# Collision

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## Content

COLLISION
SESSION – 1 & 2
AIM

To explain concept of Collision.

To explain types of Collision.

Coefficient of restitution.

To study the energy transmission

COLLISION:
Collision is said to occur if two bodies interact strongly for very
short interval of time. e.g., Hitting the ball, accident of two vehicles
etc.
m1

m2

m1

m2

u1

u2

v1

v2

Large force is imparted in short interval of time
Impulse on A = Impulse on B
F × t = −F × t
(m v − m u ) = −(m v − m u )
m u +m u =m v +m v
Momentum before collision = Momentum after collision

TYPES OF COLLISIONS
Elastic 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. Thus in addition to the linear momentum, kinetic energy
also remains conserved before and after collision for Elastic
collision.
m1

m2

m1

m2

u1

u2

v1

v2

m u +m u =m v +m v
m u + m u = m v + m v
Inelastic collision:
Collision is said to be inelastic if both the bodies do not retain their
original shape and size after the collision, i.e. some

fraction

of

mechanical energy remains stored as deformation potential energy
in the bodies. Thus only linear momentum remains conserved but
kinetic energy never remains conserved before and after collision
for Inelastic collision.
m u + m u = m v + m v
1
1
1
1
m u + m u ≠ m v + m v
2
2
2
2

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.
COEFFICIENT OF RESTITUTION (NEWTONS EXPERIMENTAL
LAW OF RESTITUTION):
The ratio of relative speed of separation to relative speed of
approach is constant for two given set of objects. This ratio “e” is
called the coefficient of restitution and is constant for two
particular objects.
Relativ speed of separation
v −v
=e=
Relative speed of approach
u −u
1] For perfect Elastic collision e = 1
2] For perfect In–elastic collision e = 0 (bodies stick together)
Value of e changes between 0 ≤ e ≤ 1
m1

m2

m1

m2

u1

u2

v1

v2

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 u1
and u2 as shown in figure. The velocities become v1 and v2 of the
spheres after collision as shown in figure.

Conserving momentum of the colliding bodies before and after the
collision
m1u1 + m2u2 = m1v1 + m2v2

........(i)

Applying Newton’s experimental aw
We have v − v = −(u − u ) {e = 1 for Elastic Collision}
v = v − (u − u )

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

Putting (ii) in (i) we obtain
m u + m u = m v + m { v − (u − u )}
(m − m )u + 2m u = (m + m )v
v =
v =

(

)
(

(

)

+(

)

)

+(

)

)
(

SPECIAL CASES
1]

Identical bodies
If m1 = m2 =m v1 = u2 and v2 = u1
i.e., when two particles of equal mass collide elastically and the
collision is head on, then they exchange their velocities.
=
v =

(

v =
v =u
v =u

)

u +

u +

=
u =0+u
u =u +0

2]

Lighter body into heavier body from behind
v =

u +

v =

u +

v = −u + 2u
u = 0 + u v = −u + 2u

velocity of the lighter body
v =u

Heavier body contains to more with almost the same

velocity

3]

,

=

v =

u +

u v = −u → Lighter body

rebounds with same velocity
v =

+

u v = u = 0 →

Heaviour body remains at rest

4]

A heavy body of the hits a light body from behind With these
approximation the final velocity of the bodies are v = u and
v = 2u − u
Heavy body contains to more with almost same velocity

v =

u +

u = u ⇒ v = u

2m u
m −m
v =
+
m +m
m +m

5]

,

u = 2u − u ⇒ v = 2u − u

=

If the lighter body were kept at rest
Then v = u it will contains to to move almost same velocity
v = 2u lighter body will fly a way with a velocity donate the
velocity o fthe heavier body
v =
v =

u +
+

u ⇒ v = u

(

)

u ⇒ v = 2u

TRANSFER OF KINETIC ENERGY IN A HEAD –ON ELASTIC
COLLISION:
v =

u +

u ⇒ v =

u

v → velocity retained after collision
v =

u +

v → Velocity transfered

u ⇒v =

u

Fraction of velocity transferred

=

Momentum retained by m = p = m v = m
p =

u

p

Momentum transferred by m to m p = m v =
=

u

p

Transfer of kinetic energy in a head –on elastic collision
i]

v1 =

u ,v =

ii]

KE of m1 after collision.

u ; K=

K1 = m v = m

m u

u ;

∴ Fraction of KE retained by m1 is

K1 = K
=

iii] KE of m2 after collision, K = m v = m
K =(

)

m u ; K2 =

(

)

