1. CONSERVATION OF LINEAR MOMENTUM
CENTRE OF MASS
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.
POSITION OF CENTRE OF MASS
(a) System of two particles
Consider first a system of two particles m1
and m2
at distances x1
and x2
respectively, from some
origin O. We define a point C, the centre of mass the system, as a distance xcm
from the origin O,
where xcm
is defined by
xcm
=
2
1
2
2
1
1
m
m
x
m
x
m


...(1)
xcm
can be regarded as mass -weighted mean of x1
and x2
.
(b) System of many particles
(i) If m1
, m2
. . . . . , mn
, are along a straight line, by definition,
xcm
=









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
 ...(2)
where M is total mass of the system.
(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 xcm
and ycm
,
where xcm
=









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

ycm
=









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
 ...(3)
X
m1
m2
x2 x1
Y
O xn
mn
(iii) If the particles are distributed in space,
xcm
=
M
x
m i
i
 , ycm
=
M
y
m i
i
 , zcm
=
M
z
m i
i
 ...(4)
So, position vector of C is given by
xcm
m1 m2
x1
x2
X
Y
O C

2. k
z
j
y
i
x
r cm
cm
cm
cm
ˆ
ˆ
ˆ 

 =


i
i
i
m
r
m
=
M
r
m i
i
 ...(5)
A



O1
O
O2
Illustration: A circular plate of uniform thickness has a diameter of 56
cm. Acircular 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 O1
, the center of
circular portion removed from the plate. Let O2
be the
center of mass of the remaining part.
Area of original plate = R2
= 
2
2
56






= 282
 cm2
Area removed from circular part =  r2
= 
2
2
2
)
21
(
2
42
cm








×
× ×
m
m1
m2
Let  be the mass per cm2
. Then
mass of original plate, m = (28)2

mass of the removed part, m1
= (21)2

mass of remaining part, m2
= (28)2
 - (21)2
 = 343 
Now the masses m1
and m2
may be supposed to be concentrated at O1
and O2
respectively.
Their combined center of mass is at O. Taking O as origin we have from definition of center
of mass,
xcm
=
2
1
2
2
1
1
m
m
x
m
x
m


x1
= OO1
= OA - O1
A = 28 - 21 = 7 cm
x2
= OO2
= ?, xcm
= 0
 0 =
)
(
343
7
)
21
(
2
1
2
2
m
m
x



 

x2
= cm
9
343
7
441
343
7
)
21
( 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:
xcm
= 



 





dm
x
M
dm
dm
x
m
x
m
Lt
i
i
i
mi
1
0
Similarly ycm
= 

  dm
y
M
dm
ydm 1
and zcm
= 

  dm
z
M
dm
zdm 1

3.  

 
 dm
r
M
dm
dm
r
rcm
1

. . . . . (6)
Illustration: Show that the centre of mass of a rod of mass M and length L lies midway between its ends,
assuming the rod has a uniform mass per unit length.
Solution: By symmetry, we see that yCM
= zCM
= 0 if the rod is placed along the x axis. Furthermore,
if we call the mass per unit length λ (the linear mass density), then λ = M/L for a uniform
rod. If we divide the rod into elements of length dx, then the mass of each element is dm =
λ dx. Since an arbitrary element of each element is at a distance x from the origin, equation
gives
xCM
=
M
L
dx
x
M
dm
x
M
L
L
2
1
1 2
0
0

 
 

x
O x
dx
dm
Because λ = M/L, this reduces to xCM
=
2
2
2
L
L
M
M
L







One can also argue that by symmetry, xCM
= L/2.
Brain Teaser: Can the centre of mass of a body be at a point outside the body.
DISTINCTION BETWEEN 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., CM
i
i
i
m r
r
m





...(1)
The location of G, the 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 gi
is the acceleration vector due to gravity of a particle P, the position vector rG
of the
center of gravity of the system is given by
( )
G i i i i i
r m g r m g
  
 
   
...(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 CM
i i
G
i
m r
r r
m
 



 
In this case, therefore the center of gravity and the center of mass coincide.
If, however the gravitational field is not uniform and gi
is not constant then, in general equation (2)
cannot be simplified and G CM
r r

