Collision:
Terminology –
1. Elastic collision – a) Momentum is conserved, b) KE is conserved. But in
dealing with questions we don’t write these two conservation equations, but
we do this a) equation for conservation of P, b) equation for ‘e’ – coefficient
of restitution, putting e = 1. (See ‘e’ below)
2. Inelastic Collision – Only Momentum is conserved. The two equations which
are written are a) equation for conservation of P, b) equation for ‘e’ –
coefficient of restitution.
3. Perfectly inelastic collision – the two bodies join together. Only one equation
(conservation of ‘p’) is enough.
4. Coefficient of restitution
For two bodies define it like thisv v
velocity.of .separation
e  ( )
 ( ) 2 1
velocity.of .approach
u2  u1
For a single body striking a wall etc., define it like thisvelocity.of .separation
v
e  ( )
 ( )
velocity.of .approach
u
Just put the terms with proper sign convention now. Putting v = 0 and v = u, we
can get value of e for perfectly inelastic and elastic collision respectively.
5. Line of Impact – The common  r to the surfaces in contact during impact.
Line of impact (common perpendicular)
Common tangent to the surfaces which are colliding.
6. Central Impact – If centre of mass the two colliding bodies are located on line
of impact. (We will study this case only). Else the collision is called eccentric.
7. Direct central Impact – velocity of two colliding bodies is directed along line
of impact. It is also called head-on collision.
VB
VA
8. Oblique central impact – If the velocity are not directed along the line of
impact.
VB
VA
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The 3 categories of Collision [the way we will do it] –
Category C (e = 0) – Perfectly Inelastic collision –There is only one variable, and
conservation of ‘p’ is enough. In it, e = 0.
m1u1 + m2u2 = (m1 + m2) v
Category A (e = 1) – Elastic Collision (We will study it in detail, but later. Its
significance is only for CBSE. In solving problems note that a question is on
Elastic collision if it is specifically written so in the question]
Category B (0 < e < 1) – Inelastic collision. This is the general case of collisions.
There are 2 equation to be written a) conservation of ‘p’, b) equation for ‘e’. The
category A is a particular case of this category (by putting e = 1)
Later on we will study Collision in the frame of C.O.M.
Let us study them one by one –
Category C - PERFECTLY INELASTIC Collision of two bodies (Head-on) –
As the 2 bodies become one after collision, so there is only one variable. So only one
equation is enough to solve the question –
Conservation law of momentum – m1u1 + m2u2 = (m1 + m2) V’
1
From this we can find V’ and also ( M 1  M 2 )V '2 which is the available KE
2
immediately after the collision.
Cases arising on what happens to this available KE –
Work done against Friction
Body compresses a spring
Ballistic pendulum
And combination of these.
Loss of KE during Impact –
From above methods, find velocities after impact. KE before collision and
after collision can be easily calculated.
Basic Questions on bullet hitting a block (part of theory) –
1. A bullet of mass m moving with speed u strikes a stationary block of mass M
and gets embedded in it. Calculate a) the speed of block after collision, b) the
kinetic energy before and after collision and loss of KE.
2. If the block in question 1 is attached to a spring of spring constant K, the find
the compression in the spring.
3. If the block in question 2 is kept on a rough surface with coefficient of friction
μ, then find the distance that the block travels before coming to rest.
4. If the block in question 1 is attached to a string of length L, then find a) the
height to which the block rises, b) the maximum angle that the string make
with the horizontal.
5. If the bullet in question 1 comes out of the block with speed v, then what are
the answers to all the questions above? [Now no more category C question –
it becomes a category B question]
Find change in momentum.
If time of interaction is given, find force of impact.
If coeff of friction is given find if the block slides or not!
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Questions from HCV etc. on category C –
1. HCV(1) Pg. 162-163, Q 31, 32, 46, 47,
2. Q 55 – Good one. Also using concept of vertical circle and projectile.
3.
