Elastic Collisions Head-on

Two laws, two unknowns

In an elastic head-on collision two laws hold at once: momentum is conserved and kinetic energy is conserved. There are two unknowns, the final velocities \(v_1\) and \(v_2\). Two equations for two unknowns settle the collision completely.

Take the simplest case first. Body 1, of mass \(m_1\), moves with velocity \(u\) towards body 2, of mass \(m_2\), at rest. Body 2 is the target. The two laws read as follows; the derivation below collects the terms of each body on one side.

\[ m_1u = m_1v_1 + m_2v_2, \qquad \tfrac12 m_1u^2 = \tfrac12 m_1v_1^2 + \tfrac12 m_2v_2^2 \]

Line 4 says something simple. Before the collision the gap between the bodies closed at \(u\). After it the gap opens at \(v_2 - v_1\), which is also \(u\). The bodies separate exactly as fast, relative to each other, as they approached.

The rate at which the gap closes is the velocity of approach, \(u_1 - u_2\). The rate at which it opens is the velocity of separation, \(v_2 - v_1\). These are relative velocities, as in the chapter on Relative Velocity in One Dimension.

Check Fig 3.1 with both laws. Momentum before: \(1 \times 6 + 2 \times (-3) = 0\). Momentum after: \(1 \times (-6) + 2 \times 3 = 0\). Kinetic energy: \(18 + 9 = 27\,\text{J}\) both times. Each block has simply reversed its velocity. That happens whenever the total momentum is zero.

The two results

Solve line 1 of the derivation together with \(v_2 = u + v_1\), and the two final velocities follow. These two results settle every elastic head-on collision with the target at rest.

The sign of \(v_1\) carries the story. If \(m_1 > m_2\), \(v_1\) is positive and the moving body continues forward. If \(m_1 < m_2\), \(v_1\) is negative and it rebounds. The target always moves forward, since \(v_2\) is always positive.

Three cases to know by heart

The two results give three pictures that are worth remembering. Move the slider below from a light target to a heavy one, and press play each time.

Equal masses swap

Put \(m_1 = m_2\) in the results: \(v_1 = 0\) and \(v_2 = u\). The moving body stops dead, and the target leaves with its whole velocity. A Newton's cradle on an office desk shows this. The struck ball carries away all the momentum and all the kinetic energy of the ball that struck it.

A heavy body flings a light one

Let \(m_1\) be much larger than \(m_2\). Then \(v_1\) is nearly \(u\) and \(v_2\) nearly \(2u\). The heavy body hardly notices the collision. The light body must separate from it at \(u\). The heavy body still moves at \(u\), so the light body moves at nearly \(2u\) relative to the ground.

A light body rebounds from a heavy one

Let \(m_1\) be much smaller than \(m_2\). Then \(v_1\) is nearly \(-u\) and \(v_2\) nearly zero. The heavy target barely moves, and the light body comes back with nearly its own speed. A ball thrown hard at a wall is this case, with the wall as the heavy body.

Reading a mass from a collision

The results can be run backwards. If the speed of the striker after the collision is measured, the ratio \(v_1/u\) fixes the mass of the target. This is one way to weigh a body that cannot be put on a scale, such as a particle too small to see.

How much energy passes to the target

The target at rest receives the kinetic energy \(\tfrac12 m_2v_2^2\). Divide it by the striker's \(\tfrac12 m_1u^2\) and put in \(v_2\). The fraction passed on is below.

\[ \dfrac{K_2}{K_1} = \dfrac{4m_1m_2}{(m_1 + m_2)^2} \]

It equals \(1\) when \(m_1 = m_2\), and it is small when the masses are very different. The slider shows it: at \(m_2 = 3\,\text{kg}\) the target takes \(75\,\%\), and at \(0.1\,\text{kg}\) or \(10\,\text{kg}\) it takes only \(33\,\%\).