆t = t2 − t1 , the change in momentum of the bullet t2 $ $ ∆p1 = ∫ F1 ⋅ dt t1

and the change in momentum of the block is t2 $ $ $ ∆p2 = ∫ ( F2 + f k ) ⋅ dt t1

$ $ The magnitude of f k is µ m2 g but F2 has a relatively very high magnitude because it is arising due to high speed $ $ $ impact, therefore, the sum of f k and F2 would be the nearly same as F2 itself, as suggested figure 6.52. Therefore, the change in momentum of the block can be approximated as t2 $ $ ∆p2 = ∫ F2 ⋅ dt t1

Therefore, the change in the total momentum of the system is $ $ $ ∆Psys = ∆p1 + ∆p2

! b

$ $ = ∫ F1 ⋅ dt + ∫ F2 ⋅ dt t2

t2

t1

t1

! a ! ! If a >>>b, then, a + b can ! be approximated by a

! ! a+b

fig. 6.52

$ $ = ∫ ( F1 + F2 ) ⋅ dt t2 t1

$ =0

$ $ $ [∵ F1 + F2 = 0]

Hence, we can assume that linear momentum of the system to be conserved in this case also. But you should not forget that it is possible only because the friction force has relatively very less magnitude as compared to the forces arising due to the collision. Here, you must notice that if we continue the observation of the system after the collision for a duration of length much larger than that of the collision, then, the change in momentum caused by the frictional force (or we can also say the impulse of the frictional force) can not be neglected. Generally, the same approach is followed for the most of the cases of collision. External forces are present during collision but the impulse contributed by them during the interval of the collision is negligible as compared to that contributed by the forces arising due to the collision. Hence, we forget them for the duration of the collision. To understand the collision in a more comprehensive way let us first analyze the situation shown in figure 6.53. Two small balls having masses m1 are moving at speeds u1 and u2, respectively, along the same direction. An ideal spring is connected to m2 as shown in the same figure. In this case m1 is moving with a speed greater than that of m2 and hence m1 is approaching to wards m2 at a relative speed of u1– u2 Centre of Mass

u1 > u2 u1

u2

m1

m2 smooth fig. 6.53

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At a certain moment m1 touches the spring connected with m2 and starts compressing it owing to its higher speed, as shown in figure 6.54 (a). As the compression in the spring increases, spring force on the each ball also u2

u1

u1> u2; compression = 0.

(a)

m1

m2 u2+

u1–

(b)

FSP

m1

m2

FSP

FSP

u2+++ =v

u1– – =v

(c)

u–1 > u+2 ; compression ≠ 0 and is increasing

m1

m2

FSP

FSP v+

v–

(d)

u–1 – = u+2 + = v(say); vrel = 0; compression is maximum.

m1

m2 v1

v – < v +; compression ≠ 0 and is decreasing

FSP v2

v1> v2; compression = 0.

(e)

m1

m2

fig. 6.54

increases, as shown in figure 6.54(b). Spring force on m1 decreases its speed and that on m2 increases its speed. Hence, the rate at which m1 is approaching m2 is decreasing, but compression in the spring keeps on increasing as long as the speed of m1 is higher than that of m2 . Then comes a moment when the speed of m1 becomes equal to that of m2 or we can say the relative speed of m1 and m2 becomes zero, as shown in figure 6.54(c). Now, m1 can not cause more compression in the spring and hence, at this moment compression in the spring is maximum. As the compression in the spring is maximum at this instant, the potential energy stored in it is maximum and the system “ m1 + m2 + spring ” has minimum kinetic energy at this moment. In figure 6.54(c), m1 and m2 have common velocity but spring forces are still acting on them. And due to maximum compression state of the spring, the spring force on each ball has maximum magnitude at this moment. Due to this reason the speed of the ball 1 suffers further decrease and that of the ball 2 suffers further increase in its magnitude, as shown in figure 6.54(d). Now, m2 is separating from m1 because it has a higher speed and hence, compression in the spring decreases thereafter. As the compression decreases, magnitude of the spring force on each ball also decreases but it continues to retard the motion of m1 and to accelerate the motion of m2 until the spring has regained its natural length. Figure 6.54(e) shows the system when the spring has regained its natural length. The speed of ball the ball m2 , v2 , is higher than that of m1 , v1 , and hence, m2 is separating from m1 at a rate of v2 − v1. A more detailed analysis of the figure 6.54(d) would led you to the fact that it is possible that when the spring regains its natural length, m1 has a velocity along the direction opposite to the direction of its initial velocity, as shown in figure 6.55. In such a case the speed of separation of m1 and m2 is v1 + v2 . Centre of Mass

v2

v1 m1

m2

fig. 6.55

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Here we see that there is no external force acting on the system “ m1 + m2 + spring ” along the horizontal direction. Therefore, linear momentum of the system is conserved along the horizontal direction and the centre of mass of the system moves with constant velocity during the course of collision. Therefore, we can write, = Psystem, at some intermediate instant Psystem,initially

( m1u1 + m2 )

(m1u1− + m2 m2+ ) = Psystem,when compression is maximum (m1v + m2v) = Psystem, finally ( m1v1 + m2 v2 )

In this case normal contact forces acting on the system from the horizontal surface balance the weight of the system and hence the momentum of the system is conserved along the vertical direction too. As the work done by nonconservative forces (normal contact forces from the horizontal surface in this case) is zero, mechanical energy of the system is also conserved. First some part of the kinetic energy is converted into potential energy and , then, it is again converted back into kinetic energy. The case we just discussed is very similar to the instances of collisions we have already discussed (figures 6.48, 6.49, 6.51, 6.53) and the instances we will discuss later in this section only. Here the spring is getting deformed while in the real collisions, colliding bodies get deformed. Here, the spring is an ideal one, therefore, there is no loss of energy in the form of left deformation in the spring while in the cases involving real collisions, bodies may or may not regain there actual shape after the collision. Bodies which are able to regain their original shape are called elastic bodies and bodies which overcome the deformation only partially or we can say they are not able to regain their original shape are called inelastic bodies. Collisions involving elastic bodies are defined as elastic collisions. In an elastic collision there is no loss of kinetic energy of the system. The kinetic energy of the system just after the collision is equal to the kinetic energy of the system just before the collision. Collisions involving at least one inelastic body are defined as inelastic collisions. In an inelastic collision, as the colliding bodies do not regain their actual shape after the collision, a part of initial kinetic energy is lost in the form of deformations in the colliding bodies. Therefore, kinetic energy of the system of the colliding bodies just after an inelastic collision is smaller than its value just before the collision. Materials which are incapable of regaining their actual shape, even partially, once they are deformed are called plastics. When bodies involved in a collision are made up of plastic materials, such a collision is called plastic or completely inelastic collision. In a completely inelastic collision, once the colliding bodies get maximum deformation they continue to move in this state only. Alternatively, we can also say that during the collision bodies stick together and they move together with a common velocity even after the collision. In this case duration of collision is spread from the momentum when the bodies touch each other to the moment when deformation in each body has become maximum. As the bodies do not separate from each other after the collision, velocity of separation after the collision is zero in plastic collisions. It is also obvious that the loss of kinetic energy due to the collision is maximum in such collisions. ONE DIMENSIONAL (OR HEAD ON) COLLISION: Consider first an elastic one-dimensional collision. We can imagine two smooth nonrotating spheres moving initially along the line joining their centres, then colliding head-on and moving along the same straight line without rotating after the collision, as shown in figure 6.56. These bodies exert forces on each other during the collision that are along the initial line of motion, therefore, u2

m2

m1

just before collision vapproach = u2 – u1

u1

N

N During collision

m2

v2

m1

v1

just after collision vseparation = v1 – v2

fig. 6.56

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the final motion is also along this same line. We take the positive direction of the momentum and the velocity to be to the right, then, from conservation of linear momentum we have Psys , in = Psys ,

⇒

fin

m1u1 + m2 u2 = m1v1 + m2v2

Because the collision is elastic, kinetic energy is conserved and we have K .E.in = K .E. fin

⇒

1 1 1 1 m1 u12 + m2 u22 = m1v12 + m2 v22 2 2 2 2

It is clear at once that if we know the masses and initial velocities, we can calculate the two final velocities from these two equations. The momentum equation can be written as

m1 (u1 − v1 ) = m2 (v2 − u2 )

...(6.37a)

and the energy equation can be written as m1 (u12 − v12 ) = m2 (v22 − u22 )

...(6.37b)

Dividing equation (6.37 b) by equation (6.37a) and assuming v2 ≠ u2 and v1 ≠ u1 , we obtain

v1 + u1 = v2 + u2 ⇒

v1 – v2 = u2 – u1

...(6.38)

⇒ velocity of separation = velocity of approach This tells us that in an elastic one-dimensional collision, the relative velocity of approach before the collision is equal to the relative velocity of separation after the collision. Solving equations (6.37a) and (6.37b), we find that v1 =

and

m1 − m2 2m 2 u1 + u2 m1 + m2 m1 + m2

 2m1   m − m1  v2 =  u1 +  2   u2 m + m m + m  1  1 2  2 

There are several cases of special interest. For example, when the colliding particles have same mass, we have,

v1 = u2 and v2 = u1 That is, in a one-dimensional elastic collision of two particles of equal masses, the particles simply exchange velocities during the collision, as shown in figure 6.57.

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u2 m

74

v2= u1

u1

A

B

m

m A

m

just before collision

(b)

B

just after collision

u m A

A

m

m B

REST

u m

A

REST

m

B

just after collision

REST

m

u

m B

REST

just before collision

(c)

v1= u2

C

u

REST

m

A

before collision

m

B

REST

m

C

after collision between A and B

REST

m

A

u

REST

m

B

m

C

after collision between B and C

fig. 6.57

If m1 is much greater than m2 and u1 = 0, we obtain v1 =

2m2 u2 % 0 m1 + m2

and

v2 =

m2 − m1 − m1 u2 % u2 m1 + m2 m2

⇒

v2 % −u2

[∵ m1 »m2 ]

That is, when a light particle collides with a massive particle at rest, the velocity of the light particle is approximately reversed and the massive particle remains at rest. For example, suppose that we drop a ball vertically onto a horizontal surface attached to the earth. This is in effect a collision between the ball and the earth. If the collision is elastic, the ball will rebound with a reversed velocity and will reach the same height from which it fell. If a collision is inelastic then, by definition, the kinetic is not conserved. The final kinetic energy may be less than the initial value, the difference being ultimately converted to heat or to potential energy of deformation in the collision but the conservation of momentum still holds (if there is an external force, its impulse for the duration of collision must be negligible for this to be true), as does the conservation of total energy. Let us consider finally a completely inelastic collision. The two particles stick together after the collision, so that there will be a final common velocity v$. It is not necessary to restrict the discussion to one-dimensional motion. Using only the conservation of momentum principle, we find $ $ $ ...(6.39) m1u1 + m2 u2 = ( m1 + m2 )v $ $ This gives v$ when u1 and u2 are known. In such a case velocity of separation after the collision is zero.

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The ballistic Pendulum. The ballistic pendulum is used to measure bullet speeds. The pendulum is a large wooden block of mass M hanging vertically by two cords. A bullet of mass m, traveling with a horizontal speed vi , strikes the pendulum and remains embedded in it. If the collision time (the time required for the bullet to come to rest with respect to the block) is very small compared to the time of swing of the pendulum, the supporting cords remain approximately vertical during the collision. Therefore, no external horizontal force acts on the system (bullet + pendulum) during collision, and the horizontal component of momentum is conserved. The speed of the system after collision v f can be easily determined, so that the original speed of the bullet can be calculated from momentum conservation.

M m y vi fig. 6.58

The initial momentum of the system is that of the bullet mvi , and the momentum of the system just after the collision is ( m + M )v f , so that mvi = ( m + M )v f .

...(i)

After the collision is over, the pendulum and bullet swing up to a maximum height y, where the kinetic energy left after impact is converted into gravitational potential energy. Then, using the conservation of mechanical energy for this part of the motion, we obtain 1 ( m + M )v 2 f 2

= (m + M ) gy.

...(ii)

Solving these two equations for vi , we obtain vi =

m+M m

2 gy .

Hence, we can find the initial sped of the bullet by measuring m, M, and y. The kinetic energy of the bullet initially is 12 mvi2 and the kinetic energy of the system (bullet + pendulum) just after collision is 12 (m + M )v 2f . The ratio is 1 ( m + M )v 2 f 2 1 mv 2 i 2

=

m . m+M

For example, if the bullet has a mass m = 5 gm and the block has a mass M = 2000 gm, only about one-fourth of 1% of the original kinetic energy remains; over 99% is converted to other forms of energy, such as heat. Centre of Mass

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The velocity of the centre of mass of two particles is not changed by their collision, as the collision does not change the total momentum of the system of two particles, it changes only the distribution of momentum between the two $ particles. The momentum of the system can be written as P = (m1 + m2 )vcm . If no external forces act on the system, $ then P is constant before and after the collision, and the centre of mass moves with uniform velocity throughout. $ $ If we$choose a reference frame attached to the centre of mass, then in this center-of-mass reference frame, vcm = 0 and P = 0 . There is a great simplicity and symmetry in describing collisions with respect to the center of mass, and it is customary to do so in nuclear physics. For whether collisions are elastic or inelastic, momentum is conserved, and in the centre of mass reference frame the total momentum is zero. These results hold in two and three dimensions as well as in one because momentum is a vector quantity.

A very large mass m1 moving with speed u1 collides elastically with a very small mass m2 initially at rest (Figure 6.59 ). What is the speed of the small mass after the collision assuming that m1 is much greater than m2 ?

m1

m2 u2= 0

m1

u1

v1 u1 Before collision

m2 v2

2u1

After collision fig. 6.59.