 
.
Brain Teaser: The weight mg of an extended body is generally shownin a diagramto act through the
centre ofmass does it mean that the earthdoes not attract other particles.
Y

4. MOTION OF THE CENTRE OF MASS
Assume that the total mass M of the system remains constant with time, then, for our fixed system of
particles
,
........
2
2
1
1 n
n
cm r
m
r
m
r
m
r
M 



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
M
dt
r
d
m
dt
r
d
m
dt
r
d
m
dt
r
d n
n
cm



 ......
2
2
1
1 ...(7)
or M cm
V

= n
n v
m
v
m
v
m 

 .....
2
2
1
1
where v1
is the velocity of the first particle, etc., and drcm
/dt (= vcm
) is the velocity of the centre of mass.
Differentiating equation (7) with respect to time, we obtain
M
dt
v
d
m
dt
v
d
m
dt
v
d
m
dt
v
d n
n
cm



 ........
2
2
1
1 ...(8)
= n
n a
m
a
m
a
m 

 ..........
2
2
1
1
,
where a1
is the acceleration of the first particle, etc., and dvcm
/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
= m1
a1
. Likewise, F2
= m2
a2
, etc. We can then write equation (8) as
M cm
a

= 1 2 n
F F ....... F F F
internal external
    
    
...(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 that they contribute nothing to the sum.
 ext
cm F
a
M 
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 maybe the rearrangement ofthe bodies inasystem, due to internalforces (such
as one part moving awayfromthe other or aninternalexplosion taking place, breaking a
bodyinto 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 par-
ticular trajectory, the C.M. will continue to move in the same trajectory, with the same accelera-
tion as if it had never experienced any explosion only if there is no change in external force.
Illustration : Two blocks of masses m1
and m2
connected by a weightless spring of stiffness k rest on a
smooth horizontal plane. Block 2 is shifted a small distance x to the left and released. Find
the velocity of the centre of mass of the system after block 1 breaks off the wall.
Solution: We know that the potential energy of compression

5. =
2
2
1
kx
When the block m1
breaks off from the wall the spring has its unstretched length and the
kinetic energy of the block m2
is given by
2
2
2
2
2
1
2
1
kx
v
m 
2
2
2
2
m
kx
v  m1 m2
k
v2
= x
2
m
k
For center of mass
xcm
=
2
1
2
2
1
1
m
m
x
m
x
m


The distances x1
and x2
are measured from the wall.
dt
dx
m
m
m
dt
dx
m
m
m
dt
dxcm 2
2
1
2
1
2
1
1




At start 0
1

dt
dx

2
2
1
2
2
2
1
2
m
k
m
m
x
m
v
m
m
m
dt
dxcm




Velocity of center of mass of system =
2
1
2
m
m
km
x

LINEAR MOMENTUM
The momentum of a single particle is a vector P defined as the product of its mass and its velocity
v . That is
P = mv

...(10)
From Newton's second law of motion
dt
P
d
v
m
dt
d
dt
v
d
m
a
m
F




 )
(
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 m1
, m2
. . . etc,
and their velocities v1
, v2
etc respectively then the total momentum P in a particular reference frame is,

6. P = P 1
+ P 2
+ . . . . . + P n
n
n
2
2
1
1 v
m
......
v
m
v
m







P = cm
V
M

Also, ext
cm
cm
F
a
M
dt
V
d
M
dt
P
d 





 ext
F

=
dt
P
d
CONSERVATION OF LINEAR MOMENTUM
If the sum of the external forces acting on a system is zero. Then,
ext
F