(

)

u

K

Fraction of KE transferred by m1 to m2 is

=

(

)

Velocity

2u
V =
1+n

For

v

to

be Target

maximum n must should
be minimum
n=

be

very light

→0
m ≪m

Momentum

p
2m u
=
1
1+n

For p to be

Target

be
maximum → must should
very heavy
be minimum ,
n→maximum
n=

→∞

m ≫m
Kinetic
energy

k
For k to be Both should
4nk
maximum (1 − n) be of same
=
(1 − n) + 4n must be minimum mass
1−n=0
n=1
m =m

=1

CLASS EXERCISE
1]

A body of mass m moving with speed v collides elastically head
on with another body of mass 2m which is initially at rest. The
ratio of kinetic energy of colliding body before and after
collision will be
a) 1: 1

2]

b) 2: 1

c) 4: 1

d) 9: 1

A body of mass 6kg travelling with a velocity 10m/s collides
head – on and elastically ----- a body of mass 4kg travelling at a
spe……. 5m/s in opposite direction. The velocity of second
body after the collision is
a) 0 m/s.

3]

b) 6 m/s

c) 8 m/s

d) 13 ms – 1

If a ball of mass 0.4 kg moving with a velocity of 3 ms–1 collides
elastically with another b….. of mass 0.6 kg which is at rest, find
the velocities after collision.

4]

a) – 0.6ms–1, 2.4ms–1

b) 2.4 ms–1, –0.6 ms….

c) 0.6 ms–1, –2.4 ms–1

d) –24.4 ms–1, 0.6 m……

A particle experiences a perfectly elastically collision with a
stationary particle. Then --- ratio of their masses, if after this
head–collision the particles fly apart in opposite directions
with equal velocities.
a) 1:3

b) 3 :1

c) 3 : 2

d) 2 : 3

5]

A body of mass 1 kg makes an elastic collision with another
body at rest and continues move in the

original

direction

after collision with a velocity equal to 1/5th of its origin
velocity. Find the mass of the second body.
a)
6]

b)

c) kg

d)

A sphere of mass 0.3 kg moving with a velocity of 4 m/s
collides with another sphere of mass 0.5 kg which is at rest.
Assuming the collision to be elastic, their velocities after the
impact are

7]

a) 4 m/s and 0 ms–1

b) –1 m/s and 3 m/s

c) 2 m/s and 2 m/s

d) 4 m/s and 8 ms–1

A steel sphere of mass 100 gm moving with a velocity of 4 m/s
collides with a dust particle elastically moving in the same
direction with a velocity of 1 m/s. The velocity of the dust
particle
after the collision is (dust particle is of mass neglisiable)
a) 8 m/s

8]

b) 7 m/s

c) 6 m/s

d) 9 m/s

A neutron moving with a certain kinetic energy collides head
on with an atom of mass number A.
The fractional inetic
energy retained by it is
a)

b)

c)

d)

9]

A perfectly elastic ball P1 of mass m moving with velocity v
collides elastically with three exactly similar balls P2, P3, P4
lying on a smooth table as shown. Velocities of the four balls
after the

collision are

a) v, v, v, v,

b) 0, 2v, 3v, 4v c) 0, 0, 0, v

d) 0, 0, 0, 0

10] Two spheres A and B moving in opposite directions have
velocities of 10ms–1 and 20ms–1. The two spheres collide with
each other elastically. If A continues to move in the same
direction at 4 ms–1, the velocity of sphere B just after the
collision is
a) 34 m/s in the same direction
b) 34 m/s in the opposite direction
c) 26 m/s in the same direction
d) 26 m/s in the opposite direction
11] A massive ball moving with speed v collides head on with a fine
ball having mass very much smaller than the mass of the first
ball. The collision is elastic. Then immediately after the impact,
the second ball will move with a speed approximately equal to
a) v

b) 2v

c) v/3

d) Infinite

SESSION - 3
AIM

To study In–elastic 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 u1
and u2 as shown in figure. The velocities become v1 and v2 of the
spheres after collision as shown in figure. If coefficient of rertitution
is “e”
m1

m2

m1

m2

u1

u2

v1

v2

Conserving momentum of the colliding bodies before and after the
collision
m1u1 + m2u2 = m1v1 + m2v2

........(i)

Applying Newton’s experimental aw
We have

= −e

v2 = v1 – e (u − u )

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

Putting (ii) in (i) we obtain
m u + m u = m v + m {v − e}
(m − em )u + (1 + e)m u = (m + m )v
v =

u +

m u

v =

u +

m u

Special cases
1]