Solution: It should be intuitively clear that the large mass will not be affected very much by a very small mass. A cannonball will hardly be slowed down if it collides with a stationary beach ball. Before the collision, the relative velocity of approach is u1 . Then after the collision the velocity of separation must be u1 . For a first approximation, we neglect any change in the velocity of m1 . Since it continues to move with velocity u1 , the velocity of the small mass m2 must be 2u1

A baseball weighing 0.35 kg is struck by a bat while it is in horizontal flight with a speed of 90 m/sec. After leaving the bat the ball travels with a speed of 110 m/sec in a direction opposite to its original motion. Determine the impulse of the collision Solution:

$ $ We cannot calculate the impulse from the definition J = ∫ F dt because we do not know the force exerted on the ball as a function of time. However, we have seen that the change in momentum of a particle acted on by an impulsive force is equal to the impulse. Hence $ $ $ J = change in momentum = p f − pi $ $ $ $ = mv f − mvi = m(v f − vi ). $ Assuming arbitrarily that the direction of vi is positive, the impulse is then $ J = 0.35 ( −110 − 90 )

= –70 N-sec. The minus sign shows that the direction of the impulse acting on the ball is opposite to that of the original velocity of the ball. Centre of Mass

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We cannot determine the force of the collision from the data we are given. Actually, any force whose impulse is –70 N-sec will produce the same change in momentum. For example, if the bat and ball were in contact for 0.0010 sec, the average force during this time would be $ ∆p −70 N-sec F= = = − − 70, 000 N. ∆t 0.0010 sec

For a shorter contact time the average force would be greater. The actual force would have a maximum value greater than this average value. COEFFICIENT OF RESTITUTION: In general, a collision is somewhere between the extreme cases of perfectly elastic, in which case the relative velocities are reversed, and perfectly inelastic, in which case there is no relative velocity after the collision. The coefficient of restitution e is defined as the ratio of the relative velocity of separation and the relative velocity of approach: ⇒ velocity of separation = e velocity of approach ...(6.40) For a perfectly elastic collision e = 1; for a perfectly inelastic collision e = 0. The coefficient of restitution thus measures the elasticity of the collision. In solving collision problems it is often easiest to use conservation of linear momentum principle and equation (6.40) to find the final velocities, thus avoiding the quadratic terms in the conservation of energy equation.

A sphere of mass m moving with a constant velocity u hits another stationary sphere of the same mass. If e be the coefficient of restitution, find the ratio of velocities of the two spheres after the collision. Solution: Let v1 and v2 be the velocities of the spheres after the collision and the direction of initial motion of the sphere A be the positive direction of motion. just before collision u m

A

just after collision v1

B

m

m

A

v2 m

B

fig. 6.60

As there is no external impulsive force acting on the system “sphere A + sphere B”, applying conservation of linear momentum, we have Psys , before = Psys , after collision

⇒

collision

mu = mv1 + mv2

⇒

[from figure 6. ..]

u = v1 + v2

...(i)

Using equation (6.40), we have  Velocity of separation   Velocity of approach   after collision  = e.  before collision      ⇒ Centre of Mass

v2 − v1 = e ⋅ u

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Solving equation (i) & (ii), we get

and

Therefore,

v1 = (1 − e)

u 2

v2 = (1 + e)

u 2

v1 (1 – e ) = v2 (1 + e )

A particle of mass m1 moves with speed u and collides head-on with a stationary particle of mass m2 . After the collision, the velocities of the particles are v1 and v2 . Prove that the condition for v1 to be positive, i.e., for the first particle to continue moving in the same direction, is m1 m2 > e , where e is the coefficient of restitution. Solution: The situation just before the collision and the situation just after the collision are shown in figure 6. 60. Applying conservation of linear momentum, we get Psys , in = Psys ,

fin

⇒

mAu = mAv1 + mB v2

⇒

m1u = m1v1 + m2v2

...(i)

[mA = m1; mB = m2 ]

Using equation (6.40), we get velocity of separation = e. velocity of approach ⇒

v2 − v1 = e.u

...(ii)

Solving equations (i) and (ii), we get  m2  m1  v1 =  − eu   m1 + m2  m2  For the first particle to continue moving along the direction of its initial velocity

v1 > 0 ⇒

m1 −e > 0 m2

⇒

m1 >e. m2

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A ball is dropped on a floor from a height h. If coefficient of restitution is e, find: (a) the height to which the ball will rise (b) the time it will take to come to rest again (c) the magnitude of the impulse given to the ball by the floor. Solution: The ball reaches the floor with speed u = 2 gh . Just after the collision, let its upward speed by v. Just before the collision, during the collision and just after the collision, the ball is shown in figure 6.61. As the coefficient Just before collision

During collision N

u = 2gh

Just after collision v +ve direction

mg N fig. 6.61

of restitution is e, we have velocity of separation = e. velocity of approach ⇒ v = e.u [There is no benefit of applying conservation of linear momentum to the system “ball + earth” why ?] [In an elastic collision e = 1, hence, speed after the collision is same as the speed before collision.] (a) The height to which the ball will rise after the impact is h1 =

v 2 e2u 2 e2 ⋅ 2 gh = = 2g 2g 2g

⇒

h′ = e 2 ⋅ h [In an elastic collision e = 1, hence, the ball would rise to the same height from which it was dropped]

(b)

Consider a situation when the ball is resting on the floor, as N(=mg) shown in the figure 6.61. In such a situation the magnitude REST of the normal contact force from the floor is equal to the A ball placed on a horizontal weight of the ball. But when the ball hits the floor, then, the surface. normal contact force arising due to the impact has a very mg high magnitude. It must not be confused with the normal fig. 6.62 contact force shown in figure 6.62. Therefore, for the duration of impact impulse of gravity can be considered negligible with respect to the impulse of the normal contact force from the floor. This could also be explained by saying that the normal contact force arising due to the impact is impulsive. Hence,

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impulse given to the impulse of the normal = ball by the floor, J contact force, N % change in the momentum of the ball (neglecting the impulse of the gravity for the duration of the collision) = p f − pi = + mv − (−mu ) = emu + mu [Using v = e.u ) = mu(1 + e ) [In an elastic collision e = 1, hence, the magnitude of the impulse given by the floor is 2mu.] (c)

When a ball is dropped from a height h, then its time of fall is 2h g

t1 =

When a ball is thrown vertically upwards with speed v, then its time of rise is

t2 = v g Therefore, the time taken by the given ball to come to rest again is

T = t1 + t2 =

2h v 2h eu + = + g g g g

[Using v = eu ]

=

2h e ⋅ 2 gh + g g

 Using u = 2 gh 

=

2h 2h + e⋅ g g

=

2h ⋅ (1 + e) g

ALTERNATE METHOD: Let the ball comes to rest after a time T, then, for this duration,

change in the momentum = impulse given by the gravity of the ball + impulse given by the floor ⇒ ⇒ ⇒ ⇒ Centre of Mass

p f − pi = − mg ⋅ T + mu (1 + e)

 Upward direction is chosen as positive direction  and the result obtained in part (b) is used   

0 − 0 = −mgT + m ⋅ 2 gh ⋅ (1 + e) gT = 2 gh (1 + e) T=

2h ⋅ (1 + e) g

 Impulse of gravity is negligible for   the duration of impact but it cannot   be neglected for the duration of fall   or rise.  Web: http://www.locuseducation.org

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An elevator platform is going up at a speed 20 m/s and during its upward motion a small ball of mass 50 gm falling in downward direction at a speed of 5 m/s. Find the speed with which the ball rebounds after an elastic collision with the platform. Solution: As the mass of the platform is very large compared to that of the ball, we can assume that the speed of the platform remains unchanged. If v be the velocity of the ball just after the impact, as show in figure 6.63. v 5 m/s 20 m/s

20 m/s Just before the collision

Just after the collision fig. 6.63

Using equation (6.40), we have ⇒

velocity of separation = e.velocity of approach velocity of separation = velocity of approach

⇒

v − 20 = 20 + 5

⇒

v = 45 m/s

[ ∵ e = 1]

A neutron moving at a speed u undergoes a head-on elastic collision with a nucleus of mass number A at rest. Find the ratio of kinetic energies of neutron after and before collision. Solution: If mbe the mass of neutron then mass of the nuclei will be mA. Let v1 be the speed of the neutron and v2 be that of the nuclei after the collision as shown in figure 6.64. v1

u m

mA

m

Before collision

v2 mA

After collision fig. 6.64

Applying conservation of linear momentum, we have Pin = Pfin

⇒ ⇒

mu = mv1 + mAv2 u = v1 + Av2

...(i)

Using equation (6.40), we have vsep. = vapproach

⇒ Centre of Mass

v2 − v1 = u

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Solving equations (i) and (ii) , we get  1− A  v1 =  u  1+ A  Thus KE of neutron after collision is 1 2 mv1 2

kf =

=

1  1− A  mu 2   2  1+ A 

2

The KE of electron before collision was ki =

1 mu 2 2

Therefore, required ratio is

 1− A  =  ki  1 + A 

kf

2

OBLIQUE COLLISION: collision in two or three dimensions: When colliding bodies exert forces on each other which are not along the line parallel to the direction of their initial motion, directions of motion of bodies change due to the collision. Such type of collisions are called oblique collisions. Consider the situations shown in figures 6.65. In figure 6.65(a) two smooth spherical balls A and B, initially ! v1

!

v2

!

A B

!

J= •

A

!

!

P1, f =P1, j+J1

!

•

dt

!

N

–N

!

! N

P1, i

•

A fig. 6.65(b)

B

!

u1

A

!

u2

m1

!

P2, i

!

!

P2, f =P2, j m2

B

!

•

!

J = –• N • dt fig. 6.65(a)

fig. 6.65(c)

moving with speeds u1 and u2 , collide obliquely and then move with speed v1 and v2 , respectively, along directions different from the directions of their initial motion. During collision (i.e., when the spheres are in contact) normal Centre of Mass

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contact force on each sphere act perpendicularly away from the surface of contact, as shown in figure 6. , and due to the spherical geometry of each body, normal contact force on each body passes through its centre. Normal contact force on each body imparts impulse to the body along the direction of its action. If impulse on a body is along the line parallel to the direction of initial momentum of the body then final momentum is also along the same line but if the impulse is not along that line, as in this case, the final momentum of the body is along a different direction. In figures 6. 65(b) and 6.65 (c) final momenta of the bodies A and B, respectively, are obtained by adding impulse on each body to the initial momentum of the body. ntial tange s n o m Com e, called a t surfac e of contac surfac

Common tangential surface, called as surface of contact

N

N

C1

C2

fig. 6.66(a)

fig. 6.66(b)

A similar case is illustrated in figure 6.67. The ball A strikes the horizontal surface obliquely and receives v A

B u

!

N

!

•

!

J =• N • dt

!

Pf Fixed horizontal surface (smooth)

A

A

!

Pi

!

–N

fig. 6.67(b)

fig. 6.67(a)

impulse along the direction perpendicularly away from the surface which changes the direction of the motion of the ball as shown in figures 6.67(a) and 6.67(b). Now, let us discuss the oblique collision in a different way which makes the solution of problems involving such collisions very simple. The situation shown in figure 6.65 is analyzed again, as shown in figure 6.68 Situations of balls A and B just before the collision (figures 6.68(a), (b), (c)), during the collision (figure 6.68(d)) and just after the collision

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u1

u1 sin

u1 u2

θ1 u2 sin

u2 u1 cos

C1 θ1

C1

θ2 C2

C2 (a)

(b) u1sin

θ1

θ1

u2 cos

θ2

θ2

(c) u1 sin θ1

θ1

u2 sin

θ2

u2 sinθ

2

v1

N u1 cos

84

N

u2 cos

(d)

v2 θ2

(e) fig. 6.68

(figure6.68 (e)) are shown. This time we are considering the motion of the balls along the line joining their centres C1 C2 , and the line perpendicular to C1C2 . During the collision impulsive normal contact force on each ball acts along C1C2 and hence the components of velocities of the balls along this direction undergo a change while the components perpendicular to this line remain unchanged even after the collision (for this to be happen the balls must be smooth). If there is no net external impulsive force acting on the system of colliding bodies, then, conservation of linear momentum is also applicable to this case. But equation (6.36) (i.e., velocity of sep. = e.velocity. of app.) is applicable along the line joining their centres, which is also called line of impact (because during the collision they exert impulsive normal contact force on each other along this line only). It is obvious from 6.68(c) that the components of velocities of the balls perpendicular to the line C1C2 have no role in bringing them together, only the components parallel to the line C1C2 are responsible for the collision.

Consider the situation shown in figure 6.69. A small ball of mass m hits the fixed horizontal surface with speed u at an angle of θ with the vertical direction. If friction between the horizontal surface and the ball be negligible and the coefficient of restitution be e, find the magnitude and the direction of the velocity of the ball just after the collision.

