One arrival, three endings
Momentum is conserved in every collision, as the last lesson showed. Kinetic energy is a different matter. Some of it may become heat, sound and a lasting dent while the bodies are squashed. So collisions are sorted by one question: what happens to the total kinetic energy of the pair?
In an elastic collision the total kinetic energy after the collision equals the total before it. The bodies deform during contact, but they regain their shape fully, and every joule stored in the deformation returns as motion. Two steel balls or two billiard balls come very close to this.
In an inelastic collision the total kinetic energy after is less than before. The bodies do not fully regain their shape, or they turn some energy into heat and sound. Almost every real collision is of this kind, from a cricket ball on a bat to two cars in traffic.
In a completely inelastic collision the bodies stick together and continue with one common velocity. They never pass beyond the stage of greatest deformation. A lump of wet clay on a trolley, and a bullet that stays inside a wooden block, are the standard examples.
Compare the three chips of Fig 2.1. The arrival is the same each time, and the momentum after is the same each time. Only the sharing of the motion changes. Elastic: A continues at \(1\,\text{m/s}\) and B at \(4\,\text{m/s}\). Inelastic: A at \(1.5\,\text{m/s}\) and B at \(3\,\text{m/s}\). Sticking: both at \(2\,\text{m/s}\).
Elastic does not mean that the bodies never deform. Two steel balls do squash a little at contact, and for that instant part of the kinetic energy is stored in the squashed metal. The collision is elastic because all of it returns before the balls separate.
A tennis ball hits a racket and leaves with less total kinetic energy than the ball and racket had before. What can you say about their total momentum?
Which statements about collisions are true?
When bodies stick
A completely inelastic collision needs only momentum. The bodies continue as one, so the common velocity of the last lesson is the final velocity.
\[ v = \dfrac{m_1u_1 + m_2u_2}{m_1 + m_2} \]
The kinetic energy before is \(\tfrac12 m_1u_1^2 + \tfrac12 m_2u_2^2\), and after it is \(\tfrac12 (m_1 + m_2)v^2\). The difference is the kinetic energy lost in the collision.
A car of \(1000\,\text{kg}\) moving at \(20\,\text{m/s}\) runs into the back of a car of \(1500\,\text{kg}\) standing at a signal. The two lock together. Find their speed just after the crash and the kinetic energy lost.
Take the direction of the moving car as positive. The cars stick, so this is a completely inelastic collision.
Momentum: \(1000 \times 20 + 1500 \times 0 = 2500\,v\).
\(v = 20000/2500 = 8\,\text{m/s}\). Kinetic energy before: \(\tfrac12 \times 1000 \times 20^2 = 200\,\text{kJ}\). After: \(\tfrac12 \times 2500 \times 8^2 = 80\,\text{kJ}\).
The cars move together at \(8\,\text{m/s}\), and \(120\,\text{kJ}\), three-fifths of the kinetic energy, goes into crushed metal, heat and sound.
One formula for the loss
Work the same steps with symbols, and the loss takes one neat form. The steps are in the derivation below. The loss depends only on the two masses and on the relative velocity \(u_1 - u_2\). It does not depend on how fast the pair as a whole moves.
\[ \Delta K = \tfrac12 m_1u_1^2 + \tfrac12 m_2u_2^2 - \tfrac12 (m_1 + m_2)v^2 \]
Why. the kinetic energy before minus the kinetic energy after.
\[ (m_1 + m_2)v^2 = \dfrac{(m_1u_1 + m_2u_2)^2}{m_1 + m_2} \]
Why. put in the common velocity \(v = (m_1u_1 + m_2u_2)/(m_1 + m_2)\).
\[ \Delta K = \dfrac{(m_1 + m_2)(m_1u_1^2 + m_2u_2^2) - (m_1u_1 + m_2u_2)^2}{2(m_1 + m_2)} \]
Why. bring both terms over the common denominator \(2(m_1 + m_2)\).
\[ \Delta K = \dfrac{m_1m_2\,(u_1^2 - 2u_1u_2 + u_2^2)}{2(m_1 + m_2)} \]
Why. open the brackets. The terms \(m_1^2u_1^2\) and \(m_2^2u_2^2\) cancel.
\[ \Delta K = \dfrac{m_1m_2}{2(m_1 + m_2)}(u_1 - u_2)^2 \]
Read it. the bracket is a perfect square.
Check it on the cars. \(m_1m_2/(m_1 + m_2) = 1000 \times 1500/2500 = 600\,\text{kg}\), and \(u_1 - u_2 = 20\,\text{m/s}\). So \(\Delta K = \tfrac12 \times 600 \times 400 = 120\,\text{kJ}\), as before. With the target at rest, the fraction lost is \(m_2/(m_1 + m_2)\). A light body that sticks to a heavy one loses almost all its energy.
The combination \(\mu = m_1m_2/(m_1 + m_2)\) is called the reduced mass of the pair. The loss is \(\tfrac12 \mu (u_1 - u_2)^2\), the kinetic energy of the relative motion. Seen from the centre of mass, the pair's motion is only this relative motion, and sticking removes all of it. The motion of the centre of mass itself can never be removed, because the momentum must survive.
A \(0.2\,\text{kg}\) lump of clay moving at \(10\,\text{m/s}\) hits a \(0.8\,\text{kg}\) trolley at rest on a smooth track and sticks to it. How much kinetic energy is lost, in joules?
Two identical cars moving at the same speed in opposite directions collide head-on and lock together. What fraction of their kinetic energy is lost?
A ball moving on a smooth floor hits a ball at rest and sticks to it. Explain why this collision cannot turn all of the kinetic energy into heat.
Momentum is conserved in the collision. Before it, the moving ball carries momentum, so after it the pair must carry the same momentum. A pair with momentum is moving, and a moving pair has kinetic energy. So some kinetic energy must remain, \(\tfrac12(m_1 + m_2)v^2\) with \(v\) the common velocity.
In 1668 the Royal Society in London asked its members for the laws of colliding bodies. Three papers arrived within a few weeks. John Wallis described bodies that stay together after the impact, the inelastic case. Christopher Wren and Christiaan Huygens described perfectly springy bodies, the elastic case. Together the papers showed that mass times velocity, counted with its direction, is the same before and after an impact.
- Treating "inelastic" as "momentum not conserved".The word is about kinetic energy only. Momentum is conserved in every collision.
- Taking "completely inelastic" to mean that the bodies stop.They stick and continue with the common velocity. They stop only if the total momentum was zero.
- Writing \(\tfrac12 m_1u_1^2 = \tfrac12 (m_1 + m_2)v^2\) for bodies that stick.Kinetic energy is not conserved when bodies stick. Use momentum for \(v\), and energy only to find the loss.