COLLISION SPP.pptx

Prof. Samirsinh P. Parmar
samirddu@gmail.com, spp.cl@ddu.ac.in
Asst. Prof. Dept. of Civil Engineering
Dharmasinh Desai University, Nadiad, Gujarat, India
COLLISIONS
CL-101 ENGINEERING MECHANICS
B. Tech Semester-I

Momentum
Impulse
Direct collisions
Collision with a smooth plane
Examination-style questions
Contents

Conservation of momentum
Total momentum before collision = Total momentum after
p = mv
The principle of conservation of momentum states that:
Moving objects are said to have momentum, p, which is the
product of mass and velocity, measured in kg ms-1 or
Newton seconds (Ns), which are equivalent.
Momentum is a vector quantity with direction as well as
magnitude. The momentum of a stationary object is zero.
When two objects collide, their velocity and momentum may
be affected.

0.2
0.2
Before
impact
After
impact
0.5
0.5
6 8
v
7
A B
Two smooth spheres A and B, of masses 0.2 kg and 0.5 kg
respectively, are projected towards each other with speeds of 6 ms-1
and 8 ms-1. After they collide sphere A rebounds with a speed of 7
ms-1.
Calculate the speed with which sphere B moves after the collision
and state its direction.

Therefore the speed of B after the collision is 2.8 ms-1 and
the particle continues moving in the same direction as before
the collision (i.e. from right to left).
v = –2.8
0.5v = –1.4
1.2 – 4 = –1.4 + 0.5v
Taking the positive direction of motion to be from
left to right:
(0.2 × 6) – (0.5 × 8) = (0.2 × –7) + (0.5 × v)

Impulse
Momentum
Impulse
Direct collisions

Impulse
The change to a body’s momentum caused by a collision is
called impulse, measured in Ns. Impulse is a vector quantity
with magnitude and direction.
I = mv – mu
where u is the initial velocity and v is the final velocity.
Impulse = change in momentum
The impulse that a body A exerts on a body B is of equal
magnitude to the impulse that B exerts on A, but it acts in
the opposite direction.

Impulse
A ball of mass 0.75 kg hits the floor with a speed of 10 ms-1.
It rebounds with a speed of 7 ms-1.
Find the impulse exerted by the floor on the ball.
Taking the upwards direction to be positive:
Impulse = (0.75 × 7) – (0.75 × –10)
= 5.25 + 7.5
= 12.75 Ns

Momentum & impulse as vectors
Sometimes a collision alters the line along which a particle is
travelling, as in the example below.
In these cases, we might need to find the magnitude and
direction of the impulse by resolving a vector quantity into its
horizontal and vertical components.

Impulse vector question
A ball of mass 2 kg is travelling horizontally at a speed of
6 ms-1 when it is struck by a bat. After the collision the ball
is travelling upwards at 30° to the horizontal in the opposite
direction at a speed of 8 ms-1.
Find the impulse exerted on the ball by the bat.
2
6 ms-1
Before 2 After
8 ms-1
30°

The velocity of the ball before and after the collision can be
expressed in vector notation as follows:
I = mv – mu = 2 × (–8cos30°i + 8cos60°j) – 2 × 6i
u = 6i
v = –8cos30°i + 8cos60°j
(where i is horizontal and j is vertical). After
2
8ms-1
30°
I = 2 × (–43i + 4j) – 12i

 
2
2
8 3 12 8 (3 s.f.)
    27.1 Ns
(–83 – 12)
8
θ
I
I =
θ = (3 s.f.) to the negative i direction.
1 8
8 3 12
tan 7
=1 .2


 
 
 
I = (–83 – 12)i + 8j

Direct collisions
Momentum
Impulse
Direct collisions

Direct collisions
Two bodies are said to have been in a direct collision when
the velocities before and after impact are in a straight line.
The outcome depends partly on the constitution of the bodies.
This means that the bodies either rebound, continue in the
same direction with reduced speed, or continue with the
same velocity.

Newton’s experimental law
In direct collisions the relative velocity of the two bodies is reversed and
the speed of separation is e times the speed of approach
It follows that the speed of separation is always less than or equal to the speed
of approach.
e depends on the nature of the two surfaces in contact.
For example, between a tennis ball and a concrete floor e will be close to 1, but
e will be much lower between a tennis ball and a carpeted floor.
Newton’s Experimental Law (NEL) states that:
where e is a constant for two surfaces in contact called the coefficient of
restitution, and 0 ≤ e ≤1.

