COM,Collisions Master (1)

CONSERVATION OF LINEAR MOMENTUM AND

COLLISIONS

TOPICS TO BE COVERED

Centre of Mass

Linear Momentum

Conservation of Linear momentum

System of Variable Mass

Impulse and momentum

Collision

Types of Collision

Line of Impact

CONSERVATION OF LINEAR MOMENTUM

C

ENTRE OF

M

ASS

In translational motion each point on a body undergoes the same displacement as any other point

as time goes on. In this way the motion of one particle represents the motion of the whole body.

P

OSITION OF

C

ENTRE OF

M

ASS

(a)

System of two particles

Consider first a system of two particles m

₁

and m

₂

at distances x

₁

and x

₂

respectively, from some

point origin O. We define a point C, the centre of mass the system, as a distance x

_cm

from the

origin O, where x

_cm

is defined by

x

_cm

=

...(1)

x

_cm

can be regarded as mass -weighted mean of x

₁

and x

₂

.

(b)

System of many particles

(i) If m

₁

, m

₂

. . . , m

_n

, are along a straight line by definition,

x

_cm

=

=

...(2)

where M is total mass of the system.

2 1 2 2 1 1

m

m

x

m

x

m























i i i n n n

m

x

m

m

m

m

x

m

x

m

x

m

...

.

...

2 1 2 2 1 1

M

x

m

_{i i}



(ii) If particles do not lie in a straight line but lie in a plane, (suppose x-y plane) the centre of

mass C is defined and located by the coordinates x

_cm

and y

_cm

,

where x

_cm

=

=

y

_cm

=

=

...(3)

(iii) If the particles are distributed in space,

x

_cm

=

, y

_cm

=

, z

_cm

=

.(4)

So, position vector of C is given by

=

=

=

...(5)

Example-1

A circular plate of uniform thickness has a diameter of

56 cm. A circular portion of diameter 42 cm is removed

from one edge of the plate as shown in figure. Find the

centre of mass of the remaining portion.

Solution:

Let O be the center of circular plate and O

₁

, the center of

circular portion removed from the plate. Let O

₂

be the

center of mass of the remaining part.

Area of original plate =R

2

₌

_{= 28}

2

_cm

2

Area removed from circular part = r

2



















i i i n n n

m

x

m

m

m

m

x

m

x

m

x

m

...

...

2 1 2 2 1 1

M

x

m

_{i i}





















i i i n n n

m

y

m

m

m

m

y

m

y

m

y

m

...

...

2 1 2 2 1 1

M

y

m

_{i i}



X

m

1

m

2

x

2

x

1

Y

O

x

n

m

n

M

x

m

_{i i}



M

y

m

_{i i}



M

z

m

_{i i}



k

z

j

y

i

x

r

cm



cm

ˆ



cm

ˆ



cm

ˆ





i i i

m

r

m

M

r

m

_ii



A    O1 O O2 2

2

56















=

Let be the mass per cm

2

_{. Then}

mass of original plate, m = (28)

2

mass of the removed part, m

₁

= (21)

2

mass of remaining part, m

₂

= (28)

2

_{- (21)}

2

_{= 343}

Now the masses m

₁

and m

₂

may be supposed to be concentrated at O

₁

and O

₂

respectively. Their combined center of mass is at O. Taking O as origin we have from

definition of center of mass,

x

_cm

=

x

₁

= OO

₁

= OA - O

₁

A = 28 - 21 = 7 cm

x

₂

= OO

₂

= ?, x

_cm

= 0

0 =

x

₂

=

This means that center of mass of the remaining plate is at a distance 9 cm from

the center of given circular plate opposite to the removed portion.

(c)

Continuous bodies:

x

_cm

=

Similarly

y

_cm

=

and z

_cm

=

. . . (6)

Example-2

The density of a thin rod of length l varies with the distance x from one end as

. Find the position of centre of mass of rod.

Solution:

Here, s = area of cross section of rod.

2 2 2

)

21

(

2

42

cm

















×

×

×

m

m

1

m

2   





2 1 2 2 1 1

m

m

x

m

x

m





)

(

343

7

)

21

(

2 1 2 2

m

m

x













cm

9

343

7

441

343

7

)

21

(

2



_

_



_

_

















_{ }     _{dm M} xdm dm x m x m Lt i i i mi 1 0







_{ ydm} M dm ydm ₁







_{ zdm} M dm zdm ₁







_  rdm M dm dm r rcm 1  2 2 0

l

x

ρ

ρ

















































l l l l

l

x

x

x

l

x

x

x

0 2 2 0 0 2 2 0 0 0 cm

d

s

d

s

dm

dm

x

)

(

)

(

)

(

)

(

Therefore,

DISTINCTION B/W CENTER OF MASS AND CENTER OF GRAVITY

The position of the center of mass of a system depends only upon the mass and position of each

constituent particles,

i.e.,

...(1)

The location of G, center of gravity of the system, depends however upon the moment of the

gravitational force acting on each particle in the system (about any point, the sum of the moments for all

the constituent particles is equal to the moment for the whole system concentrated at G).