=
dt
P
d
= 0
or P = constant.
So, when the resultant external force acting on the system is zero, the total vector momentum of the
system remains constant. This is called as principle of the conservation of linear momentum. The momen-
tum of the individual particles may change, but their sum remains constant if there is no net external force.
Brain Teaser: Ameteorite burnsinthe atmosphere beforeit reaches earth’s surface. What happens to its
momentum?
Brain Teaser: When a ballis thrown up, the magnitude ofits momentumdecreases and then increases.
Does thisviolate the conservationofmomentumprinciple.
Illustration: A man of mass m climbs a rope of length L suspended below a balloon of mass M. The
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.
 The conservation of linear momentum of the system
(balloon + man) is valid
 M 0

 m
b v
m
v , where
m
v
 = b
mb v
v 
balloon
vm
M
vb
m
man
 M 0
]
[ 

 b
mb
b v
v
m
v
where vmb
= velocity of man relative to the balloon (rope)

m
M
v
m
v
mb
b



where vmb
= v  vb
=
m
M
mv

and directed opposite to that of the motion of the man.

7. Illustration: 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 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 afterwards.
Solution: Initial momentum of rear buggy = (M + m)v0
. The momentum of man when he jumps = m(v1
+ u), where v1
is the velocity of buggy as he jumps.
By the conservation of linear momentum
(M + m)v0
= Mv1
+ m(v1
+ u)
 v1
(M + m) = (M + m)v0
- mu
v1
= v0
- u
m
M
m

v1
M v2
M
After jumping
Initial momentum of front buggy = Mv0
Mv0
+ m(v1
+ u) = (M + m)v2
Mv0
+ m 0 2
mu
v u u (M m)v
M m
 
   
 

 
v0
M v0
M
Before jumping
 Mv0
+ m 2
0 )
( v
m
M
m
M
Mu
v 









 (M + m)v0
+ 2
)
( v
m
M
m
M
mMu



 v2
= v0
+
 2
m
M
mMu

Brain Teaser: Where is the energy source ?
In order to lift a body above the ground work must be done to increase the body's potential
energy. This work comes from different sources. For example, an elevator draws its en-
ergy from the main power network, an aircraft takes off utilizing the energy developed when
fuel is oxidized (burnt) is its engine.
But what is the source of the energy used to raise stratospheric and meteorological sounding
balloons having no engines ?
SYSTEM OF VARIABLE MASS
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 F

ext
acts on the system.

8. x
y
cm
M
v
(1)
O
t
x
y
cm
M- M

v+ v

(2)
O
t+ t

M
u
cm
At a time  t later the configuration has changed to that shown in figure 2. A mass 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 - M and the velocity of the centre of mass of the system is changed to v +  v.
.
ext
F = dP/dt
  
 ext
F
P
t





{ for the finite time interval  t}
=
f i
P P
t


 
[ )( ) ] [ ]
ext
M M v v Mu Mv
F
t
      


   

= [ ( )]
v M
M u v v
t t
 
   
 

  
as t  0;
v
t



approaches
dv
dt

t
M


approaches -
dt
dM
and  v
 is negligible as compared to v.
.
 ( )
ext
dv dM
M F u v
dt dt
  
   
(as M is decreasing with time)
ext rel
dv dM
M F u
dt dt
 
  
= ext thrust
F F

 
where rel
u

, is the relative velocity of the ejected mass with respect to main body.
.
v
Illustration: A rocket of initial mass m0
(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, neglecting the resistance of
the air and variation of acceleration due to gravity.
Solution: At time t, the mass of the rocket shell and remaining fuel is m = m0
- 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 ve
the absolute
velocity of expelled fuel, we apply the principle of impulse and

9. we write
(m0 – qt)v Wt =
(m0 – qt - qt)(v + v)
mve
[Wt = g(m0 – qt)t]
[mve = qt(u – v)]
+
(m0
- qt)v - g(m0
- qt)  t = (m0
- qt - q  t)(v +  v) - q  t(u - v)
Dividing throughout by  t and letting  t approach zero, we obtain
-g(m0
- qt) = (m0
- qt) qu
dt
dv

Separating variables and integrating from t = 0, v = 0 to t = t, v = v
dv = dt
g
qt
m
du










0
dt
g
qt
m
au
dv
t
v

 










0 0
0
 v = [u ln(m0
- qt) - t
0
gt]
 v = u ln gt
qt
m
m










0
0
IMPULSE AND MOMENTUM
We know that the force is related to momentum as
dt
P
d
F   P
d
dt
F 
We can find the change in momentum of the body during a
collision (from P i
to P f
) by integrating over the time of collision and
assuming that the force during collision has a constant direction,