Category B – INELASTIC Collision of two bodies (Head-on) –
We start with two equations –
Conservation law of momentum – m1u1 + m2u2 = m1v1 + m2v2
Equation for ‘e’ e = (-) [v2 – v1] / [u2 – u1]
As there are only 2 variables, so solving these two equations is enough.
Question from HCV etc. on Category B –
1. A 5 kg mass moving with a velocity of 10 m/s from left to right collides with
another body of mass 2 kg moving in the same direction with a velocity of 5
m/s. If after the impact, the 2 kg mass has a velocity of 12 m/s in the same
direction, find the coefficient of restitution.
2. A glass marble of mass 20 g is dropped on to a horizontal glass plate from a
height of 2 m. Assuming the coefficient of restitution to be 0.9, calculate the
height to which the marble rise after impact.
It’s result is good enough to be kept for later use –
H2/H1 = e2
So after second collision, height gained is – H3/H2 = e4
3. HCV(1) Pg 162 Q 39 – A ball falls on the ground from a height of 2 m and
rebounds to a height of 1.5 m. Find the co-efficient of restitution.
4. A sphere of mass M moving with velocity u hits another stationary sphere of
same mass. If ‘e’ is the coefficient of restitution, what is the ratio of velocities
of two spheres just after collision?
5. HCV(1), pg. 162, Q 35, 36, 37.
6. HCV(1) pg 162, Q 42 (Bullet emerges out of block), pg. 163, Q 48, 54,
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Category A) ELASTIC Collision of two bodies (Head-on) –
u1
M1
u2
v1
M2
M1
v2
M2
Conceptually, there are two variables v1 and v2, so we need two equations, which are
the conservation laws of momentum and kinetic energy (It is only in Elastic Collision
that KE is conserved). But actually we write the following two eauations Conservation law of momentum – m1u1 + m2u2 = m1v1 + m2v2
Equation for ‘e’
e = (-) [v2 – v1] / [u2 – u1]
In this case e = 1
This equation for ‘e’ can be derived from normal process as under –
Cons of P -
m1u1 + m2u2 = m1v1 + m2v2
OR
m1(u1 – v1) = m2(v2 – u2)
- (1)
2
2
2
2
(1/2)m1u1 + (1/2)m2u2 = (1/2)m1v1 + (1/2)m2v2
OR
m1(u12 – v12) = m2(v22 – u22)
- (2)
Cons of KE -
Dividing (2) by (1)
OR
OR
u1 + v1 = v2 + u2
(-) (u2 – u1) = (v2 – v1)
(1) = (-)(v2 – v1) / (u2 – u1)
- (3)
In actual problems we start from equation (1) and equation (3) directly.
Solving (1) and (3) for v1 and v2, we get
v1  (
m1  m2
2m 2
)u1 
u2
m1  m2
m1  m2
v2  (
m2  m1
2m1
)u 2 
u1
m1  m2
m1  m2
[Mug up one equation for v1 and then to get equation for v2, simply interchange all 1
and 2]
The 3 cases:a) m1 = m2,  v1 = u2 and v2 = u1. So the velocities are interchanged.
If m2 was at rest, then after collision, m1 will come to rest and m2 will
move with the initial velocity of m1.
b) m2 >> m1 (Target particle is very massive)  v1 = - u1 + 2u2, v2 = u2
If the target particle was at rest (like wall) then v1 = - u1 and v2 = 0.
That is the particle rebounds with the same speed.
c) If m1 >> m2 (Like truck hitting a scooter)  v1 = u1 , v2 = u2 + 2u1. If the
scooter is at rest initially, then v2 = 2u1
Q 1. A body of mass 1 kg makes an elastic collision with another body at rest and
continues to move in the original direction with a speed equal to 1/4th of its original
speed. Find the mass of the second body.
Hint: Better use result for v1 in an elastic collision.
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Q 2. A ball is moving with velocity 2 m/s towards a heavy wall which is moving
towards the ball with speed 1 m/s. Assuming the collision to be elastic, find the
velocity of ball immediately after the collision.