Hence, if g

_i

is the acceleration vector due to gravity of a particle P, the position vector r

_G

of the

center of gravity of the system is given by

...(2)

It is only when the system is in a uniform gravitational field, where the acceleration due to gravity

(g) is the same for all particles, that equation (2)

Becomes

In this case, therefore the center of gravity and the center of mass coincide.

If, however the gravitational field is not uniform and g

_i

is not constant then, in general equation

(2) cannot be simplified and

.

M

OTION OF THE

C

ENTRE OF

M

ASS

Assume that the total mass M of the system remains constant with time, then, for our fixed

system of particles

where r

_cm

is the position vector identifying the centre of mass in a particular reference frame.

4

3

x

cm

l



CM i i i

m r

r

m







(

)

G i i i i i

r





m g





r



m g

CM i i G i

m r

r

r

m









G CM

r



r

,

...

2 2 1 1 n n cm

m

r

m

r

m

r

r

M









Differentiating this equation W.R.T. time we get

M

...(7)

or

M

=

where v

₁

is the velocity of the first particle, etc., and dr

_cm

/dt (= v

_cm

) is the velocity of the centre of mass.

Differentiating equation (7) with respect to time we obtain

M

...(8)

=

,

where a

₁

is the acceleration of the first particle, etc., and dv

_cm

/dt (= a

_cm

) is the acceleration of the

centre of mass of system. Now, from Newton's second law, the force F

₁

acting on the first particle is

given by F

₁

= m

₁

a

₁

. Likewise, F

₂

= m

₂

a

₂

, etc. We can then write equation (8) as

M

=

...(9)

Internal forces are forces exerted by the particles on each other. However, from Newton's third

law, these internal forces will occur in equal and opposite pairs, so they contribute nothing to the sum.

This states that the centre of mass of a system of particles moves as through all the mass of the

system were concentrated at the centre of mass and all the external forces were applied at that point.

Concept:

Whatever may be the rearrangement of the bodies in a system, due to internal

forces (such as one part moving away from the other or an internal explosion

taking place, breaking a body into pieces).

(a) If the body was originally at rest, the C.M. will continue to be at rest.

(b) If before the change, the body had been moving with a constant velocity, it will continue to

move with a constant velocity and

In presence of external force if body had been moving with constant acceleration in a

particular trajectory, the C.M. will continue to move in the same trajectory, with the same

dt

r

d

m

dt

r

d

m

dt

r

d

m

dt

r

d

n n cm

_

_

_

_

...

2 2 1 1 cm V

m

1

v

1



m

2

v

2



...



m

n

v

n

dt

v

d

m

dt

v

d

m

dt

v

d

m

dt

v

d

n n cm

_

_

_

_

...

2 2 1 1 n n

a

m

a

m

a

m

₁ 1



₂ 2



...



cm

a



_F

₁



_F

₂



_{... F}



_n



_F

_internal



_F

_external ext cm

F

a

M



acceleration as if it had never experienced any explosion only if there is no change in

external force.

Example-3

In the arrangement shown in figure, m

_A

= 2 kg and m

_B

= 1kg. String is

light and inextensible. Find the acceleration of centre of mass of both the

blocks. Neglect friction everywhere.

Solution :

Net pulling force on the system is

or

Total mass being pulled is

or 3 kg

Now,

downwards

L

INEAR

M

OMENTUM

The momentum of a single particle is a vector defined as the product of its mass and its velocity .

That is

P

_{= m}

_...(10)

From Newton's second law of motion

A

B



m

_A



m

_B



g



2



1



g



g

B A m m 

A

B

a

a

3

g

mass

T otal

force

pulling

Net

a





     

3

a

2

1

a

1

a

2

m

m

a

m

a

m

a

B A B B A A COM

_



















9

g



v v

v



Thus, if m is constant, 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.

Suppose that instead of a single particle we have a system of n particles with masses m

₁

, m

₂

. . . etc, and

their velocities v

₁

, v

₂

etc respectively then the total momentum

in a particular reference frame is,

P

=

₁

+

₂

+ . . . +

_n

P

=

Also,

\

=

C

ONSERVATION OF

L

INEAR

M

OMENTUM

If the sum of the external forces acting on a system is 0. Then,

F

ext



=

= 0

or

= constant.