 


f
f
i
t
t
P
P
i
f dt
F
P
d
P
P
1
;
F(t)
O t
ti tf
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, dt
F
f
i
t
t
 , is represented in magnitude by area under the force - time curve.
0
[W t g(m qt) t]
   
e
[ mv q t(u v)]
   
0
(m qt q t)(v v)
    

10. Brain Teaser: Can the impulse ofa force be zero, even ifthe force is not zero? Explain whyor whynot
Brain Teaser: A man who falls froma height on a cement floor receives more injurythanwhen he falls
fromthe sameheight on a heap ofsand. Explain.
COLLISIONS
In a collision a relatively large force acts on each colliding particle for a relatively short time.
CONSERVATION OF MOMENTUM DURING COLLISIONS
m1
m2
1
F 2
F
Consider now a collision between two particles, such as
those of masses m1
and m2
, shown in figure. During the brief
collision these particles exert large forces on one another. At any
instant F1
is the force exerted on particle 1 by particle 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 oppositely
directed.
The change in momentum of particle 1 resulting from the collision is
1
p


=
f
i
t
1
1
t
F dt F t
 



in which 1
F is the average value of the force F1
during the time interval of the collision t = tf
- ti
.
The change in momentum of particles 2 resulting from the collision is
2
p


=
f
i
t
2
2
t
F dt F t
 



in which 2
F is the average value of the force F2
during the time interval of the collision t = tf
- ti
.
If no other forces act on the particles, then p1
and p2
gives the total change in momentum for each
particle. But we have seen that at each instant 1
F = - 2
F , and therefore
1 2
p p
  
 
If we consider the two particles as an isolated system, the total momentum of the system is
1 2
P p p
 
  
,
And the total change in momentum of the system as a result of the collision is zero, that is,
1 2
P p p 0
     
  
 1 2
p p
 
 
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.

11. TYPES OF COLLISION:
Collision between two bodies may be classified in two ways:
1. Elastic collision and inelastic collision.
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 collision. Thus in addition to the linear momentum, kinetic energy also remains con-
served 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 the instant of collision then it is called as head-on or direct collision. Otherwise the impact is said to be
oblique or indirect or eccentric.
NEWTON’S LAW OF RESTITUTION:
Experimental evidence suggests that the ratio of relative speed of separation to relative speed of
approach is constant for two given set of objects.
e
approach
of
speed
lative
separation
of
speed
lative

Re
Re
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 the objects stick together.
e = 1, for an elastic collision.
HEAD-ON COLLISION
Consider two spheres A and B of mass m1
and m2
, which are moving in the same straight line and to
the right with known velocities v1
and v2
as shown in figure. If v2
is larger than v1
, particle B will eventually
strike the sphere A. Under the impact, the two spheres will deform and at the end of the period of deforma-
tion, 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).
m1
v1
+ m2
v2
= m1 1
v + m2 2
v ...(i)
m2
v2
m1
v1
(1)
B A
m2
v2
m1
v1
(2)
B A
In an elastic collision kinetic energy before and after collision is also conserved. Hence,
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 

 ...(ii)

12. Solving eqs. (i) and (ii) for 1
'
v and 2
'
v , we get
1
'
v = 2
2
1
2
1
2
1
2
1 2
v
m
m
m
v
m
m
m
m




















...(iii)
and 2
'
v = 1
2
1
1
2
2
1
1
2 2
v
m
m
m
v
m
m
m
m




















...(iv)
SPECIAL CASES :
1. If m1
= m2
, then from eqs. (iii) and (iv), we can see that
v1
 = v2
and v2
 = v1
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 m1
>> m2
and v1
= 0.
Then 0
1
2

