The cars in Figure 6.7 have been badly damaged by a collision. The front of a car is designed to absorb the impact of the crash. It has a ‘crumple zone’, which collapses on impact. This absorbs most of the kinetic energy that the car had before the collision. It is better that the car’s kinetic energy should be transferred to the crumple zone than to the driver and passengers.
Motor manufacturers make use of test labs to investigate how their cars respond to impacts. When a car is designed, the manufacturers combine soft, compressible materials that absorb energy with rigid structures that protect the people in the car. Old-fashioned cars had much more rigid structures. In a collision, they were more likely to bounce back and the violent forces involved were much more likely to prove fatal.

Two types of collision

When two objects collide, they may crumple and deform. Their kinetic energy may also disappear completely as they come to a halt. This is an example of an inelastic collision. Alternatively, they may spring apart, retaining all of their kinetic energy. This is a perfectly elastic collision. In practice, in most collisions, some kinetic energy is transformed into other forms (such as heat or sound) and the collision is inelastic. Previously we described the collisions as being ‘springy’ or ‘sticky’. We should now use the correct scientific terms, perfectly elastic and inelastic.
We will look at examples of these two types of collision and consider what happens to linear momentum and kinetic energy in each.

A perfectly elastic collision

Two identical objects, A and B, moving at the same speed but in opposite directions, have a head-on collision, as shown in Figure 6.8. Each object bounces back with its velocity reversed. This is a perfectly elastic collision.

You should be able to see that, in this collision, both momentum and kinetic energy are conserved. Before the collision, object A of mass m is moving to the right at speed v and object B of mass m is moving to the left at speed v. Afterwards, we still have two masses m moving with speed v, but now object A is moving to the left and object B is moving to the right. We can express this mathematically as follows.

Before the collision

The relative speed of approach is the speed of one object measured relative to another. If two objects are travelling directly towards each other with speed v, as measured by someone stationary on the ground, then each object ‘sees’ the other one approaching with a speed of $2v$. Thus, if objects are travelling in opposite directions we add their speeds to find the relative speed. If the objects are travelling in the same direction then we subtract their speeds to find the relative speed.
To find the relative speed of two objects you subtract the velocity of one from the velocity of the other. This is the same as adding on a velocity in the opposite direction; so, if two objects approach each other in exactly opposite directions with velocities of ${v_1}$ and $ - {v_2}$, their relative speed $ = {v_1} - ( - {v_2}) = {v_1} + {v_2}$.

An inelastic collision

In Figure 6.9, the same two objects collide, but this time they stick together after the collision and come to a halt. Clearly, the total momentum and the total kinetic energy are both zero after the collision, since neither mass is moving. We have:

Again we see that momentum is conserved. However, kinetic energy is not conserved. It is lost because work is done in deforming the two objects.
In fact, momentum is always conserved in all collisions. There is nothing else into which momentum can be converted. Kinetic energy is usually not conserved in a collision, because it can be transformed into other forms of energy – sound energy if the collision is noisy, and the energy involved in deforming the objects (which usually ends up as internal energy – they get warmer). Of course, the total amount of energy remains constant, as stated in the principle of conservation of energy.

Solving collision problems

We can use the idea of conservation of momentum to solve numerical problems, as shown in Worked example 2.

WORKED EXAMPLE

2) In the game of bowls, a player rolls a large ball towards a smaller, stationary ball. A large ball of mass $5.0 kg$ moving at $10.0\,m\,{s^{ - 1}}$ strikes a stationary ball of mass $1.0 kg$. The smaller ball flies off at $10.0\,m\,{s^{ - 1}}$.
a: Determine the final velocity of the large ball after the impact.
b: Calculate the kinetic energy ‘lost’ in the impact.
Step 1: Draw two diagrams, showing the situations before and after the collision. Figure 6.10 shows the values of masses and velocities; since we don’t know the velocity of the large ball after the collision, this is shown as v. The direction from left to right has been assigned the ‘positive’ direction.
Step 2: Using the principle of conservation of momentum, set up an equation and solve for the value of v:
$\begin{array}{l}
total\,momentum\,before\,collison\, = \,total\,momentum\,after\,collison\\
(5.0\, \times \,10)\, + \,(1.0\, \times \,0)\, = \,(5.0\, \times \,v)\, + \,(1.0\, \times \,10)\\
50\, + \,0\, = \,5.0v\, + 10\\
v\, = \,\frac{{40}}{{5.0}}\\
v\, = \,8.0\,m\,{s^{ - 1}}
\end{array}$
 

So the speed of the large ball decreases to $8.0\,m\,{s^{ - 1}}$ after the collision. Its direction of motion is unchanged – the velocity remains positive.
Step 3: Knowing the large ball’s final velocity, calculate the change in kinetic energy during the collision:
$\begin{array}{l}
total\,k.e\,before\,collision\, = \,\frac{1}{2} \times 5.0 \times {10^2} + 0\\
 = 250J
\end{array}$ 
$\begin{array}{l}
total\,k.e.\,'lost'\,in\,the\,collision\, = \,250J\, - \,210J\\
 = 40J
\end{array}$
This ‘lost’ kinetic energy will appear as internal energy (the two balls get warmer) and as sound energy (we hear the collision between the balls)

Questions

 

5) Figure 6.11 shows two identical balls A and B about to make a head-on collision. After the collision, ball A rebounds at a speed of $1.5\,m\,{s^{ - 1}}$ and ball B rebounds at a speed of $2.5\,m\,{s^{ - 1}}$. The mass of each
ball is $4.0 kg$.

a: Calculate the momentum of each ball before the collision.
b: Calculate the momentum of each ball after the collision.
c: Is the momentum conserved in the collision?
d: Show that the total kinetic energy of the two balls is conserved in the collision.
e: Show that the relative speed of the balls is the same before and after the collision.
6) A trolley of mass $1.0 kg$ is moving at $2.0\,m\,{s^{ - 1}}$. It collides with a stationary trolley of mass $2.0 kg$. This second trolley moves off at $1.2\,m\,{s^{ - 1}}$.
a: Draw ‘before’ and ‘after’ diagrams to show the situation.
b: Use the principle of conservation of momentum to calculate the speed of the first trolley after the collision. In what direction does it move?