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 this- e
.
.
.
.
v
2
 v
1 u
2
 u
1
For a single body striking a wall etc., define it like this-
.
.
v e
.
.
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.
V
B
V
A
8.
Oblique central impact – If the velocity are not directed along the line of impact.
V
B
V
A
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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. m
1 u
1
+ m
2 u
2
= (m
1
+ m
2
) 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 – m
1 u
1
+ m
2 u
2
= (m
1
+ m
2
)
V’
From this we can find V’ and also
1
2
( M
1

2
) ' 2 which is the available KE 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 – m
1 u
1
+ m
2 u
2
= m
1 v
1
+ m
2 v
2
Equation for ‘e’ - e = (-) [v
2
– v
1
] / [u
2
– u
1
]
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 –
H
2
/H
1
= e
2
So after second collision, height gained is – H
3
/H
2
= e 4
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) –
u
1 u
2
v
1 v
2
M
1
M
2
M
1
M
2
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 – m
1 u
1
+ m
2 u
2
= m
1 v
1
+ m
2 v
2
Equation for ‘e’ e = (-) [v
2
– v
1
] / [u
2
– u
1
]
In this case e = 1
This equation for ‘e’ can be derived from normal process as under –
Cons of P -
Cons of KE - m
1 u
1
+ m
2 u
2
= m
1 v
1
+ m
2 v
2
OR m
1
(u
1
– v
1
) = m
2
(v
2
– u
2
)
(1/2)m
1 u
1
2
+ (1/2)m
2 u
2
2
= (1/2)m
1 v
1
2
+ (1/2)m
2 v
2
2
- (1)
OR m
1
(u
1
2
– v
1
2
) = m
2
(v
2
2
– u
2
2
) - (2)
Dividing (2) by (1) u
1
+ v
1
= v
2
+ u
2
OR (-) (u
2
– u
1
) = (v
2
– v
1
)
OR (1) = (-)(v
2
– v
1
) / (u
2
– u
1
) - (3)
In actual problems we start from equation (1) and equation (3) directly .
Solving (1) and (3) for v
1 and v
2
, we get v
1

( m m
1
1

 m m
2
2
) u
1
 m
1
2 m
2
 m
2 u
2 v
2
 m
( m
1
2

 m m
2
1 ) u
2
 m
1
2 m
1
 m
2 u
1
[Mug up one equation for v
1
and then to get equation for v
2
, simply interchange all 1 and 2]
The 3 cases:- a) m
1
= m
2
,

v
1
= u
2
and v
2
= u
1
. So the velocities are interchanged.
If m
2
was at rest, then after collision, m
1
will come to rest and m
2
will move with the initial velocity of m
1
. b) m
2
>> m
1
(Target particle is very massive)

v
1
= - u
1
+ 2u
2
, v
2
= u
2
If the target particle was at rest (like wall) then v
1
= - u
1
and v
2
= 0.
That is the particle rebounds with the same speed. c) If m
1
>> m
2
(Like truck hitting a scooter)

v
1
= u
1
, v
2
= u
2
+ 2u
1
. If the scooter is at rest initially, then v
2
= 2u
1
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 1/4 th
of its original speed. Find the mass of the second body.
Hint: Better use result for v
1
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 –
M
1 u
1
α
1
β
1
M v
1
1
α
2
β
2
M
2 u
2
v
2
M
2
“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 v
1x
= u
1x
, and v
2x
= u
2x
. 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 X- there 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 30 o 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.
M
1
moves with Vo, strikes M
2
. Spring between them. Find max. compression.
2.
M
1
Vo
Vo
M
2
3.
M
1
Vo M
1
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

m
1 u
1
+ m
2 u
2
= m
1 v
1
+ m
2 v
2
Conservation of energy

½ m
1 u
1
2 + ½ m
2 u
2
2 = ½ m
1 v
1
2 + ½ m
2 v
2
2 + ½ kx 2
At the time of maximum compression / expansion, v
1
= v
2
B.
1.
The two blocks are pulled by force F1 and F2 [Q. 52, pg. 163, HCV1]
F
1
F
2 acc of COM, a com
= (F
2
– F
1
)/(M
1
+ M
2
)
In the COM frame, the COM is itself accelerated wrt ground, so a pseudo force acts towards left on both M
1
and M
2
Force on M
1
, F
1
’=F
1
+M
1 a =
M F
2 1

M F
1 2
M
1

M
2
, that on M
2
, F
2
’=F
2
–M
2 a =
M F
2 1

M F
1 2
M
1

M
2
When blocks stop, suppose the left block moved x
1
, and right block moved x
2
The work done by F
1
’ and F
2
’ is, W = F
1
’x
1
+ F
2
’x
2
=
M F
1

M F
2
M
1

M
2
(x
1
+ x
2
)
This should be equal to ½ k(x
1
+ x
2
) 2 . Equating them we get value of x
1
+ x
2
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. F t
= Wt of mass on floor + (M/L)v
2
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 wall-
Let a ball be thrown with speed ‘u’ at an angle θ. Let v x
and v y
be the components of its velocity at the time of impact.
Along Y- v y
will remain unchanged.
Along X- v x
becomes ev x
Since v y
does not change so time of flight T, and maximum height attained H remains unchanged.
T = t
OAB
= t
CD
+ t
DEF
= 2u sin θ / g, h
A
= h
E
= u
2 sin
2

2 g
But as v x
has become ev x
, 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 dp
We had F = .
dt
Like wise,
τ = dL dt
.
=> dp = Fdt
If Impulse = 0

=> p is conserved
dp = 0
Angular Impulse, τ dt = dL
So if Angular Impulse = 0

‘L’ is conserved.
ABOUT? – The axis about which external τ is zero.
It happens if either F ext
(net) = 0 or dt

0, F

v. small
Calculating Ang. Impulse -
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.

L
1
+


[Just as in case of p
L
2
= 0
, it changes individually for both particles, but the sum of
 p
1
+
 p
2
= 0] In the case of translation
M
1 u
1
+ M
2 u
2
= M
1 v
1
+ M
2 v
2
M
1
(v
1
– u
1
) = -M
2
(v
2
– u
2
)
M
1
 p
1
 v
1
= -M
2
 u
= -
 p
2
2 or
 p
1
+
 p
2
= 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]
O
M
1
, L
m
2
V
IF THE BALL STOPS AFTER COLLISION -
On ball + rod system, external force is zero
Cons. ‘p’  m
2
V = M
1
V
0
[How can you say that it will be the V of C.M. only?]
Also external torque is zero –
Cons. ‘L’ 
m
2
V(L/2) = I
CM
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 V
0
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 v
0 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 / 3V
0 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’ 
m
2
V = (M
1
+ M
2
) V
0
Also external torque is zero –
Cons. ‘L’ (about the centre of the rod) 
m
2
V(L/2) = I
CM
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 v
0
and eventually acquires a speed of 9v
0
/7. Show that h = 4R/5.
When the ball leaves the cue (Given v = v
0
. Let w = w
0
) ..1
When the ball starts pure rolling (Given v = 9v
0
/7, As v = Rw, so w = 9v
0
/7R) ..2
Conservation of L in 1 and 2 (about poc)
2
5 mR
2
 
0

2 mv R mR
0
5
2
(
9
7 v
R
0 )
 m (
9 V
7
0 ) R
This gives
 
2 v
0
R
Now the initial momentum of mv
0
and the angular momentum of mv
0
R are given by the force F on the ball.
So, F dt = mv
0
…3 and F h dt =
2
5 mR
2 
0
…4 from 3 and 4, h = 4R/5
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