So, when the resultant external force acting on the system is zero, total vector momentum of

system remains constant. This is called as principle of the conservation of linear momentum. The

momentum of the individual particles may change but their sum remain constant if there is no net external

force.

dt

P

d

v

m

dt

d

dt

v

d

m

a

m

F











(

)

P

P

P

P

n n 2 2 1 1

v

m

v

...

m

v

m















cm

V

M



ext cm cm

_M

_a

_F

dt

V

d

M

dt

P

d













ext

F



_dt

P

d

dt

P

d

P

Example-4

A man of mass m climbs a rope of length L suspended below a balloon of mass M.

balloon is stationary with respect to ground. If the man begins to climb up the rope at a

speed v (relative to rope) in upward direction and with what speed (relative to ground)

with the balloon move?

Solution:

Balloon is stationary

No net external force acts on it.

Conservation of linear momentum of a system

(balloon + man) is valid

M

, where

=

M

where v

_mb

= velocity of man relative to the balloon (rope)

where v

_mb

= v v

_b

=

and directed opposite to that of the motion of the

man.

Example-5

Two identical buggies move one after the other due to inertia (without friction) with the

same velocity v

₀

. A man of mass m rides the rear buggy. At a certain moment man jumps

into the front buggy with a velocity u relative to his buggy. The mass of each buggy is M.

Find the velocities with which the buggies will move lster.

balloon

v

m

M

v

b

m

man

0





m b

m

v

v

v

m

v

mb



v

b

0

]

[







mb b b

m

v

v

v

m

M

v

m

v

b mb







m

M

mv



Solution:

Initial momentum of rear buggy = (M + m)v

₀

. The momentum of man when he jumps =

m(v

₁

+ u), where v

₁

is the velocity of buggy as he jumps.

By conservation of linear momentum

(M + m)v

₀

= Mv

₁

+ m(v

₁

+ u)

v

₁

(M + m) = (M + m)v

₀

- mu

v

₁

= v

₀

-

Initial momentum of front buggy = Mv

₀

Mv

₀

+ m(v

₁

+ u) = (M + m)v

₂

Mv

₀

+ m

Mv

₀

+ m

(M + m)v

₀

+

v

₂

= v

₀

+

S

YSTEM OF

V

ARIABLE

M

ASS

Till now we have studied system with constant mass. Now, here, we will study about the system

which gains or loses mass during its motion, e.g. in case of Rocket propulsion, its motion depends upon

the constant ejection of fuel from it.

Let us choose system of mass M (as shown in figure 1) whose centre of mass is moving with

velocity as seen from a particular reference frame. An external force

_ext

acts on the system.

v

1

M

M

v

2

After jumping

v

0

M

M

v

0

Before jumping

u

m

M

m



0 2

mu

v

u

u

(M

m)v

M

m



_

_



_

_



_







2 0

(

M

m

)

v

m

M

Mu

v





















2

)

(

M

m

v

m

M

mMu









M

m



2

mMu



At a time t later the configuration has changed to that shown in figure 2. A mass

D

M has been

ejected from the system, its centre of mass moving with velocity as seen by our observer. The mass of

the body is reduced to M -

D

M and the velocity of the centre of mass of the system is changed to v +

v.

\

{ for finite time interval t}

=

=

as

D

t 0;

approaches

approaches

-and is negligible as compared to v.

\

(as M is decreasing with time)

=

where

, is the relative velocity of the ejected mass with respect to main body.

Example-6

A rocket of initial mass m

₀

(including shell and fuel) is fired vertically at

time t = 0. The fuel is consumed at a constant rate q = dm/dt and is

expelled at a constant speed u relative to the rocket. Derive an

expression for the magnitude of the velocity of the rocket at time t,

x

y

cm

M

v

(1)

O

t

x

y

cm

M- M



v+ v



(2)

O

t+ t





M

u

cm

  ext

F = dP/dt

ext

F

P

t







 f i

P

P

t





[

)(

)

] [

]

ext

M

M v

v

Mu

Mv

F

t

 

   







[

(

)]

v

M

M

u

v

v

t

t



_{   }







v

t





dv

dt

t M   dt dM

(

)

ext

dv

dM

M

F

u

v

dt







dt

ext rel

dv

dM

M

F

u

dt





dt

F

ext



F

thrust rel

u

neglecting the resistance of the air and variation of acceleration due to

the gravity(g).

