Elastic and inelastic collisions in one dimension

What a collision keeps

In a collision, the bodies push on each other with huge forces for a short time. By the third law these forces are equal and opposite, so the total momentum is conserved:

Kinetic energy is another matter:

• Elastic: kinetic energy is conserved too (hard steel balls, atoms).
• Inelastic: some kinetic energy becomes heat, sound or a dent.
• Perfectly inelastic: the bodies stick together. This loses the most kinetic energy that momentum allows.

Elastic, head-on, one body at rest

Solving both conservation laws together gives

Three cases to remember: equal masses swap velocities; a light ball bounces back off a heavy one; a heavy ball sends a light one off at nearly $2u_1$. Play with them.

You should find that with $e = 1$ no kinetic energy is lost, and with $e = 0$ the balls move off together.

Watch out. Momentum is a vector. In one dimension, give every velocity its sign before you add.

A worked example

A $4.0\ \text{kg}$ trolley at $3.0\ \text{m/s}$ hits a $2.0\ \text{kg}$ trolley at rest.

1. Elastic. $v_1 = \dfrac{4.0\ \text{kg} - 2.0\ \text{kg}}{6.0\ \text{kg}} \times 3.0\ \text{m/s} = 1.0\ \text{m/s}$ and $v_2 = \dfrac{8.0\ \text{kg}}{6.0\ \text{kg}} \times 3.0\ \text{m/s} = 4.0\ \text{m/s}$.
2. Check. Momentum: $4.0\ \text{kg m/s} + 8.0\ \text{kg m/s} = 12\ \text{kg m/s}$, as before. Energy: $\tfrac12(4.0\ \text{kg} \times 1\ \text{m}^2/\text{s}^2 + 2.0\ \text{kg} \times 16\ \text{m}^2/\text{s}^2) = 18\ \text{J}$, as before.
3. If they stick. $v = \dfrac{12\ \text{kg m/s}}{6.0\ \text{kg}} = 2.0\ \text{m/s}$, and the energy falls from $18\ \text{J}$ to $12\ \text{J}$: $6.0\ \text{J}$ is lost.

Think first. Momentum first, always. Then decide what you know about the kinetic energy.

Practise with new numbers

Try it in Python

The program collides the same two trolleys three times, from elastic to sticking.

Remember these

• Momentum is conserved in every collision with no outside force.
• Elastic: kinetic energy conserved too. Perfectly inelastic: the bodies stick.
• Elastic, target at rest: $v_1 = \dfrac{m_1 - m_2}{m_1 + m_2}u_1$, $v_2 = \dfrac{2m_1}{m_1 + m_2}u_1$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A $0.17\ \text{kg}$ pool ball at $2.4\ \text{m/s}$ hits an identical ball at rest, head-on and elastically. What are the velocities afterwards?

2. A $3.0\ \text{kg}$ cart at $5.0\ \text{m/s}$ couples with a $2.0\ \text{kg}$ cart at rest. What is their speed?

3. Two lumps of clay of equal mass, moving at equal speeds in opposite directions, hit and stick. What happens?

4. Why does a tennis ball bounce higher off a hard floor than a lump of clay does?