Centre of Mass

m u θ

smooth, e

fig. 6.69

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Solution: In figure 6.70 the ball is shown just before the collision during the collision and just after the collision. During collision

Just before collision u sinθ θ

N(impulsive)

u u cosθ fig. 6.70(b)

fig. 6.70(a)

Just after collision v´= e u cosθ

eu cosθ

v u

u sinθ

θ α

u sinθ fig. 6.70(c)

fig. 6.70(d)

From figure it is clear that the ball approaches the horizontal surface due to the vertical component of its velocity u cos θ . As during the collision the impulsive normal contact force on the ball from the horizontal surface acts along vertical direction, as shown in figure 6.70(b), the horizontal component of the ball’s velocity just after the collision is same as it was just before the collision. We have assumed that just after the collision, the vertical component of the ball’s velocity is v ', directed in the upward direction. Using equation (6.36), we have velocity of separation = e.velocity of approach ⇒

v ' = e.u cos θ Therefore, the speed of the ball just after the collision is v = (v ') 2 + (u sin θ ) 2

[see figure 6.70(c)]

= (e.u cos θ )2 + (u sin θ )2 = u sin 2 θ + e 2 .cos 2 θ

...(i)

If α be the angle made by the ball’s velocity with the vertical direction, just after the collision, from figure 6.(c) we have tan α = =

Centre of Mass

u sin θ e.u cos θ tan θ e

...(ii)

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NOTE: • In an elastic collision, e = 1. ∴

•

v = u sin 2 θ + cos 2 θ

⇒ v=u ⇒ speed just after collision = speed just before collision In an elastic collision, e = 1. tan α =

∴

•

[Using (i)]

tan θ e

[Using (ii)]

= tan θ ⇒ α=θ ⇒ angle of reflection = angle of incidence. In an inelastic collision, e < 1. tan θ e

∴

tan α =

⇒

tan α 1 = tan θ e

⇒

tan α >1 tan θ

⇒

tan α > tan θ

⇒

α >θ angle of reflection > angle of incidence

⇒

1   as e < 1, e > 1

Suppose in the last example, the horizontal surface is not smooth, as shown in figure 6.71. If the coefficient of friction between the ball and the horizontal surface be µ, find the velocity of the ball just after its collision with the horizontal surface.

m u θ

µ, e

fig. 6.71

Solution: Conditions just before the collision, during the ollision and just after the collision are shown in figures 6.72(a), 6.72(b) and 6.72(c), respectively. Just before collision u sinθ θ

u u cosθ fig. 6.72(a)

During collision +ve Y N +ve X µN fig. 6.72(b)

Just after collision v2 v = v12 + v22 α v1

fig. 6.72(c)

Here, the major difference from the previous example is that the horizontal surface is rough due to which it exerts a Centre of Mass

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87

kinetic frictional force, denoted by µN in figure 6.72(b), on the ball during the collision because the velocity of the ball has a component along the horizontal direction which causes slipping between the ball and the horizontal surface. As the normal contact on the ball from the horizontal surface during the impact, N, is impulsive, as discussed in the last example, the kinetic frictional force from the same surface, µN, is also impulsive. Hence, in this case the momentum of the ball along the horizontal direction during the collision can not remain conserved. Therefore, let us assume that just after the impact the horizontal component of the ball’s velocity has changed from u sin θ to v1 and the vertical component has changed from u cos θ in the downward direction to the v2 in the upward direction, as shown in figure 6.72(c). Using equation 6.36, we get

speed of separation just e.speed of approach just = after the collision before the collision ⇒

...(i) v2 = e.u cos θ Using impulse equation along y direction for the duration of the collision, we have tf

∫

N .dt = p y , f − p y ,i

ti

 Neglecting the impulse of gravity with   respect to the impulse of N for the  duration of the collision   

tf

⇒

∫ N .dt = +mv2 − (−mu cos θ )

ti tf

⇒

∫ N .dt = emu cosθ + mu cosθ

[Using (i)]

ti tf

⇒

∫ N .dt = mu cos θ (1 + e)

...(ii)

ti

Using impulse equation along x direction for the duration of collision, we have tf

− ∫ µ N .dt = px, f − px,i ti

⇒

− µ [mu cos θ (1 + e) ] = mv1 − mu sin θ

⇒

v1 = u sin θ − µ u cos θ (1 + e)

⇒

v1 = u [sin θ − µ cos θ (1 + e) ]

[Using (ii)]

...(iii)

Now, using (i) and (iii) speed of the ball just after the collision, v, and the angle made by the velocity of the ball with the vertical direction just after the collision, α, can be found.

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PHYSICS

LOCUS

88

Figure 6. shows the results of a collision of two objects of unequal mass. (a) Find the speed v2 of the larger mass after the collision and the angle θ2 . (b) Show that this collision is perfectly elastic. After Before m

5v0

3v0

θ1

tan θ1= 2

m

2m

2m

θ2 v2

fig. 6.73

Solution: (a) Let us select +ve X direction along the direction of the initial motion of the smaller mass and +ve Y direction along a direction perpendicular to X direction in the plane of motion of balls, as shown in figure 6.74 . Y v1= 5v0 X

θ1 m 2m θ2 v2

fig. 6.74

Considering the two balls as a single system, we can apply the conservation of linear momentum along both X and Y direction directions because there is no external impulsive force is acting on the system during the collision. Applying conservation of linear momentum along X direction we get, psys , X , fin = psys , X ,in z

⇒ ⇒ ⇒

m ⋅ v1 cos θ1 + 2m ⋅ v2 cos θ 2 = m ⋅ 3v0 5v0 ⋅

1 + 2v2 cos θ 2 = 3v0 5

v2 cos θ 2 = v0

 ∵ tan θ1 = 2    ∴ cos θ1 = 1 5 

...(i)

Applying the same along Y direction, we get, Psys ,Y , fin = Psys ,Y ,in

⇒ ⇒ ⇒

+ m ⋅ v1 sin θ1 − 2m ⋅ v2 sin θ 2 = 0 5v0 ⋅

2 = 2v2 sin θ 2 5

v2 ⋅ sin θ 2 = v0

...(ii)

Solving (i) and (ii), we get,

θ 2 = 45° and Centre of Mass

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PHYSICS

(b)

LOCUS

89

Now, we have K.E. before collision

1 2 = × m × (3v0 ) 2 9 2 = mv0 2

2 2 1 1 and K.E. after collision = × m × ( 5v0 ) + × 2m × ( 2v0 ) 2 2

5 2 4 2 9 2 = mv0 + mv0 = mv0 2 2 2

Hence the collision was an elastic one.

A ball moving translationally at speed u with another stationary, ball of the same mass. At the moment of impact the angle between the line joining the centres of the balls and the direction of the initial motion of the striking ball is θ. Assuming the balls to be smooth, find the final velocity of the each ball. Coefficient of restitution between the balls is e. Solution: The two balls just before the collision are shown in figure 6.75(a). If we consider the line of impact, the line joining the centres of the ball, coinciding with X-axis and a perpendicular direction in the same plane as Y-direction, then,

us in θ

Y usinθ uc os θ

θ

N

u

v1 N v2

X fig. 6.75(a)

fig. 6.75(b)

fig. 6.75(c)

fig. 6.75(d)

components of incident ball’s velocity along X and Y directions are shown in figure 6.75(b). Impulsive normal contact forces acting on each ball during the collision are shown in figure 6.75(c). As during the collision the force on each body acts along the X-axis, components of their velocities along the X-axis do change but components of the same along the Y-axis remain unchanged, as shown in figure 6.75(d). As there is no external force acting on the system of the two balls during the collision, we have, Px ,sys , fin = Px ,sys ,in

⇒

mv1 + mv2 = mu cos θ

⇒

v1 + v2 = u cos θ

...(i)

Using equation (6.36), we have,

speed of separation e. speed of approach = (along the line of impact) (along the line of impact) ⇒ Centre of Mass

v2 − v1 = e ⋅ u cos θ

...(ii) Web: http://www.locuseducation.org

PHYSICS

LOCUS

90

Solving (i) and (ii), we get, u cos θ (1 + e) 2 u cos θ (1 − e) v1 = 2 v2 =

and NOTE: •

Py , sys , fin = Py ,sys ,in

•

Final speed of the incident ball = v12 + (u sin θ )2

•

If e = 1, i.e. collision is elastic, then,

v1 = 0 and

v2 = u cos θ

Therefore, in this case the balls exchange velocities along the line of impact, which is in accordance with previously established concepts. In this case the angle of divergence, the angle between the final velocities of the balls, is 90:

A small ball of mass m hits the inclined surface of a wedge of mass M perpendicularly with speed u, as shown in figure 6.76(a). Before the impact, the wedge was at rest and friction is negligible at all contact surfaces. If the collision between the ball and the wedge be elastic, find the velocity of each body just after the collision.

m u θ

M

fig. 6.76(a)

Solution: At some instant during the collision the system “ball + wedge” is shown in figure 6.76(c). During the impact the impulsive normal contact force acting on the wedge from the ball, N, has a component in the vertically downward direction, due to which the normal contact force acting on the wedge from the fixed horizontal surface, N ', also Just Before Collision

During Collision

Just After Collision

SYSTEM BOUNDARY

m

v2

N u θ

(b)

M

θ

N'

(c)

N

N'

v1 θ

(d)

fig. 6.76

becomes impulsive. Hence, there is a net impulsive force acting on the system along the vertical direction due to which the momentum of the system is not conserved along the vertical direction. As there is no external force acting on the system along the horizontal direction, the momentum of the system along the horizontal direction is conserved. Just after the collision the wedge moves on the horizontal surface, let with speed v1 and the ball moves along the line of its initial motion, let with speed v2 , as shown in figure 6.76(d) Centre of Mass

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PHYSICS

LOCUS

Just Before Collision

91

Just After Collision v2 v 1 cos θ

u

θ Rest

v1

v 1 sin θ fig. 6.76(e)

Applying conservation of linear momentum along the horizontal direction, we have, Psys , H , fin = Psys , H ,in

⇒

Mv1 − mv2 sin θ = mu sin θ

...(i)

Applying equation (6.36), along the line of impact, we get, velocity of separation = velocity of approach ⇒

v1 sin θ + v2 = u

...(ii)

Now, solving equations (i) and (ii), we can solve for v1 and v2 .

Centre of Mass

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PHYSICS

1.

3.

92

Choose the correct statement(s) from the following: (i) In an elastic collision of two bodies, the momentum and energy of each body is conserved. (ii) The work done by a force on a body in nature, over a closed loop is always zero. (iii) In an inelastic collision of two bodies, the final kinetic energy is less than the initial kinetic energy of the system. (a) (c)

2.

LOCUS

(i) (iii)

(b) (d)

(ii) all.

For a system consisting of two particles that undergo an elastic collision, (a) momentum is conserved but the total energy is not conserved (b) neither the kinetic energy nor the momentum is conserved (c) neither the total energy nor the momentum is necessarily conserved (d) the mechanical energy is conserved but momentum is not conserved (e) both kinetic energy and momentum are conserved. A block of mass 2M, moving with constant velocity 3v$ collides with another block of mass M which is at rest and sticks to it. the velocity of the compound block after the collision is : (a) 3v$ (b) v$ (c) 2v$ (d) 3v$/2 .

4.

A ball hits the floor and rebounds after an inelastic collision. In this case, choose the correct alternative; (a) the momentum of the ball just after the collision is the same as that just before the collision (b) the total energy of the ball and the earth is conserved (c) the mechanical energy of the ball remains the same in the collision (d) the total momentum of the ball and the earth is conserved.

5.

A man in a spacecraft that is moving with a velocity V0 in free space runs forward and then stops. The center of mass of the system (that is, the ship and the man) (a) slows while the man is running, then speeds up when he stops (b) slows while the man is running, then resumes its original velocity when he stops (c) moves faster while the man is running, then slows when he stops (d) moves faster while the man is running, then resumes its original velocity when he stops (e) continues all the while at its original velocity. $ $ Two bodies having masses m1 and m2 and velocities u1 and u2 collide and form a composite system of $ $ momentum m1u1 + m2u2 = 0(m1 ≠ m2 ). The velocity of the composite system is : $ $ $ (a) 0 (b) u1 + u2 $ $ $ $ (c) u1 − u2 (d) (u1 + u2 )/2 .

6.

7.

A car having a mass of 200 kg is rolling at a speed of 0.1 m/s towards a spring-stop system. If the spring is non-linear such that it develops 300x²N force for a deflection of x m. The maximum deceleration that the car A undergoes; (a) 1 m/s² k A (b) 1.5 m/s² (c) 2 m/s² (d) 2.5 m/s².

Centre of Mass

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LOCUS

93

8.

In head-on elastic collision of two bodies of equal masses: (a) the velocities are interchanged (b) the speeds are interchanged (c) the momentum are interchanged (d) the faster body slows down and the slower body speeds up.

9.

A mass m1 moves with a great velocity. It strikes another mass m2 at rest in a head on collision. It comes back along its path with low speed after collision. Then: (a) m1 > m2 (b) m1 < m2 (c)

10.

11.

m1 = m2

(d)

There is no specific relation between m1 and m2 .

Two perfectly elastic particles A and B of equal masses traveling along the line joining them with velocity 15 m/s and 10 m/s respectively collide. Their velocities after the elastic collision will be (in m/s) respectively: (a) 0 and 25 (b) 5 and 20 (c) 10 and 15 (d) 20 and 5. $ A steel ball moving with velocity v collides elastically with an identical ball originally a rest. The velocity of the first ball after the collision is : (a) v$ (b) – v$ (c) (–1/2) v$ (d) Zero.

12.

A 10-g bullet is fired with velocity of 300 m/s into a pendulum bob which has a mass of 990 g. How high does the pendulum bob (plus bullet) swing after the collision?

13.

A bullet of mass m1 is fired with speed v into the bob of a ballistic pendulum of mass m2 . Find the maximum height attained by the bob if the bullet passes through the bob and emerges with speed 14 v.

14.

A 1-kg block of wood is attached to a spring of force constant 200 N/m and rests on a smooth 20 g surface, as shown in figure. A 20-g bullet is fired into the block, and the spring compresses 13.3 cm. (a) Find the original velocity of the bullet before the collision. (b) What fraction of the original mechanical energy is lost in this collision?

k=200 N/m

15.

A 4-kg object moving at 5 m/s makes a perfectly elastic collision with a 1-kg object initially at rest. (a) Find the final velocities of each object. (b) Find the energy transferred to the 1-kg object.

