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AS-Level Maths: Mechanics 2for Edexcel

M2.5 Collisions


Con

tent

s


Momentum

Momentum

Impulse

Direct collisions

Collision with a smooth plane

Examination-style questions


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

v7A 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)


Con

tent

s


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°



Impulse = change in momentum

The velocity of the ball before and after the collision can be expressed in vector notation as follows:

I = mv – mu = 2 × (–8 cos 30°i + 8 cos 60°j) – 2 × 6i

u = 6iv = –8 cos 30°i + 8 cos 60°j (where i is horizontal and j is vertical).

After2

8ms-1

30°

I = 2 × (–43i + 4j) – 12i



22

8 3 12 8 (3 s.f.) 27.1 Ns

(–83 – 12)

8

θ

I

I =

θ = (3 s.f.) to the negative i direction.

1 88 3 12

tan 7=1 .2

I = (–83 – 12)i + 8j


Con

tent

s


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.

( )m m e Xm m

2 21 2

1 21

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 = 9e79

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.


13



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.

67

674

Adding gives: 7v2 = 34 674 v2 =

v2 – v1 = 4

v2 – v1 = 4

3v2 – 3v1 = 12

3v1 + 4v2 = 22

67 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

v10

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


Con

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s



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.

14

14



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=414ve 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 .32

Loss in K.E. = (½ × m × u2) – (½ × m × (eu)2)

½m(u2 – e2u2) = ¼(½mu2)

34e2 =

4mu2 – 4me2u2 = mu2

4 – 4e2 = 132 e =


Con

tent

s



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 20001000

100 125v

Impulse = (2000 × 125) – (2000 × 100)

Therefore the impulse exerted on the front part of the rocket is 50 kNs.

Impulse = change in momentumThe 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

0.52

05

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 13

c)

u – 3eu < 0

13 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.

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