Solution:

At time t, the mass of the rocket shell and remaining fuel is m = m

₀

- qt,

and the velocity is v. During the time interval t, a mass of fuel m = q

t is expelled with a speed u relative to the rocket. Denoting by v

_e

the

absolute velocity of expelled fuel, we apply principle of impulse and

we write

(m

₀

- qt)v - g(m

₀

- qt) t = (m

₀

- qt - q t)(v + v) - q t(u - v)

Dividing throughout by t and letting t approach zero we obtain

-g(m

₀

- qt) = (m

₀

- qt)

Separating variables and integrating from t = 0, v = 0 to t = t, v = v

dv =

v = [u ln(m

₀

- qt) -

v = u ln

v

  

(m

0

– qt)v

v

W



t

=

(m

0

– qt - q



t)(v +



v)



mv

e

[W



t = g(m

0

– qt)



t]

[



mv

e

= q



t(u – v)]

+

   

qu

dt

dv

_

dt

g

qt

m

du

















0 t 0

gt]

gt

qt

m

m

_















0 0

I

MPULSE AND

M

OMENTUM

We know that force is related to momentum as

We can find the change in momentum of the body during a

collision (from

_i

to

_f

) by integrating over the time of collision

and assuming that the force during collision has a constant direction,

;

in which the subscripts i (= initial) and f(= final) refer to the times before and after the collision.

The integral of a force over the time interval during which the forces acts is called the impulse of the

force.

The impulse of this force, , is represented in magnitude by area under the force time curve.

C

OLLISIONS

In a collision a relatively large force acts on each colliding particle for a relatively short time.

CONSERVATION

OF

MONMENTUM

DURING

COLLISIONS

Consider now a collision between two particles, such as

those of masses m

₁

and m

₂

, shown in figure. During the brief

collision these particles exert large forces on one another. At

any instant F

₁

is the force exerted on particle 1 by particle 2 and

F

₂

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

oppositely directed.

The change in momentum of particle 1 results in from the collision is

dt

P

d

F



P

d

dt

F



P

_P

m

1

m

2

1

F

1

2

F

1

1

p



=

in which

is the average value of the force F

₁

during the time interval of the collision

D

t = t

_f

- t

_i

.

The change in momentum of particles 2 resulting from the collision is

2

p



₌

in which is the average value of the force F

₂

during the time interval of the collision

D

t = t

_f

- t

_i

.

If no other forces act on the particles, then

D

p

₁

and

D

p

₂

gives the total change in momentum for

each particle. But we have found that at each instant

= -

, and therefore

If we consider the two particles as an isolated system, total momentum of system is

,

And the total change in momentum of the system as a result of the collision is zero that is,

constant

Hence, if there are no external forces acting on the system, 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.

T

YPES OF

C

OLLISION

:

Collision b/w two bodies may be classified in two ways:

1.

Elastic collision and inelastic collisionn.

Collision is said to be elastic if both the bodies come to their original shape and size after the

collision, i.e., no fraction of mechanical energy remains stored as deformation potential energy in the

bodies, otherwise, it is called an inelastic collisionn. Thus in addition to the linear momentum, kinetic

energy also remains conserved before and after collision.

2.

Head on collision or oblique collision.

If the directions of the velocity of colliding objects are along the line of action of the impulses,

acting at instant of collision then it is called as head-on or direct collision. Otherwise impact is said to be

oblique or indirect or eccentric.

N

EWTON

’

S

L

AW OF

R

ESTITUTION

:

Experimental evidence suggests that the ratio of relative speed of separation to relative speed of

approach is constant for two given set of objects.

f i t 1 1 t

F dt

 

F t



1

F

f i t 2 2 t

F dt

 

F

t



1

F

F

2 1 2

p

p

  

1 2

P



p



p

1 2

P

p

p

0

     

p

1



p

2



The ratio e is called the coefficient of restitution and is constant for two particular objects.

In general,

0 £ e £ 1

e = 0, for completely inelastic collision, as both objects stick together.

e = 1, for an elastic collisionn.

HEAD-ON COLLISION

Consider two spheres A and B of mass m

₁

and m

₂

, which are moving in the same straight line and

to the right with known velocities v

₁

and v

₂

as shown in figure. If v

₂

is larger than v

₁

, particle B will

eventually strike the sphere A. Under the impact, the two spheres will deform and at the end of the period

of deformation, they will have the same velocity u as shown in figure. A period of restitution will then

place, at the end of which, depending upon the magnitude of the impact forces and upon the materials

involved, the two spheres either will have regained their original shape or will stay permanently

deformed. The purpose here is to determine the velocities v’

₁

and v’

₂

of the spheres at the end of the

period of restitution as shown in figure.