16.

A 2-kg object moving at 3 m/s to the right collides with a 3-kg object moving at 2 m/s to the left. The coefficient of restitution is 0.6. Find the velocity of each object after the collision.

17.

A ball moving at 10 m/s makes an off-center perfectly elastic collision with another ball of equal mass initially at rest. The incoming ball is deflected at an angle of 30° from its original direction of motion. Find the velocity of each ball after the collision.

18.

Consider a ball falls from some height h. Let e be the coefficient of restitution between the ball and the ground and ball rebounds again and again, then find (i) velocity after nth collision (ii) height attained after nth collision (iii) total distance traveled by ball before stop (iv) total time of motion.

19.

A ball strikes a horizontal wall at an angle α with the vertical. It rebounds at an angle θ with the vertical. Calculate the coefficient of restitution between the ball and wall.

Centre of Mass

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PHYSICS

20.

LOCUS

A railway car of mass M stands on a hill with its brakes set. The brakes are released, and the car rolls down to the bottom of the hill hm below its original position (Figure). It collides with a car of same mass resting at the bottom of the track (with brakes off). The two cars couple together and roll up the track to a height H. Find H.

94

M hm

M

H

15.

A 8 kg ball moving with velocity 4 m/s collides with a 2 kg ball moving with a velocity 8 m/s in opposite direction. If the collision be perfectly elastic, what are the velocities of balls after the collision.

17.

A ball moving with a velocity v strikes a wall moving towards the ball with a velocity u. An elastic impact occurs. Determine the velocity of ball after the impact. What is the cause of change in kinetic energy of the ball? Consider the mass of the wall to be infinitely great.

24.

A small particle traveling with a velocity v collides elastically with a spherical body of equal mass and radius r initially kept at rest. The centre of the spherical body is located at a distance (ρ < r) away from the direction of motion of the particle. Find the final velocity of the particle.

26.

A block of mass m1 = 150 kg is at rest on a very long frictionless table, one end which is terminated in a wall. Another block of mass m2 is placed between the first block and the wall, and set in motion towards m1 with constant speed u2. Assume that all collisions are perfectly elastic, find the value of m2 for which both the blocks move with the same velocity after m2 once with m1 and once with the wall. The wall has effectively infinite mass.

27.

28.

m1

Wall u2

m2

A military tank whose mass together with the artillery gun is M moves at a speed of v. The gun barrel makes an angle α with the horizontal. A shell of mass m leaves the barrel at a speed v relative to the barrel in the direction of the tank’s motion. The speed of the flat car in order that it may stop after the firing is ; (a)

mv cos α M +m

(b)

mv M +m

(c)

(M + m) v cos α

(d)

mu cos α . M +m

A set of n identical cubical blocks lies at rest parallel to each other along a line on a smooth horizontal surface. The separation between the near surfaces of any two adjacent blocks is L. The blocks of one end is given a speed v towards the next one at time t = 0. All collisions are completely inelastic, then; (a)

The last block starts moving at t =

(b)

The last block starts moving a t =

( n − 1) L v

(c)

n( n − 1) L 2v The centre of mass of the system will have a final speed v

(d)

The centre of mass of the system will have a final speed

Centre of Mass

v . n

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PHYSICS

29.

30.

LOCUS

95

In a one-dimensional collision between two identical particles A and B, B is stationary and A has momentum P before impact. During impact, B gives impulse J to A; (a) The total momentum of the ‘A plus B’ system is P before and after the impact, and (P–J) during the impact. (b) During the impact, A gives impulse J to B. (c)

The coefficient of restitution is

2J − 1. P

(d)

The coefficient of restitution is

J + 1. P

The magnitude and direction of the two identical smooth balls before central oblique collision are as shown in figure. Assuming coefficient of restitution e = 0.9, determine the magnitude and direction of the velocity of each ball after the collision. Line of impact 30°

60°

9 m/s

31.

A molecule collides with another, stationary, molecule of the same mass. Demonstrate that the angle of divergence (a) equals 90° when the collision is ideally elastic; (b) differs from 90° when the collision is inelastic.

12 m/s B A C

32.

Show that for a perfectly elastic collision between two particles of equal mass with one particle initially at rest, 2 the energy transferred to the originally stationary particle is (sin θ )E0 , where E0 is the initial energy and θ is the angle of deflection of the incoming particle. 33. A 10 g bullet moving 1000 m/s strikes and passes through a 2.0 kg block initially at rest, as shown. The bullet emerges from the block with a speed of 400 m/s. To what maximum height will the block rise above its initial position? (a) 78 cm (b) 66 cm (c) 56 cm (d) 46 cm v (e) 37 cm

Centre of Mass

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PHYSICS

LOCUS

96

A closed chain connected by a thread to a rotating shaft revolves around a vertical axis with uniform angular velocity ω, as shown in figure 6.77(a). The thread forms an angle θ with the vertical. How does the centre of mass, C, of the chain move? ω θ fig. 6.77(a) C

Solution: First of all, it is clear that the centre of mass of the chain, C, does not move in the vertical direction during the uniform rotation of the chain. This means that the vertical component of the tension force acting on the chain counterbalances the weight of the chain, as shown in figure 6.77(b). T cos θ

T θ C

C

mg

mg fig. 6.77(b)

T T sin θ=mω2r

Fnet = mω2r

C

mg fig. 6.77(c)

The horizontal component of the tension force is constant in magnitude and permanently directed towards the axis of rotation. If follows from this fact that the centre of mass of the chain, C, travels along a horizontal circle. If r be the radius of this circle, then, we have,

and

T cos θ = mg

...(i)

T sin θ = mω 2 r

...(ii)

Dividing equation (ii) by (i), we get,

tan θ =

⇒

Centre of Mass

r=

ω 2r g

g tan θ ω2

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PHYSICS

LOCUS

97

Two men, each of mass m, stand on the edge of a stationary buggy of mass M. Assuming the friction to be negligible, find the velocity of the buggy after both men jump off with same horizontal velocity u$ relative to the buggy: (a) Simultaneously; (b) One after the other. In which case will the velocity of the buggy be greater and how many times? Solution: CASE (A): WHEN THE TWO MEN JUMP SIMLTANEOUSLY: Just before the jump m REST

Just after the jump ! ! u+v ! ! u+v ! v m M m smooth

m REST

M

smooth (a)

(b)

fig. 6.78

The system just before the jump is shown in figure 6.78(a) and just after the jump is shown in figure 6.78(b). The $ horizontal velocity of the buggy just after the jump is assumed to be v , therefore, the velocity of the each men with respect to the ground just after the jump is $ vmG = velocity of the man with respect to the buggy + velocity of the buggy with respect to the ground $ $ = u + v, as shown in figure 6.78(b). As there is no impulsive force is acting on the system along the horizontal direction during the jump, $ $ Psys , H , fin = Psys , H ,in ⇒

$ $ $ $ Mv + 2m (u + v ) = 0

⇒

$ v =−

2m $ u M + 2m

...(i)

Hence the buggy moves along the direction opposite to the direction motion of the men with respect to the buggy 2m u. with speed M + 2m CASE (B): WHEN THEY JUMP ONE AFTER OTHER:

$ Let after the first jump the buggy and the remaining man moves with velocity v1 , then, velocity of the man who $ $ jumped first just after he made the jump is u + v1 with respect to the ground, as shown in figure 6.78(c). Just before the first jump

Just after the first jump

!

m m REST

M

REST

v1

! !

u + v1

m

m

M

fig. 6.78(c)

Applying conservation of linear momentum along the horizontal direction, we get, $ $ Psys , H , fin = Psys , H ,in Centre of Mass

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PHYSICS

LOCUS

⇒

$ $ $ $ ( M + m)v1 + m(u + v1 ) = 0

⇒

$ v1 = −

98

m $ u M + 2m

...(ii)

$ When the second man has made the jump, let the wedge moves horizontally with velocity v2 and hence, the velocity $ $ of the second man just after the jump is u + v2 with respect to the ground. Just before the second jump

!

Just after the second jump ! ! u + v2 ! m v2 M

m

v1 M

fig. 6.78(d)

Applying conservation of linear momentum along the horizontal direction to the system “buggy + second man”, we get, $ $ Psys , H , fin = Psys , H ,in ⇒

$ $ $ $ Mv2 + m(u + v2 ) = (m + M )v1

⇒

$ $ m $  (m + M )v2 + mu = (m + M )  − u  M + 2m  =−

m 2 + mM $ u M + 2m

⇒

 m 2 + mM $ $ (m + M )v2 = −  + mu  M + 2m 

⇒

m2 + mM + mM + 2m2 $ $ v2 = − u (m + M )(2m + M ) $ v2 = −

⇒

[Using (ii)]

m(3m + 2M ) $ u (m + M )(2m + M )

...(iii)

Now, we have, v2 m(3m + 2M ) (2m + M ) = × v1 ( m + M )(2m + M ) 2m =

3m + 2M m + (2m + 2M ) = 2m + 2 M (2m + 2M )

= 1+

⇒

Centre of Mass

[Using (ii) and (iii)]

m >1 2m + 2 M

v2 > v1

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PHYSICS

LOCUS

99

Two small blocks of masses m1 and m2 connected by a weightless spring of spring constant k, as shown in figure 6.79, rest on a smooth horizontal plane. Block 2 is shifted by a small distance x0 to the left and then released. 1 m1

2

k

m2

fig. 6.79

Find the velocity of the centre of mass of the system after block 1 break offs from the vertical wall. If block 2 is released at t = 0 and block 1 breaks off from the wall at t = t0 , then find the average normal contact force acting on the block 1 from the vertical wall for the duration [0, t0 ]. Solution: The system “block 1 + block 2 + spring” is shown at t = 0, at some time t < t0 and at t = t0 in figures 6.80(a), 6.80(b) and 6.80(c), respectively. For t < t0 , there is some compression in the spring due to which it exerts a force leftwards on the block 1 and a force rightwards on the block 2, as shown in figure 6.80(b). For t < t0 , the spring force on the block 2 accelerates it rightwards but the spring force on the block 1 can not accelerate it towards the left due to the vertical wall. Here, the normal contact force on the block 1 from the vertical wall, N, balances the spring force acting on it, as shown in figure 6.80(b). The only horizontal external force acting At t = 0 on the system for the duration [0, t0 ] is N. At any instant during the mentioned period the magnitude of N is equal to the magnitude of the spring force on the block 1, hence, at t = 0 magnitude of N is

natural length (a) m1

x0 +ve

m2

kx0 and at t = t0 magnitude of N becomes zero. At t = t0 the block 1 breaks from the wall because at this moment the outward velocity of the block

At some time t < t0 .

2, v0 , starts elongating the spring and hence, spring pulls the block 1 alongwith it. Hence, for the

(b)

duration [0, t0 ] , the momentum of the system is not conserved due to the action of N but as the point of application of N does not move during this interval, the work done by N is zero. Therefore,

N

for the time interval [0, t0 ] , mechanical energy of the system is conserved. Hence, we have, Ein ( E at t = 0) = E fin ( E at t = t0 )

⇒

kin + U in = k f + U f

⇒

1 1 0 + kx02 = m2 v02 + 0 2 2

Centre of Mass

v

kx m1

x

N

+ve kx

m2

REST

+ve At t = t0 (c)

m1

v0 m2

fig. 6.80

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PHYSICS

⇒

LOCUS

v0 =

100

k x0 m2

...(i)

At t = t0 , the velocity of the centre of mass of the two blocks is vcm =

m1v1 + m2 v2 m1 + m2

=

0 + m2v0 m1 + m2

=

m2 k x0 m1 + m2

[Using (i)]

If N av be the average value of the normal contact force on the block from the vertical wall, N, then, we have,

Impulse of N av for change in momentum of the = system for the duration [0, t0 ] the duration [0, t0 ] ⇒ ⇒

N av ⋅ (t0 − 0) = (m2 v0 − 0) N av =

m2 v0 x = m2 k 0 t0 t0

NOTE: Here you should notice that even though contact force from the vertical wall N is not very large but still we are not neglecting its impulse because its duration of action is not very short.

A uniform thin rod of mass M and length L is standing vertically along the y-axis on a smooth horizontal surface, with its lower end at the origin (0, 0). A slight disturbance at t =0 causes the lower end to slip on the horizontal surface along the positive x-axis, and the rod starts falling. (a) What is the path followed by the centre of mass of the rod during its fall? (b) Find the equation of trajectory of a point on the rod located at a distance r from the lower end. What is the shape of the path of this point? Solution: (a) In figure 6.81(a), the forces acting on the rod are shown at some moment during its fall. There are only two forces acting on the rod: force of gravity along the vertically downward direction and the normal contact force from the horizontal surface along the vertically upward direction. Hence, the force on the rod along the horizontal direction is zero. As the centre of mass of the rod was initially at rest and there is no force acting on the rod along the horizontal direction, the centre of mass of the rod does not sufferany displacement along the horizontal direction. Therefore, the centre of mass of the rod falls vertically downwards in a straight line.

Centre of Mass

N Mg fig. 6.81(a)

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PHYSICS

(b)

LOCUS

101

At some moment during its fall the rod is again shown in figure 6.81(b). At this moment let the position of the point at a distance r from the lower end of the rod, P, be (x, y). If C be the centre of the rod, then, we have,

Y B C

CA = L/2; PA = r;

xy

CP = CA – PA

O

P (x,y ) θ A

X

Fig. 6.81(b) = L/2 – r. At the same moment if the rod makes an angle θ with the horizontal, then, from figure 6.81(b), we have,

x = CP ⋅ cos θ L  x =  − r  cos θ 2 

⇒

cos θ =

⇒

x L   −r 2 

and,

y = PA ⋅ sin θ

⇒

y = r ⋅ sin θ

⇒

sin θ = y r

...(i)

...(ii)

By squaring and adding (i) and (ii), we get, x2

(L

2 − r)

2

+

y2 =1 (r )2

 Equation of the path of the point   P, which represents an ellipse. 