2]

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

u +

m u ;

v =

u +

m u

If collision is perfectly inelastic, i.e., e = 0, then bodies move
together after collision.
v =

+

v =

+

v =v =V=

3]

If m1 = m2 Identical Bodies
v =

(

)

u +

(

)

u ; v =

(

)

u +

(

)

u

LOSS IN KINETIC ENERGY FOR PERFECTLY IN-ELASTIC
COLLISION
KE = m u + m u
KE = (m + m )v
Losses in KE = KE − kE
=

mu + M u

− (m + m )v

= [mu + m u ] − (m + m )
= m u +m u −
= [m u + m m u + m m u + m u − m u − m u −
2m m u u ]
=

[u +u − 2u u ]

Δ E =

[u − u ]

If the collision is not perfect inelastic collision
Loss in KE =

(u − u ) (1 − e )

Example 1:
Two bodies A and B of masses m and 2m respectively are placed on
a smooth floor. They are connected by a light spring of stiffness k. A
third body C of mass m moves with velocity v0 along the line joining
A and B and collides elastically with A. If I0 be the natural length of
the spring, then find the minimum separation between the blocks A
and B.
m
C

m
v0

A

2m

Solution:
Initially there will be collision between C and A which is elastic,
therefore, by using momentum conservation
mv0 = mvA + mvC or

v0 = vA + vC

Since e = 1, v0 = vA – vC
Solving the above two equations, vA = v0 and vc = 0
Now A will move and compress the spring which in turn accelerate
B and retard A and finally both A and B will move with same
velocity v.
Since net external force is zero, therefore momentum of the system
(A and B) is conserved.
Hence mv0 = (m + 2m)v  v = v0/3
If x0 is the maximum compression, then using energy conservation,
mv = (m + 2m)v + kx
⟹ mv = (3m)
⟹x =v

+ kx

Hence minimum distance D = I0 – x0 = I0 – v0

BALLISTIC PENDULUM
A bullet of mass m moving with a horizontal velocity u strikes a
stationary block of mass M suspended by a string of length L. The
bullet gets embedded in the block. What is the maximum angle
made by the string after impact.

L

L

h
m
u

Solution:
Let v be the combined velocity of the bullet-block system just after
collision, then by conservation of linear momentum.
mu = (m + M)V

..... (1)

After collision the KE of the bullet-block system gets converted into
potential energy.
∴ (m + M)V = (m + M)gh
Putting this value in equation (1), mu = (
or ℎ =
Form the figure, cos  =

=

1−

=1−

+

) 2 ℎ

CLASS EXERCISE
1]

When a U-238 nucleus, originally at rest decayed by emitting
an -particle, say with a speed of v m/s, the recoil speed of the
residual nucleus is (in m/s)
a)

2]

b) −

c)

d) −

A bullet of mass m is fired into a block of wood of mass M
which hangs at the end of pendulum and gets embedded in it.
When the bullet strikes, the block of wood, the pendulum
swings with a maximum rise equal to R, then the velocity of the
bullet is
a)

3]

(

)

b)

(

)

c)

(

)

d) (m + M)gR

A particle of mass 4m at rest explodes in to 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 is
a) 1/2 mv2

b) 1/3 mv2

c) 3/2 mv2

d) 2/3 mv2

4]

A shell initially at rest explodes into two pieces of equal mass.
The two pieces will:
a) At rest
b) Move with different velocities in different direction
c) Move with same velocity in opposite directions.
d) Move with the same velocity in the same direction

5]

A bullet is fired from a rifle. If the rifle were allowed to recoil
freely, its K.E. as a result of recoil would be………than that of the
bullet.
a) equal to

b) less than

c) greater than

d) Data insufficiency

SESSION – 4
AIM

To discuss special uses of coefficient of restitution

Falling of body from height H

Total Range and Time of flight of projectile

FALLING OF BODY FROM HEIGHT H AND BOUNCING.
1]

Time taken by the body to come to rest.

H

H1

H1

H 2 H3

H3

Consider the motion described in the fig. A body falls from height H
on a surface and coefficient of collision for this case is e.
Velocity of body just before 1st collision
=

2

Time taken during this free fall
=
Velocity just after 1st collision =

=

=

Maximum height attained after 1st collision

2

=

=

Time of flight to return again
=

= 2

Similarly after 2nd collision
=

=

2

, 2 =

H,

= 2

Similarly after 3rd collision
v3 = e3 u = –e3 2

; Hn =

Total time TNet =

+ 2e

=

2

H; T3 = 2en
+ 2e3

....... 2en

1+2

=

Total distance travelled:
d = H + 2H1 + 2 H2 + 2H3 ......... 2H4 .........
= H + 2e2 H + 2e4 H + 2e6 H........ 2e2n H.........
= H [1 +2 ( e2 + e4 + e6.......... e2n........)]
= [ 1 + 2
=

]

........