(i)

For elastic collision

Considering first the two spheres as a single system, we note that there is no impulsive, external

force. Applying law of conservation of linear momentum (COLM).

m

₁

v

₁

+ m

₂

v

₂

= m

₁

+ m

₂

...(i)

In an elastic collision kinetic energy before and after collision is also gets conserved. Hence,

...(ii)

Solving eqs. (i) and (ii) for

and

, we get

v

'

1

=

...(iii)

and

=

...(iv)

S

PECIAL

C

ASES

:

1.

If m

₁

= m

₂

, then from eqs. (iii) and (iv), we can see that

v

₁

¢

= v

₂

and v

₂

¢

= v

₁

e

approach

of

speed

lative

separation

of

speed

lative

_

Re

Re

1

v



v



₂

m

2

_v

2

m

1

v

1

(1)

B

A

m

2

_v



2

m

1

v



1

(2)

B

A

2 2 2 2 1 1 2 2 2 2 1 1

'

2

1

'

2

1

2

1

2

1

v

m

v

m

v

m

v

m







1

'

v

v

'

₂ 2 2 1 2 1 2 1 2 1

2

_v

m

m

m

v

m

m

m

m

































2

'

v

1 2 1 1 2 2 1 1 2

2

_v

m

m

m

v

m

m

m

m

































i.e., when two particles of equal mass collide elastically and the collision is head on, then they

exchange their velocities e.g.

2.

If m

₁

>> m

₂

and v

₁

= 0.

Then

With these two substitutions

We get the following two results:

v

₁

¢

»

0 and v

₂

¢

»

-v

₂

i.e., the particle of mass m

₁

remains while the particle of mass m

₂

bounces back with same speed

v

₂

.

3.

If

m

₂

>> m

₁

and v

₁

= 0

With the substitution and v

₁

= 0, we get the results

v

₁

¢

»

2v

₂

and v

₂

¢

»

v

₂

i.e., the mass m

₁

moves with velocity 2v

₂

while the velocity of mass m

₂

remains unchanged.

(ii)

For Inelastic Collision

The kinetic energy of particles no longer remains conserved. Suppose the velocities of two

particles of mass m

₁

and m

₂

before collision are v

₁

and v

₂

in the directions shown in figure. Let v

₁

¢

and

v

₂

¢

be the velocities after collision. Applying the law of COLM.

m

₁

v

₁

+ m

₂

v

₂

= m

₁

v

¢

₁

+ m

₂

v

¢

₂

...(i)

Applying Newton's Law of Restitution,

separation speed = e(approach speed)

v

¢

₁

- v

¢

₂

= e(v

₂

- v

₁

)

...(ii)

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

v

_{1 1}

=

...(iii)

and

v

₂

1

=

...(iv)

S

PECIAL

C

ASES

:

1.

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

v

₁

’

=

and

v

₂

’

=

which are same as eqs. (iii) and (iv).

2.

If collision is perfectly inelastic, i.e., e = 0, then

V

’

₁

= v

₂

’

=

= v

,

(say)

0

1 2



m

m

2 2 1 2 2 1 2 1 2 1

v

m

m

em

m

v

m

m

em

m



































2 2 1 2 2 1 2 1 2 1

_v

m

m

em

m

v

m

m

em

m



































2 2 1 2 1 2 1 2 1

2

_v

m

m

m

v

m

m

m

m

































1 2 1 1 2 2 1 1 2

2

_v

m

m

m

v

m

m

m

m

































2 1 2 2 1 1

m

m

v

m

v

m





3.

If m

₁

= m

₂

and v

₁

= 0, then

L

INE OF

I

MPACT

It is important to know the line of impact during

the collision. line of impact is line along which the

impulsive force act on the bodies. To find it draw tangent

at point of contact of two bodies. Draw a normal to

tangent at the point. This normal line is known to be line

of impact.

O

BLIQUE

C

OLLISION

:

Let us now consider the case when the velocities of the two colliding spheres are not directed

along the line of impact as shown in figure. As already discussed the impact is said to be oblique. Since

velocities v

¢

₁

and v

¢

₂

of the particles after impact are unknown in direction and magnitude, their

determinaion will require the use of four independent equations.

We choose as coordinate axes the n-axis along the line of impact, i.e. along the common normal

to the surfaces in contact, and the t-axis along their common tangent. Assuming that the sphere are

perfectly smooth and frictionless, we observe that the only impulses exerted on the sphere during

the impact are due to internal forces directed along the line of impact i.e. along the n axis. It follows

that

2 2 2 1

2

1

'

2

1

'

e

v

and

v

e

v

v











 













 



Line of impact v2 v2 v1 v1 n B A t m2v2 n B A t m1v1

+

B A t n _n B A t m1v1 m2v2

=

(i)

The component along the t axis of the momentum of each particle, considered separately, is

conserved; hence the t component of the velocity of each particle remains unchanged throughout.