The inclined surfaces of two movable wedges of the same mass M are smoothly conjugated with the horizontal m

h

M

M

fig. 6.82(a)

plane, as shown in figure 6.82(a). A small block of mass m slides down the left wedge from a height h. To what maximum height will the block rise along the right wedge? Friction is negligible.

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102

Solution: Since there is no friction, external forces do not act on the system under consideration in the

v

M

m

u

M

fig. 6.82(b)

horizontal direction (6.82(b)). In order to determine the velocity v of the left wedge and the velocity u of the block immediately after the descent, we can use the energy and momentum conservation laws. Hence, we have, Mv 2 mu 2 + = mgh, 2 2

Mv = mu

and

Since at the moment of maximum ascent hmax of the washer along the right wedge, the velocities of the washer and the wedge will be equal, the momentum conservation law can be written in the form.

mu = ( M + m )V , where V is the total velocity of the washer and the right wedge. Let us also use the energy conservation law: mu 2 M + m 2 = V + mghmax . 2 2 The joint solution of the last two equations leads to the expression for the maximum height hmax of the ascent of the washer along the right wedge. hmax = h

M2

( M + m )2

.

A symmetric block of mass m1 with a notch of hemispherical shape of radius r rests on a smooth horizontal surface near the wall (shown in figure 6.83(a)).

m2

r

A small washer of mass m2 slides without friction from the initial position. Find the maximum velocity of the block. m1 fig. 6.83(a)

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103

SOLUTION: The block will touch the wall until the washer comes to the lowest position. By this instant of time, the washer has acquired the velocity v which can be determined from the energy conservation law: v 2 = 2 gr. During the subsequent motion of the system, the washer will “climb” the right-hand side of the block, m2 accelerating it all the time in the rightward direction (shown in figure 6.83 (b)) until the velocity of the washer and the block become equal. then the washer will slide down the block, the block being u=√2gr acclerated until the washer passes through the lowest position. Thus, m1 the block will have the maximum velocity at the instant is at which the washer passes through the lowest position during its backward fig. 6.83(b) motion relative to the block. In order to calculate the maximum velocity of the block, we shall write the momentum conservation law from the instant at which the block is separated from the wall:

m2 2 gr = m1v1 + m2v2 , and the energy conservation law for the instants at which the washer passes through the lowest position: m2 gr =

m1v12 m2v22 + . 2 2

This system of equations has two solutions: (1)

v1 = 0, v2 = 2 gr ,

(2)

v1 =

2m2 m1 + m2

2 gr ,

v2 =

m2 − m1 2 gr . m1 + m2

solution: (1) corresponds to the instants at which the washer moves and the block is at rest. We are interested in solution (2) corresponding to the instants when the block has the maximum velocity: v1max =

2m2 2 gr . m1 + m2

A thin hoop of mass M and radius r is placed on a horizontal plane. At the initial instant, the hoop is at rest. A small washer of mass m with zero initial velocity slides from the upper point of the hoop along a smooth groove in the inner surface of the hoop. Determine the velocity u of the centre of the hoop at the moment when the washer is at a certain point A of the hoop, whose O radius vector forms an angle ϕ with the vertical (shown in r ϕ figure6.84). The friction between the hoop and the plane should A be neglected. fig. 6.84

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104

Solution: The forces acting on the hoop-washer system are the force of gravity and the normal reaction of the plane. These forces are directed along the vertical. Consequently, the centre of mass of the system does not move in the horizontal direction. Since there is no friction between the hoop and the plane, the motion of the hoop is translatory. According to the momentum conservation law, at any instant of time we have zero net momentum. Therefore, Mu + mvx = 0,

...(i)

where u and vx are the horizontal components of the velocities of the centre of the hoop and the washer. Since vx periodically changes its sign, u also changes sign “ synchronously”. The general nature of motion of the hoop is as follows: the centre of the hoop moves to the right when the washer is on segments BC and BE, and to the left when the washer is on segments BC and BE, and to the left when the washer is on segments CD and DE (shown in figure 6.85). The velocities of the washer and of the hoop v and u, respectively, are connected through the energy conservation law: mgr (1 + cos ϕ ) =

y B

C

O

E

rϕ A D

x

fig. 6.85

mv 2 Mu 2 + . 2 2

...(ii)

The motion of the washer relative to a stationary observer can be represented at any instant as the superposition of two motions: the motion relative to the centre of the hoop at a velocity vt directed along the tangent to the hoop, and the motion together with the hoop at its velocity u having the horizontal direction (shown in figure 6.86). y O r ϕ vx

u

x

A vy fig. 6.86

v

vt

The figure shows that

vy vx + v y

= tan ϕ .

...(iii)

Solving Equation (i) to (iii) together, we determine the velocity of the centre of the hoop at the instant when the radius vector of the point of location of the washer forms an angle ϕ with the vertical u = m cos ϕ

Centre of Mass

2 gr (1 + cos ϕ ) . ( M + m)( M + m sin ϕ )

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105

A smooth washer impinges at a velocity v on a group of three smooth identical cubical blocks resting on a smooth horizontal surface as shown in figure 6.87. The mass of each block is equal to the mass of the washer. The diameter of the washer and its height are equal to the edge of the block.

v

fig. 6.87

Determine the velocities of all the bodies after the impact.

Solution: It is clear that at the moment of impact, only the extreme blocks come in contact with the washer. The force acting on each such block is perpendicular to the contact surface between the washer and a block and passes through its centre (the diameter of the washer is equal to the edge of the block!), as shown in figure 6.88. Therefore, the middle block remains at rest as a result of the impact.

During Collision

Just After Collision

u N N

45°

v´

N N

fig. 6.88

u

fig. 6.89

For the extreme blocks and the washer, we can write the conservation law for the momentum in the direction of the velocity v of the washer:

mv = ⇒

2mu + m v′ 2

v = 2u + v′.

Here m is the mass of each block and the washer, v' is the velocity of the washer after the impact, and u is the velocity of each extreme block. The energy conservation law implies that 1 2 1  1 mv = 2  mu 2  + mv′2 2 2  2 ⇒

v 2 = 2u 2 + v′2 . As a result, we find that u = v 2 and v′ = 0. Consequently, the velocities of the extreme blocks after the impact form the angles of 45° with the velocity v, the washer stops, and the middle block remains at rest.

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106

Three small bodies with the mass ratio 3:4:5 (the mass of the lightest body is m) are kept at three different points on the inner surface of a smooth hemispherical cup of radius r. The cup is fixed at its lowest point on a horizontal surface. At a certain instant, the bodies are released. Determine the maximum amount of heat Q that can be liberated in such a system. At what initial arrangement of the bodies will the amount of librated heat be maximum? Assume that collisions are perfectly inelastic. Solution: For the liberated amount of heat to be maximum, the following conditions must be satisfied: (1) the potential energy of the bodies must be maximum at the initial moment; (2) the bodies must collide simultaneously at the lowest point of the cup; (3) the velocity of the bodies must be zero immediately after the collision. If these conditions are satisfied, the whole of the initial potential energy of the bodies will be transformed into heat. consequently, at the initial instant the bodies must be arranged on the rim of the cup at a height r above the lowest point. The arrangement of the bodies must be such that their total momentum before the collision is zero (in this case, the body formed as a result of collision from the bodies stuck together will remain at rest at the bottom of the cup). m a a Pc

Pb 5 m 3

c

Pa

4 m b 3

fig. 6.90

b

c

fig. 6.91

Since the values of the masses of the bodies at any instant are to one another as 3:4:5, the arrangement of the bodies at the initial instant must be as in figure 6.90. (top view). After the bodies are left to themselves, the amount of heat Q liberated in the system is maximum and equal to loss in potential energy of the system, which is 5 4 mgr + mgr + mgr = 4mgr. 3 3

A system consists of two small identical cubes, each of mass m, linked together by a compressed weigthless spring of spring constant k. The cubes are also connected by a thread which is burned through at a certain moment. Find (a) at what value ∆ l , the initial compression of the spring, the lower cube will bounce up after the thread has been burned through; (b) to what height h the centre of gravity of this system will rise if the initial compression of the spring is 7 mg/k?

m

m

fig. 6.92(a) Centre of Mass

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x 0 PE level ∆l natural length

SYSTEM IS RELEASED FROM REST

kx

N

fig. 6.92(b)

mg

v=0 m x0= mg/k 0 PE level

natural length

Solution: When the thread is burned through, the upper block accelerates in the upward direction (for this to be happen the upward spring force on the upper block must be greater than its weight). If the upper block manages to elongate the spring, there would be an upward spring force on the lower block and when this force becomes equal to the weight of the block, it lifts the block from t he horizontal surface. Now, it is obvious in the figure 6.92(b) that when kx becomes equal to mg, i.e., when x becomes equal to mg/k, the lower block breaks off from the surface. If the upper block is just able to lift the lower block then at x = mg/k its speed becomes zero, as shown in figure 6.92(c) From the moment when the system was released from rest to the moment when the lower block is just about to leave the horizontal surface, the only non conservative force acting on the system is the normal contact force, N, on the lower block from the horizontal surface. This force does change the momentum of the system but it does not change the mechanical energy of the system because work done by it is zero, as its point of application remains at rest. Hence equating the mechanical energy of the system at the two mentioned moments we get,

107

kx0= mg m

N=0 mg

E1 = E2

fig. 6.92(c)

⇒

U1 + k1 = U 2 + k2

⇒

U1, gr + U1, sp + k1 = U 2, gr + U 2, sp + k2 1 1 − mg ∆l + ⋅ k ⋅ ∆l 2 + 0 = + mg ⋅ x0 + ⋅ k ⋅ x02 + 0 2 2

⇒ ⇒

1 mg 1  mg  − mg ⋅ ∆l + ⋅ k ⋅ ∆l 2 = mg ⋅ + ⋅k ⋅  2 2 k  k 

⇒

1 3 m2 g 2 ⋅ k ⋅ ∆l 2 − mg ⋅ k ∆l = ⋅ 2 2 k

⇒

∆l 2 − 2

⇒

∆l 2 +

⇒

mg   3mg    ∆l +   ⋅∆l − =0 k  k  

Centre of Mass

Gravitational potential energy lower of the lower block at the two intances is same, if   written, it would get cancelled out from the   twosides of the equation, hence, we just left it.

      

2

mg m2 g 2 ⋅ ∆l − 3 =0 k k

mg mg m2 g 2 ⋅ ∆l − 3 ⋅ ∆l − 3 ⋅ 2 = 0 k k k

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∆l = −

⇒

mg 3mg or ∆l = + k k

As the acceptable value of ∆l = +

3mg 3mg . [What is , the required minimum initial compression in the spring is k k

the significance of initial compression = – (b)

108

mg ?] k

It is clear from the result obtained in part (a), that, if the initial compression is 3 mg/k then the upper block is just able to lift the lower block, but, if the initial compression is greater than 3 mg/k, then, at the moment of the break off of the lower block, the upper block would be moving upwards. Therefore, there would be further rise in the height of the centre of mass of the system. The situation is shown in figure 6.92(d). When the lower block is about to break off, let the upper block has an upward speed u. At the same moment the centre of mass would have an upward speed ucm = u / 2. Once the lower block has lost contact with the horizontal surface then only external force acting on the system is gravity, hence, the acceleration of the centre of mass of the system is g in vertically downward direction. u m Zero Gravitational P.E. Level

x0= mg/k

natural length

∆l =7mg/k

system

ucm= u/2

∆h =8 mg/k

m g

The lower block is about to break off

The system is just released from the rest.

m

m (i)

(ii)

(iii) fig. 6.92(d)

Therefore, further rise in the height of the centre of mass would be 2 ucm (u 2 /4) ∆h′ = = 2g 2g

u2 8g But upto the moment when the lower block leaves the surface, the upper block has already covered a distance of 8 mg/k in the upward direction as shown in figure 6.92(d)(ii) or we can say the moment at which the lower block breaks off from the horizontal surface, the rise in height of the centre of mass of the system till that moment is =

∆h1 =

1 (8.mg/k) 2

= 4 mg /k Therefore, the maximum rise in the height of the C.M. is

∆H = ∆h1 + ∆h′

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mg u 2 + k 8g

=4

.