m
m
With these two substitutions
We get the following two results:
v1
  0 and v2
  -v2
i.e., the particle of mass m1
remains while the particle of mass m2
bounces back with same speed
v2
.
3. If m2
>> m1
and v1
= 0
With the substitution and v1
= 0, we get the results
v1
  2v2
and v2
  v2
i.e., the mass m1
moves with velocity 2v2
while the velocity of mass m2
remains unchanged.
(ii) For Inelastic Collision
The kinetic energy of particles no longer remains conserved. Suppose the velocities of two particles
of mass m1
and m2
before collision are v1
and v2
in the directions shown in figure. Let v1
 and v2
 be the
velocities after collision. Applying the law of COLM.
m1
v1
+ m2
v2
= m1
v1
+ m2
v2
...(i)
Applying Newton's Law of Restitution,
separation speed = e(approach speed)
v1
- v2
= e(v2
- v1
) ...(ii)
Solving eqs. (i) and (ii), we get
v1
 = 2
2
1
2
2
1
2
1
2
1
v
m
m
em
m
v
m
m
em
m





















...(iii)
and v2
 = 1
2
1
1
1
2
2
1
1
2
v
m
m
em
m
v
m
m
em
m





















...(iv)

13. SPECIAL CASES:
1. If collision is elastic, i.e. e = 1, then
v1
 = 2
2
1
2
1
2
1
2
1 2
v
m
m
m
v
m
m
m
m




















and v2
 = 1
2
1
1
2
2
1
1
2 2
v
m
m
m
v
m
m
m
m




















which are same as eqs. (iii) and (iv).
2. If collision is perfectly inelastic, i.e., e = 0, then
v1
= v2
 =
2
1
2
2
1
1
m
m
v
m
v
m


= v (say)
3. If m1
= m2
and v1
= 0, then
2
2
2
1
2
1
'
2
1
' v
e
v
and
v
e
v 




 






 

LINE OF IMPACT
Line of impact
v2
v2
v1
v1
n
B
A
t
It is important to know the line of impact during the
collision. The lineof impact is the line along which the impul-
sive force act on the bodies. To find it draw the tangent at
the point of contact of the two bodies. Draw a normal to the
tangent at the point. This normal line is known as line of
impact.
OBLIQUE COLLISION :
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 per-
fectly 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
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 con-
served; hence the t component of the velocity of each particle remains unchanged. We can write.
(v1
)t
= (v1
)t
; (v2
)t
= (v2
)t

14. (ii) The component along the n axis of the total momentum of the two particles is conserved. We write.
m1
(v1
)n
+ m2
(v2
)n
= m1
(v1
)n
+ m2
(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 restitu-
tion.
(v2
)n
- (v1
)n
= e[(v1
)n
- (v2
)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 .
v
v
 
Illustration: A ball of mass m hits a floor with a speed v making an
angle of incidence θ with normal. The coefficient of resti-
tution 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 v. It is an oblique impact. Resolving the ve-
locity v along the normal and tangent, the components are
v cosθ and v sinθ. Similarly, resolving the velocity after
reflection along the normal and along the tangent the com-
ponents are - vcos 
 and vsin 
 .
Since there is no tangential action,
v sin  = vsin 
 ...(i)
Applying Newton's law for collision,
(- vcos 
 - 0) = -e(v cos - 0)
 vcos 
 = ev cos ...(ii)
From equations (i) and (ii),
v2
= v2
sin2
 + e2
v2
cos2

v = 
 2
2
2
2
2
cos
sin v
e
v 
v = 2 2 2
(v sin e cos )
   and tan 
 =
e

tan

 = tan-1 





e

tan
.
d
B
v
A
Illustration: A disc A of radius r moving on perfectly smooth
surface at a speed v undergoes an elastic collision
with an identical stationary disc B. Find the veloc-
ity of the disk B after collision if theimpact param-
eter is d as shown in figure.

15. Solution: One of the discs is at rest before impact. After the impact its velocity will be in the direction
of the center line at the moment of contact because this is the direction in which the force
acted on it.
Thus, sin  2
=
r
d
2
 1
+  2
=
2

Since the masses of both disks are equal, the triangle of momenta turns into triangle of
velocities.
we have V1
= v cos  1
= v sin  2
1
v1
v2
2
v1 900
v2
v
=
r
vd
2
v2
= v cos  2
= v 2
2
4
1
r
d