We can write.

(v

₁

)

_t

= (v

¢

₁

)

_t

; (v

₂

)

_t

= (v

¢

₂

)

_t

(ii)

The component along the n axis of the total momentum of the two particles gets conserved. We

write.

m

₁

(v

₁

)

_n

+ m

₂

(v

₂

)

_n

= m

₁

(v

¢

₁

)

_n

+ m

₂

(v

¢

₂

)

_n

(iii)

The component along the n axis of the relative velocity of the two particles after impact is

obtained by multiplying the n component of their relative velocity before impact by the

coefficient of restitution.

(v

¢

₂

)

_n

- (v

¢

₁

)

_n

= e[(v

₁

)

_n

- (v

₂

)

_n

]

We have thus obtained four independent equations, which can be solved for the components of

the velocities of A and B after impact.

Note: Definition of coefficient of restitution can be applied along common normal direction in the case

of oblique collisions .

Example-7

A ball of mass m hits a floor with a speed v making an

angle of incidence

with normal. The coefficient of

restitution is e. Find the speed of reflected ball and the

angle of reflection.

Solution:

Suppose the angle of reflection is

and the speed after

collision is

. It is an oblique impact. Resolving the

velocity v along the normal and tangent, the components

are v cos

and v sin

. Similarly, resolving the

velocity after reflection along the normal and along the

tangent the components are -

cos

and

sin

.

Since there is no tangential action,

v sin

= sin

...(i)

Applying Newton's law for collision,

(-

cos

- 0) = -e(v cos

- 0)

v



cos

= ev cos

...(ii)

From equations (i) and (ii),

θ

v

v

q

q



q

v



θ

θ

v

q



v



q





q

v

q



v



q



q





q

q



2

_{= v}

2

_sin

2

_{+ e}

2

_v

2

_cos

2

= and tan

=

= tan

-1

_.

Example-8

A ball of mass m, moving with a velocity v along X-axis, strike 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

velocity of the other piece ?

Solution:

The total linear momentum of the balls before the collison is mv along the X-axis. After

the collision, momentum of the first ball = 0, momentum of the first piece = mv

₁

along

gthe Y-axis and momentum of the second piece = mv

₂

along its direction of motiion

where v

₂

is the speed of the second piece. These three should add to mv along the x-axis,

which is the intiial momentum of the system.

Taking components along the X-axis,

...(1)

& taking components along Y-axis

...(2)

From (1) and (2),



q

q





q













e

q

tan

Y

2m

X

m

v

Y

q

X

m

v

v

1

v

2 2

mv cos

q 

mv

2 1

mv sin

q 

mv

v

v

₁

/

tan

q



2 2 1 2

v

v

v

v





q



cos

OBJECTIVE PROBLEMS (SOLVED)

1.

A small cube of mass m slides down a circular path of radius R

cut into a larger block of mass M, as shown in figure. M rests on

table, and both blocks move without friction. blocks are initially at

rest, & m starts from top of path. velocity v of cube as it leaves

the block is

(a)

(b)

(c)

(d)

Solution:

From COLM,

From COE

.

Ans. (d)

2.

A block of metal weighing 2 kg is resting on a frictionless

plane. It gets struck by jet releasing water at a rate of 1 kg/s

and at a speed of 5 m/s. The initial acceleration of the block is

2mgR

M

2gR

2mgR

m



M

2MgR

m



M

m m

Mv



mv

2 2 M m

1

1

mgR

MV

mv

2

2





wmgR





2 2 2 m m 2

m

M

V

mV

m





m

2MgR

V

M

m







Block

2kg

(a)

m/s

2

_(b)

_m/s

2

(c)

m/s

2

_(d)

_m/s

2

Solution:

Force

Ans. (d)

3.

A ball of mass m moving with a speed u undergoes a head-on elastic collision with a ball

of mass nm initially at rest. fraction of incident energy transferred to the heavier ball is

(a)

(b)

(c)

(d)

.

Solution:

Let the velocities of the two balls, after collision, be v

₁

and v

₂

in the direction of u

- which we take as positive

momentum conservation mu + 0 = m v

₁

+ nm v

₂

e = 1 v

₂

– v

₁

= u

Solving these equations, we have

fraction of energy transferred

Ans. (d)

4.

An insect of mass m is initially at one end of the stick of length L and mass M,

which rests on a smooth horizontal floor. coefficient of friction b/w the insect and

stick is k. The minimum time in which the insect can reach the other end of the

stick is t.