109

... (i)

Calculation for u: Equating the mechanical energies of the system at the moment shown in figure 6.92(d) (i) and at the moment shown in 6.92(d)(ii), we get,

Ei = Eii ⇒

U i , gr + U i , sp + ki = U ii , gr + U ii , sp + kii

⇒

1 1 1 − mg ⋅ ∆l + ⋅ k ⋅ ∆l 2 + 0 = + mg ⋅ x0 + ⋅ k ⋅ x02 + mu 2 2 2 2

⇒

7 mg 1 mg 1  7 mg   mg  1 2 − mg ⋅ + ⋅ k ⋅ + ⋅ k ⋅  = mg ⋅  + mu 2 2 2 k k k k    

⇒

1  m2 g 2 1  49 − 7 − 1 − = mu 2   2 k 2  2

⇒

1 m2 g 2 2 mu = 16 2 k

⇒

u 2 = 32

2

[

]

2

mg 2 k

... (ii)

Solving (i) and (ii), we get,

mg  mg 2  ∆H = 4 +  32 /8g k  k  =4

mg mg +4 p k k

=8

mg k

A chain AB of length l is located in a smooth horizontal tube so that a part of its length h hangs freely and touches the surface of the table with its end B, as shown in figure 6.93(a). At a certain moment the end A of the chain is set free. With what velocity will this end of the chain slip out of the tube? A

h

B

fig. 6.93(a) Centre of Mass

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110

Solution: Consider the moment when the part of the chain inside the tube has slipped by a distance x, as shown in figure 6.93(b). If v be the speed and a be the magnitude of the acceleration of the moving part of the chain, then, we have

A v=dx/dt a =dv/dt x

v

h a

fig. 6.93(b)

net force along the length on = mass of the moving part of the chain the moving part of the chain × magnitude of the acceleration

⇒

weight of the hanging = mass of the moving part × acceleration part fo the chain mg ⋅

⇒

⇒

⇒

gh = (l − x ) ⋅

dv dx ⋅ dx dt

v0

l −h

∫ v ⋅ dv 0

⇒

h l−x = m ⋅a l  l 

= gh ∫ 0

dx l−x

dv dv dx   ∵ a = dt = dx ⋅ dt   Let when the end A leaves the tube, i.e.,when x   becomes equal to l -h,the speed is v  0  

l −h v02 = − gh ln (l − x ) 0 2

= − gh ( ln h − ln l ) = − gh ln ( h./ l ) = gh ln (l / h ) ⇒

v02 = 2 gh ln (l / h )

⇒

v0 = 2 gh ln (l / h )

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111

NOTE : •

•

a

Theoretically it is wrong to say net force along the length.We can say net pulling force or something like that.

m1

v

T T

If we consider hanging part of the chain and the part of the chain inside the tube as different systems, as shown in figure 6.93(c), then, motion equation for these two systems would be

T = m1a

m2

v a mg. h l

fig. 6.93(c)

h mg ⋅ − T = m2 a l

By adding these two equations, we would get the same equation as we used in the dotted rectangular box. •

You should notice that upward thrust force from the surface of the table ( = λv 2 ) on the falling part of the chain is not included in the motion equation. It is simply because the surface exerts this force on the element which comes in contact with it and we were not considering that element as a part of the moving system.

A small block of mass M is suspended by a thread from a fixed support. A bullet of mass m hits the block with speed u at an angle θ with the horizontal and gets embedded in it as shown in figure 6.94(a) What is the impulse given by the thread to the block if it is known that the duration of impact is very small and the thread does not break during the impact?

m u

M Fig. 6.94(a)

Solution: Let us analyze the system “bullet + block” $ during the impact first, which is shown in figure 6.94(b). During the impact the bullet exerts an impulsive force N on the block along the direction of its initial motion and as a result $ $ a reaction force − N acts on it from the block. Now consider N $ as the sum of its horizontal and vertical components, N H and $ Nv , respectively. It is obvious that due to the high magnitude of $ ! N v the block pulls the thread with a very high force and hence, T ! tension in the thread also becomes impulsive during the impact. $ –N $ As the tension force acting on the block, T , is impulsive and N m $ M ! and − N are internal force of the system “bullet – block”, the NH θ momentum of the system remains conserved along the horizontal ! ! ! ! direction but it is not conserved along the vertical direction. If we N =NH + Nv Nv neglect the impulse of the gravity for the duration of impact, as it $ is very small as compared to that of the impulsive tension force T , Fig. 6.94(b) we have,

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112

Impulse of the = change in the momentum of the tension force system along the vertical direction J T = Pv , fin − Pv ,in

⇒

= 0 − (− mu sin θ ) = mu sin θ

[From figure 6.94(c)]

M

v

usinθ

NOTE:

∫

m+M

Just before collision

t fin

•

ucosθ m

T .dt = JT = mu sin θ .

Fig. 6.94(c)

tin

•

Just after collision

If T be the average tension force for the duration of impact and ∆t be the duration of impact, then, T ⋅ ∆t = JT = mu sin θ ⇒

T = ( musinθ) ∆t

Two small blocks, each of mass m, connected by a light inelastic string which goes over a smooth pulley, are at rest as shown in figure 6.95(a). A small ball, C, of the same mass hits the block B with speed u and sticks to it. Find the final speed of the each body just after the collisin.

m c u m

m

Fig. 6.95(a)

Solution: During the impact the bodies are shown in figure 6.95(b). The ball exerts an impulsive normal contact force, N, on the block B and a reaction force of the same magnitude acts on the ball itself. Due to the impulsive downward force N acting on the block B, the tension force acting on the block, T, also becomes impulsive.

+ve

N T A m

T

c

m B

N fig. 6.95(b)

+ve

T A m

T c m B

fig. 6.95(c)

v

m m

m

v

fig. 6.95(d)

When the block A, the block B and the ball C are considered as a single system, it becomes obvious that the string exerts an impulsive force 2T on the system during the impact, T on the each block, as shown in figure 6.95(b). Hence the momentum of the system is not conserved along the vertical direction. Centre of Mass

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113

Let us assume that just after the collision the block B and the ball C move in the vertically downward direction with speed v. As the string is inelastic, the block A must also be moving upwards with speed v. If we neglect the impulse of the gravity for the duration of collision and consider vertically downward direction as the +ve direction, then, we have, For the ball C: ⇒ For the block B: ⇒ For the block A: ⇒ ⇒

p f = pi + ∆p mv = mu − ∫ N ⋅ dt

...(i)

p f = pi + ∆p mv = 0 + ∫ N ⋅ dt − ∫ T ⋅ dt

...(ii)

p f = pi + ∆p − mv = 0 − ∫ T ⋅ dt mv = ∫ T ⋅ dt

...(iii)

Adding (i), (ii) and (iii), we get,

3mv = mu ⇒

Just before the collision

v=u 3 A

Alternate Method: If we analyze the system “ block A + block B + ball C” along the length of the thread as we did in the example and many examples in Newton’s Laws of Motion, all the impulsive forces acting on the system during the collision appear to be internal. Hence, applying conservation of the linear momentum along the same direction, from figure 6.95(e) we can write

m

m c Rest

u

•

m B REST

Just after the collision v

v m

•

m

m

fig. 6.95(e)

p fin = pin

⇒ ⇒

(3m) ⋅ v = m ⋅ u

v = u/3

Here you should note that this is just a method to get the result. Theoretically it is wrong to define a direction along the thread and to say same thing like momentum along this direction.

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114

A ball is projected from a point in one of the two smooth parallel vertical walls against the other in a plane perpendicular to both. After being reflected once at each wall the ball impinge again on the second at a point in the same horizontal plane as it started from. Show that Re 2 = l (1 + e + e 2 ),

where e = coefficient of restitution, R = free range on a horizontal plane and l = distance between the walls. Solution: The ball is projected from O, a point on the first wall with a velocity whose horizontal and vertical components are u and v respectively. It then strikes the second wall at A, then the first wall at B and strikes again the second wall at C such that O and C are in the same horizontal plane. The ball strike A with a horizontal velocity ‘u’, which after impact becomes eu. With this horizontal velocity eu, the ball strikes at B and after impact at B, the horizontal velocity of the ball is e2u and with this horizontal velocity ball strikes the second wall at C. The horizontal distance moved as the particle moves from O to A or A to B or B to C is the same i.e., ‘a’. ∴

Total time taken in moving from O to C

a a a a + + 2 = 2 (e2 + e + 1) ...(i) u eu e u e u Throughout the motion from O to C, the vertical velocity remains unaffected due to impact with walls and vertical distance moved from O to C is zero. =

h = ut −

By

eu

O

2v =

 2uv  2 2  g  e = a(e + e + 1)   Also it is given that the free range on the horizontal plane = b i.e., time of fligh × horizontal speed =

Centre of Mass

a

u

C

2

ag 2 (e + e + 1) e 2u

or

∴

u

Fig. 6.96

We have,

or

A

v

1 2 gt 2

1  a    a  0 = v  2  (e 2 + e + 1) − g  2  × ( e 2 + e + 1) 2  e u  e u 

e2u

B eu

⇒

2v ⋅u = b g

⇒

2vu =b g

...(ii)

be 2 = a (e 2 + e + 1)

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115

A pan of mass m supported by an ideal spring of stiffness k is at rest. A ball of the same mass hits the pan vertically with speed u, as shown in figure 6.97(a). If the coefficient of restitution be e, find the maximum compression in the spring. Neglect multiple collisions between the ball and the pan. m

•

e =½

u

m

k

fig. 6.97(a)

Solution: Before the collision, as the pan is at rest, the net force on the pan is zero, therefore, the compression in the spring is mg/k, as shown in figure 6.97(b). During the collision the ball exerts an impulsive normal contact force on the pan and a reaction force of the same magnitude acts on the ball itself, as shown in figure 6.97(c). During the collision as the compression in the spring is not changed largely, the spring force is also considered nonimpulsive along with

mg ∆li = k

•

m

m

N u

•

e =½

k

N

Just before the collision fig. 6.97(b)

•

m

m

k

∆lmax

v1 v2

∆lmax–

m REST

k

During the collision

Just after the collision

fig. 6.97(c)

fig. 6.97(d)

mg k

k

When the compression is maximum fig. 6.97(e)

the gravity. Hence, linear momentum of the system “ball + pan” can be considered to be conserved for the duration of collision. If v1 and v2 be the speeds of the ball and the pan, respectively, just after the collision (as shown in figure 6.97(d)). Then, we have, psys , fin = psys ,in

⇒

mv1 + mv2 = mu

⇒

v1 + v2 = u

...(i)

For an inelastic collision, we have, of separation just (speed ) after the collision

Centre of Mass

(

= e. speed of approach just before the collision

)

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LOCUS

⇒

v2 − v1 = e.u

⇒

v2 − v1 =

Now, (i) + (ii) gives, v2 =

116

u 2

...(ii)

3 u. 4

Therefore, we can assume that just after the collision the spring is compressed by mg k and the pan has a downward 3 speed v2 = u . Therefore as the pan moves downwards, the compression in the spring increases and the speed of 4 the pan decreases. When the speed of the pan becomes zero, the compression in the spring becomes maximum, as shown in figure 6.97(e). After the collision as there is no nonconservative force acting on the system “pan + spring”, the mechanical energy of this system is conserved. Therefore, mechanical energy of the system mechanical energy of the system "pan+spring" just after the collision = when pan comes to rest, i.e., when compression in the spring becomes maximum

or, we can say, loss in gravitational potential energy of = gain in the spring the pan + loss in kinetic energy of the pan potential energy

⇒

mg  1 2 1 1  mg  2  mg  ∆lmax −  + mv2 = k ( ∆lmax ) − k   k  2 2 2  k  

⇒

m2 g 2 1  3  1 m2 g 2 2 mg ⋅ ∆lmax − + m  u  = ( ∆lmax ) − k 2 4  2 2k

⇒

( ∆lmax )

2

2

2

 9mu 2 m2 g 2  2mg − − ( ∆lmax ) +  =0 k k   16k

Now, for the given values of m, g, u and k, we can solve above quadratic equation in ∆lmax to get the required maximum compression.

Centre of Mass

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PHYSICS

1.

LOCUS

117

Three identical spheres, each of mass 1 kg are placed touching each other with their centres on a straight line. Their centre are marked P, Q and R respectively. The distance of centre of mass of the system from P is: (a)

PQ + PR + QR 3

(b)

PQ + PR 3

(c)

PQ + QR 3

(d) none of these

2.

The center of mass of a system of particles is so defined that (a) it is always at rest (b) it is always at rest or moving with constant velocity (c) it always moves in a straight line even if the particles are rotating about it (d) the kinetic energy of the system is a maximum about any axis through the centre of mass (e) its location depends only on the masses of the particles and their locations.

3.

A woman on a spaceship traveling at a velocity v0 in free space runs forward and then suddenly stops. when she stops, the center of mass of the system (that is, the ship and the woman) (a) moves in the same direction as v0 with a slight increase in speed (b) moves in the same direction as v0 with a slight decrease in speed (c) comes to rest if the woman can run fast enough (d) continues unchanged at a velocity v0 (e) moves in the direction that the woman runs until she stops.

4.

A body A of mass M while falling vertically downwards under gravity breaks into two parts, a body B of mass M/3 and body C of mass 2/3 M. The centre of mass of bodies B and C taken together shifts compared to that of body A towards: (a) Body C (b) body B (c) Depends on height of breaking (d) Does not shift.

5.

A boy is standing at the stern (back) of a boat that is 8.0 m long. There is no friction between the boat and the water. The boy has a mass of 63 kg and the boat has a mass of 780 kg. The bow (front) of the boat is touching a dock and the force-and-aft axis of the boat is perpendicular to the dock. The boy walks from the stern of the boat to the bow. When he reaches the bow, his distance from the dock is (a) 7.6 m (b) 0.65 m (c) 0.51 m (d) 0.56 m (e) 1.3 m.

6.

If a body moves in such a way that its linear momentum is constant, then (a) its kinetic energy is zero (b) the sum of all the forces acting on it must be zero (c) its acceleration is greater than zero and is constant (d) its center of mass remains at rest (e) the sum of all the forces acting on the body is constant and non zero.

7.

In a centre of mass reference frame, (a) the system’s kinetic energy is zero (c) the total momentum is zero (e) non of these is correct.

Centre of Mass

(b) momentum is not conserved (d) all collisions are elastic

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PHYSICS

8.

LOCUS

118

An object of mass M1 is moving with a speed v on a straight, level, frictionless track when it collides with another mass M 2 that is at rest on the track. After the collision, M1 and M 2 stick together and move with a speed of (a) v

(b) M 1v

(c) ( M1 + M 2 )v/M1

(d) M 1v/( M 1 + M 2 )

(e) M 1v/M 2 . 9.