Elastic and inelastic collisions
A perfectly elastic collision is one in which e = 1.
In a perfectly elastic collision, the kinetic energy of the
system will be conserved. In all other collisions kinetic
energy will be lost.
A perfectly inelastic collision is one in which e = 0.
The loss in kinetic energy in the collision is equal to:
where X is the speed of approach.
( )
mm
e X
mm
 2 2
1 2
1 2
1
2

Applying Newton’s Experimental Law
When applying Newton’s Experimental Law, it is advisable
to treat each velocity as either positive or negative in the
same direction.
v2 – v1 = –e(u2 – u1)
So Newton’s Experimental Law can be expressed as:
A and B are travelling in opposite directions, so the speed
of approach is u1 + (–u2) and the speed of separation is
(–v1) + v2.
Before
After
u1 –u2
–v1 v2
A B

Particles question 1
Question 1: A particle A is moving at a speed of 4 ms-1 towards a
particle B moving at a speed of 5 ms-1 in the same line.
After the collision the direction of motion of both particles is reversed. A
moves with a speed of 5 ms-1 and B moves with a speed of 2 ms-1.
Calculate the coefficient of restitution between these particles.

Before
After
4 –5
–5 2
7 = 9e
7
9
Intuitively, in this simple example, we can see that the
speed of approach is 9 ms-1 and the speed of separation
is 7 ms-1, which also gives the required result.
Apply NEL: (2 – –5) = –e(–5 – 4)
e =
A B

Question 2: Two particles A and B of mass 3 kg and 4 kg respectively
are travelling towards each other in a straight line.
A has a speed of 10 ms-1 and B has a speed of 2 ms-1.
State the direction of motion of each particle.
Given that the coefficient of restitution between the two
particles is , find the speeds of A and B immediately after
the collision.
1
3

Use conservation of momentum:
Total momentum before collision = Total momentum after
In this example we have two
unknowns. Therefore, we
need to solve using
simultaneous equations.
30 – 8 = 3v1 + 4v2
(3 × 10) + (4 × –2) = (3 × v1) + (4 × v2)
 3v1 + 4v2 = 22
4 kg
Before
After
10 –2
v1 v2
3 kg

Apply NEL:
v2 – v1 = – (–2 – 10)
1
3
Solve the equations simultaneously:
Therefore after the collision the speed of A is ms-1 and the
speed of B is ms-1. Since the velocities are positive, both
particles are moving in the original direction of A.
6
7
6
7
4
Adding gives: 7v2 = 34 6
7
4
 v2 =
v2 – v1 = 4 
v2 – v1 = 4
3v2 – 3v1 = 12
3v1 + 4v2 = 22
6
7
 v1 =

Question 3: Three particles A, B and C of masses 2 kg, 3 kg and 5 kg
respectively are on a straight level smooth surface.
A is moving towards B and C with a speed of 15 ms-1 whilst B and C are at
rest.
If A is brought to rest after the collision with B and B is brought to rest after
the collision with C, find the coefficient of restitution between particles B and
C.

It is necessary to deal with the two collisions separately.
First, consider the collision between A and B:
Using conservation of momentum:
(2 × 15) + (3 × 0) = (2 × 0) + (3 × v1)
30 = 3v1
 v1 = 10
15 0
v1
0
Before
After
2 kg 3 kg

Now that we know the speed of B after the first collision, we
can examine the collision between B and C.
(3 × 10) + (5 × 0) = (3 × 0) + (5 × v2)
Apply NEL to the collision between B and C:
6 – 0 = – e(0 – 10)
The coefficient of restitution between B and C is 0.6.
30 = 5v2
6 = 10e
 e = 0.6
 v2 = 6
10 0
0 v2
Before
After
3kg 5kg

Momentum
Impulse
Direct collisions

v = eu
When looking at a collision between a particle and a
smooth plane we use the formula:
where u is the speed of approach and v is the speed of
separation.

Smooth plane question 1
Question 1: A particle is travelling horizontally with a speed
of 8 ms-1 when it hits a fixed smooth vertical plane. The
coefficient of restitution between the particle and the plane
is .
Calculate the speed with which the particle rebounds after
the impact.
Apply NEL:
v = eu = × 8 = 2
Therefore the particle rebounds with a speed of 2 ms-1.
1
4
1
4

Question 2: A particle falls from rest onto a fixed smooth
horizontal plane 10 m below. If it rebounds from the plane
with a speed of 3.5 ms-1, calculate the coefficient of
restitution between the particle and the plane.
We know: u = 0, s = 10, a = 9.8
v2 = 0 + 2 × 9.8 × 10 = 196
The particle strikes the plane with a speed of 14 ms-1.
Apply NEL: .
3 5 1
=
4
14
v
e u
 
Therefore the coefficient of restitution is ¼.
 v = 14
Therefore we use v2 = u2 + 2as to calculate the speed of the
particle as it collides with the plane.