[IIT98] A wedge of mass m and triangular cross section (Ab = BC = CA = 2R) is
moving with a constant velocity –V towards a sphere of radius R fixed on smooth
horizontal table as shown. The wedge makes an elastic collision with the fixed sphere
and returns along the same path without any rotation. Neglect all friction and suppose
that the wedge remains in contact with the sphere for a very short time Δt, during
which the sphere exerts a constant force F on the wedge. a) Find the force F and also
normal force N exerted by the table on the wedge during time Δt, b) Let h denote the
perpendicular distance between the centre of mass of the wedge and the line of action
of F. Find the magnitude of the torque due to the normal force N about the centre of
the wedge during time Δt.
[IIT 2009, ] Three objects A, B and C are kept in a straight line on a frictionless
horizontal surface. These have masses m, 2m and m respectively. The object A moves
towards B with a speed of 9 m/s and makes an elastic collision with it. Thereafter, B
makes completely inelastic collision with C. All motions occur on the same straight
line. Find the final speed of C ( to nearest integer value). [Ans. 4]
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OBLIQUE Collision of two bodies –
Consider the diagram below –
M1
M1
v1
u1
α1
β1
α2
β2
u2
M2
v2
M2
“Line of impact” and “Tangent to the surfaces which are colliding” are the reference
axes. The collision is along the ‘Line of Impact’ which is the common  r.
1. Normally there are two variables after the collision. So, two equations are
needed. They are conservation of momentum along X- and Y- axis. Note
momentum conservation is valid along any axes.
2. If there are 3 variables, then in addition to above two equations, use equation
for ‘e’ in Y- direction. Note that equation of ‘e’ has to be along the line of
impact only. In the diagram above, equation of ‘e’ has to be along Y-direction.
3. If there are 4 variables, then put v1x = u1x, and v2x = u2x. The other two
equations will be conservation of momentum along X- and Y- axis. Note that
the collision is along Y-direction (where we write equation for ‘e’), along Xthere is no impulse.
1. A ball moving with a speed of 9 m/s strikes a ball at rest, such that after
collision, the direction of each ball makes an angle of 30o with the original line
of motion. Find the speeds of the two balls after collision.
2. At what angle must a body be incident on a perfectly hard plane so that the
angle between the directions before and after impact may be right angle? Take
the coefficient of restitution to be 1/3.
3. A ball of mass m hits the floor with a speed v making an angle of incidence θ
with the normal. The coefficient of restitution is ‘e’. Find the speed of
reflected ball and the angle of reflection.
4. A ball of mass m falling vertically hits a wedge of mass M with speed V. All
surfaces are frictionless, wedge has angle θ. If ‘e’ is given, find velocity of
wedge and ball after collision.
5. HCV1, pg 162, Q 43, 44, 45
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Bhai – Bhai ka pyaar (Questions on similar types)
A.
1. M1 moves with Vo, strikes M2. Spring between them. Find max.
compression.
Vo
M1
M2
2.
Vo
3.
M1
Vo
M1
4. Find max. expansion in the spring if velocity of mass on right is V
V
In all the above four questions
Conservation of momentum  m1u1 + m2u2 = m1v1 + m2v2
Conservation of energy  ½ m1u12 + ½ m2u22 = ½ m1v12 + ½ m2v22 + ½ kx2
At the time of maximum compression / expansion, v1 = v2
B.
1. The two blocks are pulled by force F1 and F2 [Q. 52, pg. 163, HCV1]
F1
F2
acc of COM, acom = (F2 – F1)/(M1 + M2)
In the COM frame, the COM is itself accelerated wrt ground, so a pseudo force acts
towards left on both M1 and M2
M F  M 1 F2
M F  M 1 F2
Force on M1, F1’=F1+M1a = 2 1
, that on M2, F2’=F2–M2a = 2 1
M1  M 2
M1  M 2
When blocks stop, suppose the left block moved x1, and right block moved x2
M F  M 1 F2
The work done by F1’ and F2’ is, W = F1’x1 + F2’x2 = 2 1
(x1 + x2)
M1  M 2
This should be equal to ½ k(x1 + x2)2. Equating them we get value of x1 + x2
2. F1 = 0, F2 = F [HCV1 - pg 154, Ex. 19]
3. F1 = F, F2 = F [Q. 51, pg. 163, HCV1]
4.