A projectile with a mass 6M is fired at a speed of 400 m/s at an angle of 60° above the horizontal. At the highest point of its trajectory, the projectile is broken into two equal pieces by an internal explosion. Just after the explosion, one of the two pieces is known to be traveling vertically downward at a speed of 300 m/s. the magnitude of the velocity of the other half of the projectile is (a) 500 m/ (b) 1.50 m/s (c) 400 m/s (d) 710 m/s (e) 123 m/s

10.

A particle of mass 2m is moving to the right in projectile motion. At the top of its trajectory, an explosion breaks the particle into two equal parts. After the explosion, one part falls straight down with no horizontal motion. What is the direction of the motion of the other part just after the explosion? (a) up and to the left (b) stops moving (c) up and to the right (d) straight up (e) down and to the right.

11.

Two blocks A and B each of mass m are connected by massless spring of natural length L and spring constant k. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length as shown in Fig. A third identical block C also of mass m moves on the floor with speed v along the line joining A and B, and collides with A then: C

v

A

B

(a) The KE of the AB system at maximum compression of the spring is zero (b) The KE of he AB system at maximum compression is (1/4) mv². (c) The maximum compression of spring is v m/k (d) The maximum compression of spring is v m/2k . 12.

13.

A hemisphere of mass 4m is free to slide with its base on a smooth horizontal table. A particle of mass m is placed on the top of the hemisphere. The hemisphere has now acquired a velocity v. Then the angular velocity of the particle relative to the hemisphere at an angular distance θ is: (a)

3v a cos θ

(b)

4v a cos θ

(c)

5v a cos θ

(d)

2v . a cos θ

A bomb of mass 16 kg at rest explodes into two pieces of masses 4 kg and 12 kg. The velocity of the 12 kg mass is 4 ms −1. The kinetic energy of the other mass is : (a) 192 J (c) 144 J

Centre of Mass

(b) 96 J (d) 288 J.

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14.

LOCUS

Two blocks A and B of mass m and 2m are connected by a massless spring of force constant k. They are placed on a smooth horizontal plane. Spring is stretched by an amount x and then released. The relative velocity of the blocks when the spring comes to its natural lengths is  3k  (a)  2m  x  

(c) 15.

119

2kx m

 2k  (b)  3m  x  

(d)

B

A

3km 2x

3 Displacement of a particle of mass 2kg moving in a straight line varies with time as s = (2t + 2) m. Impulse of the force acting on the particle over a time interval between t = 0 and t = 1s is : (a) 10 N − s (b) 12 N − s

(c) 8 N − s

(d) 6 N − s

16.

A block of mass M is attached with a spring of spring constant K. The whole arrangement is placed on a vehicle as shown in the figure. If the vehicle starts moving towards right with an acceleration a (there is no friction anywhere), then K Ma a (a) maximum elongation in the spring is M K 2Ma (b) Maximum elongation in the spring is K 2Ma (c) Maximum compression in the spring is K (d) Maximum compression in the spring is zero.

17.

Two masses M and m are tied with a string and arranged as shown. The velocity of block M when it loses the contact is (a) 2 gh (c)

2m gh (m + M )

m gh (b) (m + M ) 2 M gh (d) . (m + M )

m h

M

18.

A sphere A of mass 4 kg is released from rest on a smooth hemispherical shell of radius 0.2 m. The sphere A slides down and collides elastically with another sphere B of mass 1 kg placed on the bottom of the shell. If the sphere B has to just reach the top, the height h from where the sphere A should be released is (a) 0.08 m (b) 0.02 m A (c) 0.18 m h (d) 0.10 m. B

19.

A particle of mass m moving with a speed v collides elastically with another particle of mass 2m on a horizontal circular tube of radius R, then select the correct alternative(s). m 2π R . (a) The time after which the next collision will take place is v v (b) The time after which the next collision will take 2m place is proportional to m. (c) The time after which the next collision will take place is inversely proportional to m. (d) The time after which the next collision will take place is independent of the mass of the balls.

Centre of Mass

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PHYSICS

20.

21.

22.

23.

24.

25.

26.

27.

LOCUS

120

A moving particle is stopped by a single head-on collision with a second, stationary particle, if the moving particle undergoes (a) an elastic collision with a second particle of much smaller mass (b) an elastic collision with a second particle of much greater mass (c) an elastic collision with a second particle of equal mass (d) an inelastic collision with a second particle of any mass (e) any type of collision in which the coefficient of restitution is zero Five billiard balls are in contact and at rest on a wire that passes through their centers. Two billiard balls are slammed into one end of the row of five at a velocity v$ If the balls are free to slide but not roll and if the collision is elastic, which of the following is most likely to take place? (a) One ball at each end goes off with a speed v. (b) One ball on the side opposite the striking balls goes off with a speed of 2v. (c) Two balls on the side opposite the striking balls go off with a speed of v. (d) None of these will occur. Glider A, travelling at 10 m/s on an air track, collides elastically with glider B travelling at 8.0 m/s in the same direction. The gliders are of equal mass. The final speed of glider B is (a) 8.4 m/s (b) 10 m/s (c) 8.0 m/s (d) 4.0 m/s (e) 12 m/s Two equal masses travel in opposite directions with equal speeds. They collide in a collision that is between elastic and inelastic. Just after the collision, their velocities are (a) zero (b) equal to their original velocities (c) equal in magnitude but opposite in direction to their original velocities (d) less in magnitude and in the same direction as their original velocities (e) less in magnitude and opposite in direction to their original velocities. In a real collision, (a) kinetic energy is conserved (b) linear momentum is conserved in the absence of external forces (c) both momentum and kinetic energy are conserved (d) neither momentum nor kinetic energy is conserved In an elastic collision of two objects, (a) momentum is not conserved (b) momentum is conserved, and the kinetic energy after the collision is less than its value before the collision (c) momentum is conserved, and the kinetic energy after the collision is the same as the kinetic energy before the collision (d) momentum is not conserved, and the kinetic energy of the system after the collision differs from the kinetic energy of the system before the collision (e) the kinetic energy of the system after the collision depends on the masses of the object. Two identical cars approach an intersection. One is travelling east at 18 m/s and the second is travelling north at 24 m/s. They collide violently, sticking together. Immediately after the crash they are moving (a) 30 m/s, 37° N of E (b) 30 m/s, 37° E of N (c) 15 m/s, 37° N of E (d) 15 m/s, 37° E of N (e) 42 m/s, 37° N of E Two balls of equal mass are thrown against a massive wall with equal velocities. The first rebounds with a speed equal to its striking speed, and the second sticks to the wall. The impulse that the first ball transmits to the wall, relative to the second, is (a) twice as great. (b) half as great (c) the same (d) four times as great (e) one-fourth as great.

Centre of Mass

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PHYSICS

28.

LOCUS

Which of the following potential energy curve can describe the elastic collisions of two billiards balls? Here r is the distance between centres of balls; U

U

(a)

U

(b) r

2R

29.

121

U

(c) r

2R

(d) r

2R

. 2R

r

Two particles of masses m1 and m2 in projectile motion have velocities v1 and v2 (v1 < v2 ) respectively at &$ &$ t = 0. They collide at time t0. Their velocities become v′1 and v′2 at time 2t0 while still moving in it. The &$ &&$ &$ &&$ value of (m1 v′ + m2 v2′ ) − (m1 v1 + m2 v2 ) is : (b) (m1 + m2 ) g t0 1 (c) 2 (m1 + m2 ) g t0 (d) ( m1 + m2 ) g t0 2 Two balls shown in figure are identical, the first moving with speed v towards right and the second staying at rest. The wall at the extreme right is fixed and smooth. Assuming all collisions to be elastic, which of the following statements are correct? v (a) There are only three collisions (b) The speed of first ball is reduced to zero finally after all collisions A B (c) Only two collisions are possible (d) The speeds of balls remain unchanged after all collisions have taken place. (a) zero

30.

31.

32.

33.

Six identical balls are lined up along a straight frictionless grove. Two similar balls moving with speed v along the grove collide with this row on extreme left side end. Then: (a) One ball from the right end will move on with speed 2v; all the others remain at rest (b) Two balls from the extreme right will move on with speed v each and the remaining balls will be at rest (c) All the balls will start moving to right with speed v/8 each (d) All the six balls originally at rest will move on with speed v/6 each and the two incident balls will come to rest. $ A ball of mass m moving with a velocity v collides head on elastically with another of the same mass m but $ moving with a velocity – v . After collision: (a) The velocities are exchanged between the balls (b) Both the balls come to rest (c) Both of them move at right angles to original direction of motion (d) One ball comes to rest and the other ball travels back with velocity 2v. A neutron traveling with a velocity v and kinetic energy E collides elastically head on with the nucleus of an atom of mass number A at rest. The fraction of total energy retained by the neutron is :  A −1  (a)    A +1

34.

2

 A +1 (b)    A −1 

2

 A −1  (c)    A 

2

2

 A +1 (d)   .  A 

A sphere of mass m moving with a constant velocity u hits another stationary sphere of the same mass. If e is the coefficient of restitution, then ratio of velocities of the two spheres after collision will be: 1− e  (a)   1+ e 

Centre of Mass

(b)

1+ e    1− e 

 e +1  (c)    e −1 

(d)

 e −1   .  e +1 

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PHYSICS

35.

LOCUS

122

A block of wood of mass m kg is placed on a second block of identical size but of mass 2m kg, this latter block resting on a smooth horizontal plane surface, as shown in the fig. The coefficient of static friction between the blocks is 0.3 while the coefficient of kinetic friction is 0.25. A particle of mass m kg with a constant horizontal velocity u m/s hits the upper block at midpoint P on its vertical edge and instantaneously sticks to it on impact which lasts 0.1 s. For the subsequent motion of the blocks which of the following statements will be true? (g = 10 m/s²) •

u

p

m 2m

(a) If u
1 m/s, the upper block will move a certain distance relative to the lower block before coming to relative rest, then move back over the lower block the same relative distance to return to its initial condition.

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PHYSICS

LOCUS

123

1.

There are two masses m1 and m2 placed at a distance l apart. Let the centre of mass of this system be at a point named C. If m1 is displaced by l1 towards C and m2 is displaced by l2 away from C, find the distance, from C where the new centre of mass will be located.

2.

At one instant, the centre of mass of a system of two particles is located on the x-axis at x = 3.0m and has a velocity of (6.0m / s ) ˆj. One of the particles is at the origin, the other particle has a mass of 0.10kg and is at rest on the x-axis at x = 12.0m. (a) What is the mass of the particle at the origin ? (b) Calculate the total momentum of this system. (c) What is the velocity of the particle at the origin?

3.

A system consists of two particles. At t = 0, one particle is at the origin; the other, which has a mass of 0.60 kg , is on the y-axis at y = 80m. At t = 0 the centre of mass of the system is on the y-axis at y = 24 m and has a velocity given by (6.0m / s 3 )t 2 ˆj. (a) Find the total mass of the system (b) Find the acceleration of the centre of mass at any time t. (c) Find the net external force acting on the system at t = 3.0s.

4.

Two identical buggies move one after the other due to inertia (without friction) with the same velocity v0. A man of mass m rides the rear buggy. At a certain moment the man jumps into the front. Mass of each buggy is equal to M, find the velocities with which the buggies will move after that.

5.

A stationary pulley carries a rope whose one end supports a ladder with a man and the other end the counterweight of mass M. The man of mass m climbs up a distance ‘l’ with respect to the ladder and then stops. Neglecting the mass of the rope and the friction in the pulley axle, find the displacement l of the centre of inertia of this system.

6.

A particle of mass 1.0 g moving with velocity v1 = 3.0iˆ − 2.0 ˆj experiences a perfectly inelastic collision with another particle of mass 2.0 g and velocity v2 = 4.0 ˆj − 6.0kˆ. Find the velocity of the formed particle (both the vector v and its modulus), if the components of the vectors v1 and v2 are given in the SI units.

7.

An open-topped freight car of mass 20 kg is rolling along a track at 5 m/s. Rain is falling vertically downward into the car. After the car has collected 2 kg of water, what is its speed?

8.

Two masses are held at rest initially on a horizontal frictionless surface. They are pushed together, compressing a small spring between them which is not attached to either mass. When the masses are released, the spring accelerates both, giving mass m1 a velocity 5 m/s to the left and m2 velocity 15 m/s to the right. (a) What is the total momentum of the system before the masses are released? After they are released? (b) What is the ratio m1/m2 ?

9.

A bullet of mass m1 is fired with speed v into the bob of a ballistic pendulum, which has mass m2. The bob is attached to a very light rod of length L, which is pivoted at the other end. The bullet is stopped in the bob. Find v such that the bob will swing through a complete circle.

10.

A projectile is launched at 20 m/s at an angle of 30° with the horizontal. In the course of its flight it explodes, breaking into two parts, one of which has twice the mass of the other. The two fragments land simultaneously. The lighter fragment lands 20 m from the launch point in the direction the projectile was fired. Where does the other fragment land?

11.

A particle of mass 2m is projected at an angle of 45° with horizontal with a velocity of 20 2 m / s. After 1 s explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained by the other part. Take g = 10m / s 2 .

Centre of Mass

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PHYSICS

LOCUS

124

12.

A gun fires a bullet. The barrel of the gun is inclined at an angle of 45° with horizontal. When the bullet leaves the barrel it will be travelling at an angle greater than 45° with the horizontal. Is this statement true or false ?

13.

A small sphere of mass 10 g is attached to a point of a smooth vertical wall by a light string of length 1 m. The sphere is pulled out in a vertical plane perpendicular to the wall so that the string makes an angle of 60° with the wall and is then released. It is found that after the first rebound the string makes a maximum angle of 30° with the wall. Calculate the coefficient of restitution and the loss of K.E. due to impact. If all the energy is converted into heat, find the heat produced by the impact.