Question 3: A particle is travelling horizontally when it
collides with a smooth fixed vertical plane.
As a result of this collision the particle loses ¼ of its kinetic
energy.
Find the coefficient of restitution between the particle and the
plane.
Let the speed of the particle immediately before the collision
be u ms-1.
Applying NEL, v = eu
Therefore the particle has a speed of eu ms-1
immediately after the collision.

K.E. before collision = ½ × m × u2
K.E. after collision = ½ × m × (eu)2
Since the particle loses ¼ of its kinetic energy in the
collision
(½ × m × u2) – (½ × m × (eu)2) = ¼(½ × m × u2)
Therefore the coefficient of restitution is .
3
2
Loss in K.E. = (½ × m × u2) – (½ × m × (eu)2)
½m(u2 – e2u2) = ¼(½mu2)
3
4
e2 =
4mu2 – 4me2u2 = mu2
4 – 4e2 = 1
3
2
 e =

Momentum
Impulse
Direct collisions

Exam question 1
A rocket of mass 3000 kg is travelling at 100 ms-1. It then splits into two
sections, the front and the rear, of masses 2000 kg and 1000 kg
respectively.
Both pieces continue to travel in the same direction as before the
separation.
If the speed of the front section is 125 ms-1 after the separation find:
a) The impulse exerted on the front piece as a result of the
separation.
b) The speed of the rear section after the separation.

Exam question 1
Before After
3000 2000
1000
100 125
v
Impulse = (2000 × 125) – (2000 × 100)
Therefore the impulse exerted on the front part of the rocket is 50
kNs.
The change in momentum refers to the front part of the rocket only and so
only the front part of the rocket is used when calculating the momentum
before the collision.
= 50 000

Exam question 1
3000 × 100 = 1000v + 2000 × 125
Therefore the rear section of the rocket is travelling at a speed of 50 ms-1
after separation.
300 000 = 1000v + 250 000
1000v = 50000
v = 50

Exam question 2
Two particles A and B of masses 2 kg and 0.5 kg respectively are at rest on
a smooth horizontal surface.
A is projected towards B with a speed of 5 ms-1.
As particle A collides with B they coalesce to form a single particle C which
continues to move in the same direction as A.
Particle C strikes a smooth fixed vertical barrier.
If the coefficient of restitution between C and the barrier is ½, calculate the
energy lost in the collision between C and this barrier.

Exam question 2
Before After
2.5
v1
0.5
2
0
5
v2
e = ½
To find the loss in kinetic energy after the second collision
it is first necessary to find the speed of C before and after
the collision.
The first step is to calculate the speed of C immediately
after A and B have coalesced.

Exam question 2
Using conservation of momentum,
2 × 5 + 0 = 2.5 × v1
Therefore the speed of C before the collision with the barrier is 4 ms-
1.
It is now necessary to find the speed of C after the collision with the barrier.
10 = 2.5v1
 v1 = 4

Exam question 2
Apply NEL:
v = eu = ½ × 4 = 2
Therefore the speed of C after the collision is 2 ms-1.
We can now find the loss in kinetic energy asked for.
K.E. before = ½ × 2.5 × 42 = 20
Therefore the energy lost in the collision between C and the barrier is 15
J.
Loss in K.E. = 15
K.E. after = ½ × 2.5 × 22 = 5

Exam question 3
Another sphere B of mass 3m kg is at rest on the table.
a) show that the speed of B immediately after the collision is ½(1 + e)u.
b) find the range of values of e.
c) kinetic energy is lost in the collision. What form of energy could this lost
kinetic energy be transferred to?
A collides directly with B and as a result of this collision its direction of
motion is reversed. If the coefficient of restitution is e:
A smooth sphere A of mass m kg is moving in a straight line on a smooth
horizontal surface with a speed of 2u ms-1.

Exam question 3
a) Using Conservation of Momentum:
2mu + 0 = mv1 + 3mv2
Apply NEL:
v2 – v1 = –e(0 – 2u)
Solve simultaneously by adding:
Therefore the speed of B after the collision is ½(1 + e)u ms-1.
 v1 + 3v2 = 2u
 v2 – v1 = 2eu
v2 = ½u(e + 1)
4v2 = 2eu + 2u
m 3m
2u 0
v1 v2

Exam question 3
v1 = v2 – 2eu
Since v1 < 0: ½u(1 – 3e) < 0
Therefore < e ≤ 1
1
3
c)
u – 3eu < 0
1
3
 e >
1 – 3e < 0
b) To calculate the range of values of e we use the
fact that the motion of A is reversed, i.e. v1 < 0.
v1 = (½u + ½eu) – 2eu
v1 = ½u(1 – 3e)
Kinetic energy could have been transformed into sound
energy.

COLLISION SPP.pptx

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