(a) the centre of mass of the plank has velocity magnitudes

with

respect to horizontal floor at time t.

5

3

25

4

25

8

5

2

dm

V

dt



5 Kg

m

₂

s



f

2

a

2.5 m / s

m







n

1 n



2

n

(1 n)



2

2n

(1 n)



2

4n

(1 n)



2

2u

v

n 1





KE of ball of mass nm Incident KE  2 2

1

2u

(nm)

2

n 1

1

mu

2







_







2

4n

(n 1)





kmg

t

M













(b)

(c) the magnitude of the linear momentum of the insect at time ‘t’ is (kmgt) w.r.t.

horizontal floor

(d) all the above

Solution:

Acceleration of insect and the stick will be kg and

respectively, but in

opposite direction. Therefore, relative acceleration

Ans. (b)

5.

A strip of wood of mass M and length l is placed on a smooth horizontal surface.

An insect of mass m starts at one end of the strip and walks to the other end in

time t, moving with a constant speed

(a) the speed of the insect as seen from the ground is

(b) the speed of the insect as seen from the ground is

(c) the speed of the insect as seen from the ground is

(d) the total kinetic energy of the system is

Solution:

As the insect moves from one end to the other, the strip moves in the opposite

direction. The insect, therefore, gets to cover a distance less than and its speed <

Ans. (a)

The distance covered by the insect w.r.t ground

its speed

Ans. (b) is, wrong.

Ans. (c) is, therefore wrong.

2LM

t

k(M

m)g





m

kg

M

m

kg 1

M







_



_





2

1

m

kg 1

t

2

M









_



_





2 M

t

kg(M

m)







t



















m

M

m

M

t

l

m

t M

m







_











2

t

M

m

2

1

_













l

t

M

M

m





M

t M

m





The total KE (seen from ground frame)

.

Ans. (d) is wrong.

6.

A ball strikes on the ground at an angle of 30º from the

vertical and rebound at an angle of

from the

vertical as shown in the figure. If the coefficient of

restitution between ball and ground is

, then

is

(a) 30º

(b) 45º

(c) 60º

(d) 15º

Solution:

Velocity along the plane before collision is equal to after collision v sin 30°

Velocity perpendicular to plane after collision is equal to e v cos 30°

Therefore,

= 1

Ans. (b)

7.

A ball of mass 1 kg is dropped on a horizontal ground from a height of 128 meter. If

coefficient of restitution between ball and ground is 1/2, then height gained by

ball just after 3

rd

_{impact from the ground is}

(a) 64 meter

(b) 32 meter

(c) 16 meter

(d) 2 meter

Solution:

Veloicty of ball just before collision is

Velocity of ball just after collision is

Height attained by ball after Ist collision is [at maximum height veloicty is zero]

2 2

1

M

1

m

m

M

2

t M

m

2

t M

m











_

_



_

_













2 2

1

mM(M

m)

1

(M

m)

2 t

(M

m)

2

t



 

 



_{ }





_{ }



 

 



3

1/



r  30º

1

e

3



3

e

1

30

cos

ev

30

sin

v

tan

.

















45

V

2

1

v

eV

v

/



gh

2

gh

2

e

h

e

h

2





_;

Therefore, after 2nd collision

Therefore, after 3rd collision

Ans. (d)

8.

The linear momentum P of a particle varies with time as P = a + bt

2

_{, where a and b}

are constants. The net force acting on the particle is

(a) zero

(b) constant

(c) proportional to t

(d) proportional to t

2

Solution:

Ans. (c)

9.

A ball impinges directly on a similar ball at rest. The first ball is brought to rest by

the impact.If half the kinetic energy is lost by impact, what is the value of the

coefficient of restitution ?

(a)

(b) 1

(c)

(d)

Solution: Balls exchange there velocities in the case of a perfectly elastic collisionn. Therefore,

e = 1.

Ans. (b)

10.

A shell is fired from a cannon with a velocity v (m/s) at an angle with the

horizontal direction. At highest point in its trajectory it explodes into two pieces of

equal mas. 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)

(b)

(c)

(d)

Solution: Let be the velocity of second fragment. From conservation of linear momentum

Therefore,

Ans(a).

h

e

h





(

2

)

2

h

e

h





(

2

)

3

bt

2

dt

dp

F





2

2

1

2

1

2

3

q

q

cos

v

3

2

v

cos

q

q cos v 2 3 q cos v 2 3 v



v

q





m

v





m



v

cos

q



m

2

cos

q





3

v

cos

v

SUBJECTIVE PROBLEMS (SOLVED)

1.