Total range and time of flight of projectile
u
usin 

eusin 
H

2

e usin 

V1
H1
1
 ucos 

ucos 
usin 

V2

 2 H2
1 ucos 

3

H3

 ucos 

eusin

2

e usin 

Consider the case above. Here projectile falls back on ground again
and again, continuing motion in same direction.
If initial velocity of projection is u at angle
=
=
=
The point to the noted here is that the horizontal component of
velocity never changes. The coefficient of restituion changes the
vertical component only. This change is responsible for changes of
other parameter.
So after

1st collision.

V1 y = eu sin 

hence tan 1 = e tan 

V1 x = u cos 
=

=

=

=

;

=

Similarly after 2nd collision
V2y = e2 u sin 

×

=

V2y = u cos 

tan 2 = e2 tan 

=

=

=

Just like earlier (AIM - 1) we can now conclude
Total time of F light
Tnet = T + T1 + T2............
=
=

(

[1 + +

… … … ]

)

Note same at RN = Ux (constant) × TN
=

(

)

CLASS EXERCISE

1]

When a ball is dropped on a floor from height h =20m and the
coefficient of collision is e=0.5 then calculate the time after
which it will comes to rest.
a) 6 sec

b) 4 sec

c) 3 sec

d) none

2]

A rubber ball falls from a height h and rebounds to a height
h/2. A rubber ball of double the mass falling from the same
height h rebounds to a height.
a) h

3]

b) h/2

c) zero

d) None

When a ball is dropped on a floor from height h =30 m and
coefficient of restitution e= 0.5 then calculate. The distance
travelled before coming to rest.
a) 50 m

4]

b) 30 m

c) 20 m

d) none

A ball falling freely from a height of 4.9 m hits a horizontal
surface. If =

then the ball will hit the surface second time

after
a) 0.5 sec
5]

b) 1.5 sec

c) 3.5 sec

d) 3.4 sec

A ball is allowed to fall from a height of 10 m. If there is 40%
loss of energy due to impact, then after one impact ball will go
up to
a) 10 m

6]

b) 8 m

c) 4m

d) 6m

A tennis ball is released from height ‘h’ above ground level. If
the ball makes in–leastic collision with ground, to what height
will it rise after third collision.
a) ℎ

b)

c)

d) ℎ

7]

A ball is projected vertically down with an initial velocity from
a height of 20 m onto a horizontal floor. During the impact it
loses 50% of its energy and rebounds to the same height. The
initial velocity of its projection is.
a) 20 /

b) 15 /

c) 10 /

d) 5 /

SESSION – 5
AIM

To study COLLISION in two dimensions (Oblique Collision)

LINE OF IMPACT
It is important to know the line of impact during this collision. The
line of impact is the line along which the impulsive 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 line normal
to tangent on the surface of impact 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 and v of the particles after impact are unknown in
direction and magnitude, their determination 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

t

t

t
m2v2

m1v1
A

A

B
n

m1v1

i]

B

+

n

=

A

B
n

m2v2

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.
We can write.
(v1)t = (v1)t; (v2)t = (v2)t

ii]

The component along the n axis of the total momentum of the
two particles is conserved. We write.
m1 (v1)n + m (v)n = m1 (v1)n + m (v)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)n - (v)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.

Illustration–1:
The magnitude and direction of the velocities of two identical
frictionless balls before they strike each other are as shown.
Assuming e = 0.90, determine the magnitude and direction of the
velocity of each ball after the impact.
Solution:
The impulsive force that the balls exert on each other during the
impact are directed along a line joining the centres of the balls
called the line of impact. Resolving the velocities into components
directed, respectively, along the line of impact and along the
common tangent to the surfaces in contact, we write.
(VA)n = VA cos 30° = +26 m/s
(VA)t = VA sin 30° = +15 m/s
(VB)n = –VB cos 60° = –20 m/s
(VB) t = VB sin 60° = +34.6 m/s
Since the impulsive forces are directed along the line of impact, the
t component of the momentum, and hence the t component of the
velocity of each ball, is unchanged. We have