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C. A small block moves on another mass ending into a vertical circle.
1. HCV(1), Q 29, pg. 161
2. HCV(1), Q 61, pg. 164
3. IIT, also in BT
D. Questions on Chains
1. [Q57, pg 163, HCV1, modified] A uniform chain of mass M and
length L is held vertically in such a way that its lower end is at a
distance ‘D’ from the horizontal floor. The chain is released from rest
in this position. Any portion that strikes the floor comes to rest.
Assuming that the chain does not form a heap on the floor, calculate
the force exerted by it on the floor when a length ‘x’ has reached the
floor.
Ft = Wt of mass on floor + (M/L)v2
2. A chain (M, L) begins to fall through a hole in the ceiling.
3. The end of a chain (M, L) piled on the platform is lifted vertically.
E. Questions on ball hitting the wallLet a ball be thrown with speed ‘u’ at an angle θ. Let vx and vy be the components of
its velocity at the time of impact.
Along Yvy will remain unchanged.
Along Xvx becomes evx
Since vy does not change so time of flight T, and maximum height attained H
remains unchanged.
u 2 sin 2 
T = tOAB = tCD + tDEF = 2u sin θ / g, hA = hE =
2g
But as vx has become evx, so range changes (not if e = 1, i.e. elastic collision)
F.
Collision in the frame of C.O.M.
Questions from HCV –
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
Angular Impulse & Conservation of L
Like wise,
dL
τ=
.
dt
Angular Impulse, τ dt = dL
So if Angular Impulse = 0
 ‘L’ is conserved.
We had F =
dp
.
dt
=> dp = Fdt
If Impulse = 0  dp = 0
=> p is conserved
It happens if
either Fext (net) = 0
or dt  0, F v. small
ABOUT? – The axis about which external τ is zero.
Calculating Ang. Impulse In Collision, L is conserved only for system.

For any force ‘F’ acting ‘r’
[For individual particles; L changes]
away, T = Fr
But the sum of the changes in Angular momentum
Tdt = (Fr) dt
of two components of a system is zero.
i.e. L1 + L2 = 0

[Just as in case of p , it changes individually for both particles,
but the sum of p1 + p2 = 0]
In the case of translation
M1u1 + M2u2 = M1v1 + M2v2
M1 (v1 – u1) = -M2 (v2 – u2)
M1v1 = -M2u2
p1 = - p2
or p1 + p2 = 0
Category B) 4 categories of questions on rotational collisions
B1.
Particle(s) strike a rod , particle stops, rod moves (rot. + trans.)
B2.
Rod etc. fixed at one point, particle/bullet strikes, gets embedded, both
rotate together about the fixed point
B3.
Particle strikes a rod etc. gets embedded in it, both move together.
(trans. + rot. about combined C.M)
B4.
Rod, sphere etc. hit by a cue, imparting rotational/translational impulse
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Questions on Category B –
B1. Particle strikes rod, particle stops.
Q1. [based on HCV, solved example 21 pg. 188]
M1 , L
O
m2
V
IF THE BALL STOPS AFTER COLLISION On ball + rod system, external force is zero
Cons. ‘p’  m2V = M1V0 [How can you say that it will be the V of C.M.
only?]
Also external torque is zero –
Cons. ‘L’  m2V(L/2) = ICM ω + r x p
(the value of r x p about CM is zero)( ω is about CM)
[Why is L conserved about CM only? Can’t it be about highest or lowest point etc.?]
It can be, but the calculation is longer as you have r x p term also!