Locate the centre of mass of a uniform semicircular rod of radius R and linear density

kg/m.

Solution: From the symmetry of the body we see that the CM must lie along the y axis, so x

_CM

= 0. In

this case it is convenient to express the mass element in terms of the angle , measured in

radians. The element, which subtends an angle d at the origin, has a length R d

and a

mass dm = R

d. Its y coordinate is y = R sin

.

Therefore, y

_CM

=

y

_CM

=

The total mass of the ring is M =

R

; Therefore, y

_CM

=

.

2.

Find the centre of mass of a uniform solid hemisphere of radius R and mass M with centre of

sphere at origin and the flat of the hemisphere in the x, y plane.

Solution: Let the center of the sphere be the origin and let the flat of the hemisphere lie in the x, y plane

as shown. By symmetry

Consider the hemisphere divided into a series of slices

parallel to x, y plane. Each slice is of thickness dz

The slice between z and (z + dz) is a disk of radius,

r =.

Let r be the constant density of the uniform sphere.

Mass of the slice, dm =

The value is obtained by

=

λ

θ

θ

λ

θ



ydm

_M

M R M R d R M 2 0 2 0 2 2 ] cos [ sin 1 

_

_

_q

_q

_



_

_q

 _



π

λ



R

2

.

0





y

x

2 2

z

R



 

2



2 2



r

dz

R

z

dz

 

 



y

z

x

R

r

z

z

M

dm

z

R



0

=

=

Since M =

Hence center of mass has positive coordinates as

.

3.

The balloon, the light rope and the monkey shown in figure are at rest in the

air. If monkey reaches top of rope, by what distance does balloon descend ?

Mass of the balloon – M, mass of the monkey = m and the length of the rope

ascended by the monkey = L.

Solution:

M

dz

z

z

R

R





0 3 2

)

(



R z z

z

z

R

M

 



























0 4 2 2

4

2



4 4

R

R

2

4

z

M







_



_







4

R

z

4M





3

2

V

R

3





  

_



_





4 3

R

3

z

R

2

8

4

R

3













_



_

















R

8

3

,

0

,

0

m L M

(

)

M

x

=

m L

-

x

mL

M

m

=

+

x

4.

A bullet of mass m strikes a block of mass M connected to a

light spring of stiffness k, with a speed v

₀

and gets

embedded into mass M. Find the loss of K.E. of the system

just after impact.

Solution: The process of impact of bullet and block is transient. Within a very short time of impact, the

compression of the spring is negligible. Therefore the corresponding spring force is

negligible. Even though it is external to the system (M + m), we can conserve its momentum

just before and after the impact (impact force is internal). Conservation of linear momentum

of bullet plus block just after and before impact yields

(M + m)V = mV

₀

V =

where V = common velocity of block & a bullet.

Therefore the loss of K. E. of the system



KE =

Putting V =

we obtain, KE = .

5.

A bullet of mass 10

–3

_{kg strikes an obstacle and moves at angle 60° to its original direction.}

If its speed also changes from 20 m/s to 10m/s during collision. Find the magnitude of

impulse acting on the bullet.

Solution: m = 10

–3

kg

Consider components of impulse along initial velocity

J

₁

= 10

–3

[–10cos 60° – (–20)]

J

₁

= 15 10

–3

_N.S

Similarly impulse perpendicular to initial velocity we have

J

₂

= 10

-3

_{[10 sin60 - 0] =}

The magnitude of resultant impulse is given by

m M mV  0 2 2 0

(

)

2

1

2

1

V

m

M

mV





M

m

mV



0 

2

(

)

2 0

m

M

MmV



3

5 3 10





N S

.

J =

J =

6.

A particle of mass M is attached with a string of length l and

is released from the position as shown in the figure. When

string becomes vertical particle strikes with the block of

mass M as shown in the figure. Calculate maximum

compression in the spring if collision between particle and

block is perfectly elastic

Solution: Velocity of left block just after collison

When compression is maximum then velocity of block will be same

Using conservation of energy

7.

A system of two blocks A and B are connected by an inextensible

massless string as shown. The pulley is massless and frictionless.

2 2 3 2 2 1 2

10

(15)

(5 3)









J

J

2

3 10





N S

. .

M

M

k

smooth

M

l

V

=

2g

l

2

V

V

2M

MV







2 2 2

V

M

2

2

1

MV

2

1

kx

2

1

_

_

_

2

M

MV

2g

2

4

=

l

-2

1

M2g

Mg

kx

Mg

2

=

-

2× 2

=

2

l

l

l

Mg

k

=

l

x