(VA)t = 15 m/s, (VB)t = 34.6 m/s 
In the n direction, we consider the two balls as a single system and
not that by Newton’s third law, the internal impulses are,

respectively, Ft and –Ft and cancel. We thus write that the total
momentum of the balls is conserved.
mA(VA)n + mB(VB)n = mA(VA)n + mB(VB)n
m(26) + m(–20) = m(VA)n + m(VB)n
(VA)n + (VB)n = 6.0

........(i)

Using law of restitution,
(V) n – (V)n = e[(V) n – (V) n]
(V) n – (V) n = (0.90)[26 – (–20)]
Solving equations (i) and (ii) simultaneously, we obtain
(V)n = –17.7 m/s

(V)n = +23.7 m/s

(V)n = 17.7 m/s 

(V)n = +23.7 m/s 

Resultant motion: Adding vectorially the velocity components of
each ball, we obtain

VA = 23.2 m/s

, VB = 41.9 m/s

Illustration–2:
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.



v

v

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,

 = ′

… (i)

Applying Newton's law for collision,
(−

− 0) = − (

′ =

− 0)

o

…(ii)

From equations (i) and (ii),
=

+

′ = √
′=
=

+
+

′ =

Illustration–3:
A sphere A of radius r moving on perfectly smooth surface at a
speed v undergoes an elastic collision with an identical stationary
sphere B. Find the velocity of the sphere B after collision if the
impact parameter is d as shown in figure.
A
v
d

B

Solution:
One of the spheres is at rest before impact. After the impact its
velocity will be in the direction of the centre line at the moment of
contact because this is the direction in which the force acted on it.
Thus, sin 2 =

and

1 + 2 =

Since the masses of both spheres are equal, the triangle of momenta
turns into triangle of velocities.
we have V1 = v cos 1 = v sin 2 =
A
v
d

B

2 =

2 =

1−

4

Illustration–4:
A wedge of mass M rest on a horizontal surface. The inclination of
the wedge is. A ball of mass m moving horizontally with speed u hits
the inclined face of the wedge in elastically and after hitting slides
up the inclined face of the wedge. Find the velocity of the wedge just
after collision. Neglect any friction.
u m
M

Solution:
Let velocity of the block after collision is ⃗ and that of the ball is ⃗
with respect to the wedge in directions as shown in the figure. As
the net impulse on the system in the horizontal direction = 0.
Conserving momentum along horizontal, we get
v1

v1

1
v2

2

900

v

v2

mu = m [v2cos + v1] + Mv1
 mu = mv2 cos + v1] + Mv1

....(i)

Since common normal is along y’, therefore momentum of ball
remains constant along the incline (along X’ ∴ ⃗ = 0 for the ball)
⟹  =
⟹ = ( − )

+

.....(ii)

From equations (i) and (ii), we get
mu = mu cos2  – mv1 cos2  + (M + m)v1

=

Note: e = 0 is satisfied for perfectly inelastic collision.
m
I1 I1

M

I2

CLASS EXERCISE
1]

An object at rest in space suddenly explodes into three parts of
same mass. The momentum of two parts are 2

and

⃗.

The

momentum of the third part is
a) P√3
2]

b) P√5

c) P

d) 2P

The magnitude of the impulse developed by a mass of 0.2kg
which

changes

its

velocity

from

5⃗ − 3⃗ + 7 ⃗ /

2⃗ + 3⃗ + ⃗ is
a) 2.7 N.s

b) 1.8 N.s

c) 0.9 N.s

d) 3.6 N.s

to

3]

Two vehicles of equal masses are moving with same speed v on
two roads inclined at an angle. They collide in elastically at the
junction and then move together. The speed of the combination
is

a) v Cos
4]

b) 2v Cos

c) v/2 Cos

A mass m moves with a velocity

d) v/2 Cos /2

and collides inelastically

with another identical mass. After collision the 1st mass moves
with velocity

in a direction perpendicular to the initial

direction of motion.

Find the speed of second mass after

collision.
a)

b)

c)

*****

d) √3

KEY
SESSION – 1 & 2
CLASS EXERCISE
1] d

2] d

3] a

4] a

8] d

9] c

10] a

11] b

5] a

6] b

7] b

6] a

7] a

SESSION - 3
CLASS EXERCISE
1] b

2] c

3] c

4] c

5] b

SESSION - 4
CLASS EXERCISE
1] a

2] b

3] a

4] b

5] d

SESSION - 5
CLASS EXERCISE
1] b

2] b

3] a

4] a

********

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