7 Collisions

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Collisions

Specification references

 3.5.1 a) b) c) d) e)

 3.5.2 a) b) c)

 M2.2, M2.3

 M3.1

Introduction

At GCSE you should have met the concept of momentum, given by the equation: momentum, p  mass, m  velocity, v

You should also know that, assuming no external force is acting, the total

momentum before a collision is equal to the total momentum after it. In other words, linear momentum is conserved in a collision. Kinetic energy is conserved in elastic collisions, but some of it is converted to other forms in inelastic collisions.

In this worksheet you will become more familiar with the concept of collisions and practise problems involving collisions.

Learning outcomes

After completing the worksheet you should be able to:

 understand that momentum is equal to the product of mass and velocity

 understand the concept of the conservation of linear momentum

 apply the principle of conservation of linear momentum to problems in one dimension, including elastic and inelastic collisions

 understand that force may be expressed as the rate of change of momentum,

 understand that impulse is the rate of change of momentum and be able to express it as an equation, F Δt Δ(m v)

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Background

There are three common collision scenarios that you should be familiar with. These are shown below. In each it’s assumed that the positive direction is to the right.

m1 is the mass of the first object

u1 is the velocity of the first object before collision

v1 is the velocity of the first object after collision

m2 is the mass of the second object

u2 is the velocity of the second object before collision

v2 is the velocity of the second object after collision 1 Standard collision

total momentum before collision  total momentum after collision

m1 u1 m2 u2m1 v1m2 v2

2 Objects join together on impact

Since they join together, both objects travel with the same velocity after impact.

m1 u1 m2 u2 (m1 m2) v

3 Head-on collision

Objects travel in opposite directions after impact

m1 u1 m2 u2 −m1 v1m2 v2

(In this case the term m1 v1 has a minus sign before it since the momentum is to

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Worked example 1 – Collision

Question

A truck of mass 50 kg travelling with a velocity of 3.0 m s–1_{collides with a stationary}

truck of mass 30 kg and they move on together.

a Calculate their velocity after the collision.

b Is the collision elastic or inelastic?

(Be on the lookout for keywords in the question to identify what kind of collision it is.)

Answer

Step 1

Draw a simple diagram of the collision, including the values provided.

a Step 2

Find the velocity after collision by applying the law of conservation of momentum.

m1 u1m2 u2 (m1m2) v

(50  3)  (30  0)  (50  30) v

150  0  80 v

v

v 1.9 m s−1_{(to two significant figures)} b Step 3

To decide whether the collision is elastic or inelastic, you need to calculate the kinetic energy before and after the collision, and compare the values.

EK m v2

Before collision:

EK m1 u12 m2 u22

 (  50  32₎_₍ _₃₀_₀2₎

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After collision:

EK (m1m2) v2

  (50  30)  1.92

 140 J

Since EK before collision ≠ EK after collision, the collision is inelastic.

Worked example 2 – Impulse

Question

A snooker cue hits a stationary white snooker ball of mass 0.20 kg and applies an average force of 40 N over 9.0  10–3_s.

Calculate

a the impulse

b the velocity of the snooker ball after the impact.

Answer

a Step 1

Calculate the impulse. impulse F Δt

 40  9.0  10−3

 0.36 N s (or kg m s−1₎ b Step 2

Since the impulse represents the change in momentum, you can now calculate the velocity after impact using the equation:

impulse  momentum after collision − momentum before collision

F tm v – m u

Since the snooker ball starts from stationary, u 0 m s−1_.

0.36  (0.20 v) − (0.20  0) 0.36  0.2 v

v

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Worked example 3 – force–time graphs

Question

The graph shows how the force acting on an object varies with time. If the object is moving in a straight line, calculate the change in momentum.

Answer

Since the change in momentum is equal to the impulse, you can find it by calculating the area under the force–time graph. This is because the area under a force–time graph is equal to the force multiplied by time.

You can divide the area under this particular graph into a rectangle and a triangle. total area under the graph  area of rectangle  area of triangle

 (2  6)  (  4  4)

 12  8

 20 kg m s−1

Questions

1 A car has a total mass of 2000 kg and is moving at 12.5 m s−1_{when it strikes a}

smaller car head on. The smaller car has a mass of 1200 kg and was moving at 16.7 m s−1_{before the collision.}

a Calculate the velocity of the cars if they move off together as one after the collision. (2 marks)

b Is the collision elastic or inelastic? You must show your working. (2 marks)

2 A bullet of mass 2.0  10−3_{kg moving horizontally with a velocity of 400 m s}–1

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3 A tennis player serves a ball of mass 0.057 kg at an initial horizontal speed of 55 m s−1_{. The ball is in contact with the racquet for 0.055 s.}

Calculate the average force exerted on the ball during the serve. (2 marks)

4 Figure 1 shows the first few seconds of an object’s motion. Use the graph to calculate the increase in momentum of the object during the first 4 s of its motion.

(2 marks)

Figure 1

Exam-style question

5 A cricket ball of mass 0.12 kg strikes a stationary bat at a speed of 20 m s−1_.

The contact time of the ball with the bat is 0.15 s. The ball then returns along its original path with a speed of 16 m s−1_.

Assuming the bat remains stationary, calculate

a the momentum of the ball before the impact with the bat (1 mark)

b the momentum of the ball after the impact with the bat (1 mark)

c the change of momentum of the ball during contact with the bat (1 mark)

d the average force acting on the ball while in contact with the bat(1 mark)