14.

The 1 kg sphere shown is figure is released from rest when θ = 90°. The coefficient of restitution between the sphere and the block is 0.70. If the coefficient of friction between the block and the horizontal surface is 0.3, determine how far the block will move after the impact. 1.2 m

θ

1 kg

5 kg

15.

Find the increment of the kinetic energy of the closed system comprising two spheres of masses m1 and m2 due to their perfectly inelastic collision, if the initial velocities of the spheres were equal to v1 and v2.

16.

A particle of mass m1 experienced a perfectly elastic collision with a stationary particle of mass m2. What fraction of the kinetic energy does the striking particle lose, if (a) it recoils at right angles to its original motion direction. (b) the collision is a head-on one?

17.

A particle of mass m1 collides elastically with a stationary particle of mass m2(m1 > m2). Find the maximum angle through which the striking particle may deviate as a result of the collision.

18.

A spaceship of mass m0 moves in the absence of external forces with a constant velocity v0. To change the motion direction, a jet engine is switched on. It starts injecting a gas jet with velocity u which is constant relative to the spaceship and directed at right angles to the spaceship motion. The engine is shut down when the mass of the spaceship decreases to m. Through what angle α did the motion direction of the spaceship deviate due to the jet engine operation?

19.

Two equal spheres A, B are in a smooth horizontal groove at opposite ends of a diameter. ‘A’ is projected along the groove and at the end of time ‘T’ impinges on ‘B’. Show that the second impact will occur after a further time 2T/e, where e = coefficient of restitution.

20.

A disc of mass 0.1 kg is kept floating horizontally in mid air by firing bullets of mass 0.05 kg each vertically at it, at the rate of 10 per sec. If the bullets rebound with the same speed, what is the speed of the bullets with which these are fired?

21.

If a body falls normally on a surface from height h, what will be the height regained after collision if coefficient of restitution is e?

22.

A simple pendulum is suspended from a peg on a vertical wall. The pendulum is pulled away from the wall, coefficient of restitution being (2 / 5). What is the minimum number of collisions after which the amplitude of oscillation becomes less than 60°?

Centre of Mass

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PHYSICS

LOCUS

125

23.

A ball of mass 50g is dropped from a height h = 10 m. It rebounds losing 75 per cent of its total mechanical energy. If it remains in contact with the ground for ∆t = 0.01 sec., find the impulse of the impact force.

24.

A ball falls vertically for 2s and hits a plane inclined at 30° to the horizontal. If the coefficient of restitution be 5/8, prove that the time that elapses before it again hits the plane is 2.5 sec. Find also the distance along the plane between the first and second impact. (g = 10 m/s²) A particle of mass m and velocity v$ collides elastically with a stationary particle of mass m. Calculate the angle between the velocity vectors of the two particles after the collision.

25. 26.

Consider a mass moving along x-axis with a certain velocity, hits mass m2 and sticks to it. Masses m2 and m3 are connected through a massless rigid rod. All the three particles have same mass. Find the motion of particles after impact. All particles can move without friction on the horizontal x-y plane.

27.

A cart of mass 10 kg is rolling along a level floor at a speed of 5 m/s. A 4 kg mass dropped from rest lands in the cart. (a) What is the velocity of the 2 kg piece after the explosion? (b) What is the velocity of the center of mass after the explosion?

28.

A sphere of radius r1 moving parallel to the x axis collides with a stationary sphere of radius r2 . Show that when the spheres are in contact, the angle made by the line of centers of the spheres with the x axis is given by sin θ = b /(r1 + r2 ), where b is the impact parameter.

29.

A cannon of mass M rests on wheels on a hard horizontal surface. Its barrel makes an angle θ0 with the horizontal. It fires a cannonball of mass m and recoils freely. Show that the initial angle made by the velocity of the ball with the horizontal is related to θ0 by tan θ = (1 + m/M ) tan θ 0 .

30.

Show that in a one-dimensional collision between two particles the fractional relative energy loss is related to the coefficient of restitution by ∆E − k ,rel = 1 − e2 Ek ,rel

31.

A particle has speed v0 originally. It collides with a second particle at rest and is deflected through angle φ. Its speed after the collision is v. The second particle recoils, its velocity making an angle θ with the initial direction of the first particle. Show that tan θ =

v sin φ v0 − v cos φ

Do you have to assume that this collision was elastic or inelastic to get this result? 32.

In the figure shown the hollow tube of mass ‘M’ is free to move without friction in the horizontal direction supported by two fixed vertical lugs attached to the roof. The system is released from rest. Find the velocity of the tube, when the block B has fallen, through a height ‘h’. Neglect any friction and mass of the string. m A

M

B m

Centre of Mass

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PHYSICS

33.

LOCUS

126

Two blocks A and B of masses m and 2m respectively are placed on a smooth floor. They are connected by a spring. A third block C of mass m moves with a velocity v0 along the line joining A and B and collides elastically with A, as shown in figure. At a certain instant of time t0 after collision, it is found that the instantaneous velocities of A and B are the same. Further, at this instant the compression of the spring is found to be x0 . Determine (i) the common velocity of A and B at time t0 , and (ii) the spring constant. C m

B

A v0

2m

m

34.

Two parallel vertical walls of height ‘h’ stand on a horizontal plane. A ball is projected from one foot of one wall towards the other and after impact it just clears the top of the first wall. Prove that the point of impact with the second wall is at a depth h (1 + e) 2 below its top, where e = coefficient of restitution.

35.

In the figure shown a plank of mass ‘M’ is placed on a smooth horizontal surface. Two light identical springs having stiffness ‘k’ are rigidly connected to struts at the end of the plank as shown in the figure. When the springs are in their unextended positions the distance between their free ends is 3l. A block of mass ‘m’ is placed on the plank and pressed against one of the springs so that it is compressed by a length ‘l’. To keep the block at rest it is connected to the strut by means of a light string as shown in the figure. Initially, the system is at rest and string is burnt through. Find: strut K

3l

strut

a string

l m M

(a) the maximum displacement of the plank. (b) maximum velocity of the plank. (c) the period of oscillation of the plank.(ignore any friction) 36.

Two masses m1 and m2 are connected by a spring of force constant k and are placed on a frictionless horizontal surface. Initially the spring is stretched through a distance x0, when the system is released from rest. Find the distance moved by two masses before they again come to rest.

37.

A thin disc A with a mass of 5 kg translates along a frictionless surface (see figure) at a speed of 10 m/s. It strikes a square stationary plate B with a mass of 5 kg at the centre of a side as indicated in the diagram. What will be the coefficient of restitution between disc and plate so that disc will stop after collision?

5 kg B

10 m/s

45°

A

38.

A small elastic ball is dropped from a height of 5 m onto a rigid cylindrical body (see figure) having a radius of 1.5 m. At what position on the x-axis does the ball land? y 0.6m

5m

Centre of Mass

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PHYSICS

39.

LOCUS

127

A small disc of mass m slides down a smooth hill of height h without initial velocity and gets onto a plank of mass M lying on the horizontal plane at the base of the hill (Fig.). Due to friction between the disc and the plank the disc slows down and beginning with a certain moment, moves in one piece with the plank. m

h M

(1) Find the total work performed by the friction forces in this process. (2) Can it be stated that the result obtained does not depend on the choice of the reference frame? 40.

A body of mass M (Fig.) With a small disc of mass m placed on it rests on a smooth horizontal plane. The disc is set in motion in the horizontal direction with velocity v. To what height (relative to the initial level) will the disc rise after breaking off the body M? The friction is assumed to be absent. m

v M

41.

A cart loaded with sand moves along a horizontal plane due to a constant force F coinciding in direction with the cart’s velocity vector. In the process, sand spills through a hole in the bottom with a constant velocity µ km/s. Find the acceleration and the velocity of the cart at the moment t, if at the initial moment t = 0 the cart with loaded sand had the mass m0 and its velocity was equal to zero. The friction is to be neglected.

42.

A flat car of mass ' m0 ' starts moving to the right due to a constant horizontal force F (figure). Sand spills on the flat car from a stationary hopper. The velocity of loading is constant and equal to µ kg / s. Find the time dependence of the velocity and the acceleration of the flat car in the process of loading. The friction is negligibly small.

f

43.

(a) A rocket set for vertical firing weighs 50 kg and contains 450 kg of fuel. It can have maximum exhaust velocity of 2 km/s. What should be its minimum rate of fuel consumption (i) to just lift it off the launching pad? (ii) to give it an acceleration of 20 m/s2 ? (b) What will be the speed of the rocket when the rate of consumption of fuel is 10 kg/s after whole of the fuel is consumed ? (Take g = 9.8m / s 2 ) 44. A rocket of initial mass m0 has a mass m0 (1 − t / 3) at time t. The rocket is launched from rest vertically upwards under gravity and expels burnt fuel at a speed u relative to the rocket vertically downward. Find the speed and height above the launching pad at t = 1. 45.

Find the mass of the rocket as a function of time, if it moves with a constant acceleration a, in absence of external forces. The gas escapes with a constant velocity u relative to the rocket and its mass initially was m0 .

Centre of Mass

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46.

LOCUS

128

A truck for washing down streets has a total mass of 10,000 kg when its tank is full. With the spray turned on, 40 kg of water per second issue from the nozzle with a velocity of 20 m/s relative to the truck at the 30° angle shown. If the truck is to accelerate at the rate of 0.6 m/s² when starting on a level road, determine the required attractive force P between the tires and the road when

30°

a Petrol

(a) the spray is turned on and (b) when the spray is turned off 47.

A mass ‘m’ is fastened to the end of a spring which is hanging vertically and the spring is allowed to stretch slowly. When the spring has stretched and the mass is stationary, it is noted that the spring has stretched by a distance ∆x. The gravitational potential energy transferred in the mass lowering is given by mg ∆x. The 1 1 elastic energy transferred to the spring is given by × (force × extension), i.e. mg ∆x. It would seem that 2 2 energy has not been conserved. Comment and explain.

48.

(a) In the arrangement shown below the friction coefficient between 1 . The frictional force is 3 sufficient to keep the system in equilibrium. Find how much the normal reaction between the wedge and floor is displaced form the centre of mass of wedge. All other surfaces are frictionless. Assume masses of both the bodies same and equal to ‘m’. Radius of sphere = R, length the triangular wedge and floor is µ =

60° h = 2R 60°

60°

3 R. 3 (b) If the radius of contact between the floor and the wedge also becomes frictionless, then find the speed of both the bodies when sphere hits the ground. Initial height of centre of mass of sphere = 2R as shown. of side of triangular wedge =

49.

Two identical equilateral triangular wedges of mass M rest on a smooth horizontal surface. A smooth sphere of mass m moving vertically down with a velocity v0 strikes the wedges symmetrically. If the coefficient of restitution is e find the velocities of the sphere and that of the wedges just after collision. m

v0

M 60°

Centre of Mass

60°

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50.

LOCUS

129

Two particles of equal masses 4M are initially at rest. A particle of mass M moving at speed u collides elastically with one of the larger balls. How many collisions occur? u M

4M

51.

52.

4M

A small ball is thrown between two vertical walls such that in the absence of the wall its range would have been 5d. The angle of projection is a. Given that all the collisions are perfectly elastic, find (a) maximum height attained by the ball. (b) Total number of collisions before the ball comes back to the ground, and (c) Point at which the ball falls finally. The walls are supposed to be very tall.

α O

d/2

A ball is released from rest relative to the elevator at a distance h1 above the floor. The speed of the elevator at the time of ball release is v0 . Determine the bounce height h2 of the ball with respect to the elevator.

a=g/4 h1

h2

v0

(a) if v0 is constant and (b) if an upward elevator acceleration a = g/4 begins at the instant the ball is released. The coefficient of restitution for the impact is e. 53.

Two parallel vertical walls of height ‘h’ stand on a horizontal plane. A ball is projected from one foot of one wall towards the other and after impact it just clears the top of the first wall. Prove that the point of impact with the second wall is at a depth h / (1 + e)2 below its top, where e = coefficient of restitution.

54.

A man is riding on a flat car travelling with a constant speed 4 m/ s. He wishes to throw a ball through a stationary hoop above the height of his hand in such a manner that the ball will move horizontally as it passes through the hoop. He throws the ball with a speed of −1 4 5 m/s at an angle of tan 2 with respect to himself. Find coordinates of the hoop assuming point of projection as origin and x-axis as horizontal. The mass of the system (Flatcar + man + ball) is 100 kg and mass of the ball is 1 kg. [Take g = 10 m/s²]

Centre of Mass

y

O

x

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55.

LOCUS

130

Two men A and B each of mass m sit on two light platforms at the ends of a light inextensible rope passing over a smooth pulley, A being higher than B by h. A ball of mass m/10 is placed in the hand of B. He instantly throws it to A so that it just reaches A. Calculate

A

(a) the distance moved by A before he catches the ball and

m

(b) the total distance travelled by A by the time he ceases to ascend. 56.

B m

m/10

A particle of mass m with kinetic energy K collides elastically with stationary particle of mass M. After collision both the particles diverge at an angle θ. Calculate the kinetic energy of the recoiled particle (M). m m M

θ M

57.

58.

A small particle of mass m moving with constant horizontal velocity u strikes a wedge shaped block of mass M on its inclined surface as shown in the figure. After collision particle starts moving up the inclined plane. Calculate the velocity of wedge immediately after collision. Also calculate the maximum height it can ascend on the wedge.

m

u

M

α

Two identical balls in contact on a table are in equilibrium. A third ball collides then simultaneously symmetrically and remains at rest after impact. Calculate coefficient of restitution between the balls.

Centre of Mass

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