V of any other point can be found as vector resultant of V0 and x ω.
[IIT 2000] A rod AB of mass M and length L is lying on a horizontal frictionless
surface. A particle of mass m travelling along the surface hits the end ‘A’ of the rod
with the velocity v0 in a direction perpendicular to AB. The collision is elastic. After
the collision the particle comes to rest.
a) Find the ratio m / M.
b) A point P on the rod is at rest immediately after collision. Find the
distance AP.
c) Find the linear speed of the point P after a time πL / 3V0 after the
collision.
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B2. Rod etc. fixed at one point, particle/bullet strikes, get embedded
Questions on (rod pivoted at one end) + particle strikes and gets stuck to it = find
angular velocity after collision OR height to which the rod rises
[2005, 2M] A rod of length L and mass M is hinged at point O. A small bullet of
Mass m hits the rod as shown in the figure. The bullet gets embedded in the
rod.
Find angular velocity of the system just after impact.
O
L, M
mv
Note – Linear Momentum will not be conserved!
[1994, 6M] Two uniform rods A and B of length 0.6m each and of masses 0.01 kg
and 0.02 kg respectively are rigidly joined end to end. The combination is pivoted at
the lighter end, P as shown in fig. such that it can freely rotate about point P in a
vertical plane. A small object of mass 0.05 kg, moving horizontally, hits the lower end
of the combination and sticks to it. What should be the velocity of the object, so that
the system could just be raised to the horizontal position.
P
A
B
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B3. Particle strikes a rod etc. gets stuck to it, both move (trans. + rot.)
IF THE BALL STICKS TO THE ROD AFTER COLLISION –
On ball + rod system, external force is zero
Cons. ‘p’  m2V = (M1 + M2) V0
Also external torque is zero –
Cons. ‘L’ (about the centre of the rod)  m2V(L/2) = ICM ω + r x p
where ‘r’ is the distance of C.M. of combined system from the centre of the rod.
Now if this conservation equation is applied about CM of combined system the term r
x p is zero. [But in that case the left hand side is also calculated about it]
Category B3 (Easy case)
Q. (HCV, last solved example)
Even on sticking, it is said that CM does not shifts appreciably. So centre of
sphere and C.M. of combined system is one and same thing. Only difference is that I
after collision will be the net I of two bodies.
Even in the following question, the CM of system remains in the same line as each
particle has same mass and strikes the rod symmetrically.
[1989, 8M] A thin uniform bar lies on a frictionless horizontal surface and is free
to move in any way on the surface. Its mass is 0.16 kg and length is √3m. Two
particles, each of mass 0.08 kg are moving on the same surface and towards
the bar in a direction perpendicular to the bar one with a velocity of 10m/s,
and the other with 6m/s, as shown in fig. The first particle strikes the bar at
points A and the other point B. Points A and B are at a distance of 0.5m from
the centre of the bar. The particles strike the bar at the same instant of time
and stick to the bar on collision. Calculate the loss of kinetic energy of the
system in the above collision process.
10m/s
A
…………………
6m/s
B
…………………
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B4. Rod, sphere etc. hit by cue, imparting rotational impulse
A billiard ball, initially at rest, is given a sharp impulse by a cue (which applies a
force F horizontally at a height h above the centre of the ball). The ball leaves the cue
with a speed v0 and eventually acquires a speed of 9v0/7. Show that h = 4R/5.
When the ball leaves the cue (Given v = v0. Let w = w0)
..1
When the ball starts pure rolling (Given v = 9v0/7, As v = Rw, so w = 9v0/7R)
..2
Conservation of L in 1 and 2 (about poc)
9v
9V
2
2
mR 20  mv0 R  mR 2 ( 0 )  m( 0 ) R
5
5
7R
7
v0
This gives  
2R
Now the initial momentum of mv0 and the angular momentum of mv0R are given by
the force F on the ball.
So, F dt = mv0
…3
2
and F h dt = mR 20
…4
5
from 3 and 4, h = 4R/5
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