Interactive textbook: Collisions

What is a collision?

A collision is an interaction between two particles or bodies where there is an exchange of energy and momentum.

Energy

If there is no mechanical (kinetic and potential) energy lost from the system then kinetic energy is conserved. This is the case when electrons collide since they have no internal structure and, hence, no means of gaining internal energy. Balls are made of atoms which are disturbed during the collision. This disturbance increases the kinetic and potential energy of the atoms (internal energy). The law of conservation of energy states that energy can not be created or destroyed so the total energy in a collision is always conserved.

Notice in this collision how the 'atoms' inside the balls gain energy.

The atoms gain PE because:

The atoms are held together by the interatomic force, increasing their separation require work to be done so energy is transferred.

Momentum

The principle of conservation of momentum states that, for a system of isolated bodies, the momentum is conserved. So, provided that the bodies are isolated, the momentum before the collision is equal to the momentum after it. This is the case when two electrons collide in a vacuum but is rarely true for balls in the lab; a rolling ball has external forces acting on it. In exam questions we ignore this but it is worth bearing in mind when doing practical work.

Elastic collisions

In an elastic collision, both momentum and kinetic energy are conserved.

First we will look at the simplest case of one ball hitting another identical ball (mass \(m\)) that is stationary:

We don't know what will happen after the collision, so let's guess it's something like this:

Applying the principle of conservation of momentum:

\(mv = mv_1 + mv_2\)

\(v = v_1 + v_2\)

If kinetic energy is also conserved:

\({1\over2}mv^2 = {1\over2}m{v_1}^2 + {1\over2}m{v_2}^2\)

\(v^2 = {v_1}^2 + {v_2}^2\)

Squaring the first equation gives:

\(v^2 = ({v_1}^2 + {v_2}^2)^2 = {v_1}^2 + {v_2}^2 +2v_1v_2\)

Putting the two equations together gives:

\({v_1}^2 + {v_2}^2 +2v_1v_2 = {v_1}^2 + {v_2}^2\)

\(\Rightarrow 2v_1v_2 = 0\)

So either \(v_1\) or \(v_2\) is \(0\). If \(v_2\) is \(0\) then no collision could have taken place, so \(v_1\) must be \(0\). The first equation \(v = v_1 + v_2\) then tells us that \(v_2 = v\):

Now let's try with two identical balls moving with \(u_1>u_2\):

Again, we will assume some result:

Applying the principle of conservation of momentum:

\(mu_1 + mu_2 = mv_1 + mv_2\)

\(u_1 + u_2 = v_1 + v_2\)

Kinetic energy is also conserved so:

\({1\over2}m{u_1}^2 + {1\over2}m{u_2}^2 = {1\over2}m{v_1}^2 + {1\over2}m{v_2}^2\)

\({u_1}^2 + {u_2}^2 = {v_1}^2 + {v_2}^2\)

The sum of the velocities and their squares are equal before and after the collision. This can only be true if either:

  •  \(u_1 = v_1\) and \(u_2 = v_2\)
  • \(u_1 = v_2\) and \(u_2 = v_1\) 

The first option would imply that there was no collision so the second must be the solution; the velocities swap.

Two balls of 1 kg mass collide elastically.

What is the velocity of the blue ball after the collision?

The velocities swap.

Two 1 kg balls collide elastically.

What is the change in KE of the red ball?

1/2 x 1 x 32 - 1/2 x 1 x 12

Inelastic collisions

During an inelastic collision, only momentum is conserved.

Sticky collisions

There are many possible outcomes of an inelastic collision. One of those is when the bodies stick together (coalesce):

Applying the principle of conservation of momentum:

\(mu =2mv\)

\(v = {u\over2}\)

We can see that kinetic energy has been lost:

\(\text{loss of }E_\text{k} = {1\over2}mv^2 - {1\over2}(2m)({v\over2})^2 = {1\over2}mv^2 - {1\over4}mv^2={1\over4}mv^2\)

This kinetic energy has been given to the internal energy of the balls.

Explosions

Momentum is also conserved in explosions:

If the initial velocity of the two balls is \(u\) and the final velocity of the single ball is \(v\)...

...applying the principle of conservation of momentum:

\(2mu = mv\)

\(v = 2u\)

...and the gain in kinetic energy is therefore:

\({1\over2}m(2u)^2 - {1\over2}(2m)u^2 = 2mu^2 - mu^2=mu^2\)

Two balls of equal mass collide inelastically sticking together after the collision.

What is the velocity of the balls after the collision?

Momentum before = 3 -1 = 2

Momentum after = 2 x v

Momentum conserved so 2v = 2

Two balls of equal mass collide inelastically.

What is the total loss of KE?

KE before = 4.5 + 0.5 = 5 J

KE after = 1/2 x 2 x 12 = 1 J

KE lost = 4 J

Two-dimensional collisions

Two-dimensional collisions are a bit beyond the Subject Guide, but it's worth taking a moment to consider the simplest case of an elastic collision between two bodies with equal mass:

When dealing with one-dimensional collisions, the direction of velocity and momentum is given by the sign. However, in two dimensions we need to draw the vectors when adding momenta. Applying the principle of conservation of momentum leads to the vector equation:

\(\overrightarrow{u}=\overrightarrow{v_1}+\overrightarrow{v_2}\)

This can be represented by the following vectors:

If the collision is elastic then kinetic energy is conserved. Since energy is a scalar quantity, we can simply use the magnitudes of the velocities:

\( u^2 = {v_1}^2 + {v_2}^2\)

This relationship implies that the balls always bounce off each other at \(90°\). The reason? This condition only applies for right-angled triangles, as per Pythagoras' theorem.

In the collision above the balls are:

velocities only add like this if the bodies have the same mass and KE is only conserved in elastic collisions.

Total Score:

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