In one line
The coefficient of restitution e is the ratio of the relative velocity of separation to the relative velocity of approach along the line of impact.
Explained simply
Drop a cricket ball and a tennis ball from the same height: the tennis ball bounces higher. The coefficient of restitution measures this 'bounciness'. A value of 1 means a perfect bounce with no energy lost; 0 means the bodies stick together.
Technically, along the line of impact, \( e = \dfrac{v_2 - v_1}{u_1 - u_2} \), where u are velocities before and v after impact. It depends on the materials of both bodies.
How it works
- Take the line of impact as the axis and choose a positive direction.
- Velocity of approach \( = u_1 - u_2 \); velocity of separation \( = v_2 - v_1 \).
- Use only components along the line of impact.
- Ball dropped on a fixed floor: impact speed \( \sqrt{2gh_1} \), rebound speed \( \sqrt{2gh_2} \), so \( e = \sqrt{h_2/h_1} \).
- After n bounces, the rebound height is \( e^{2n} \) times the drop height \( h_1 \) (so the first rebound is \( h_2 = e^{2} h_1 \)).
Key formula
\( e = \dfrac{v_2 - v_1}{u_1 - u_2} \); for bounce: \( e = \sqrt{h_2/h_1} \). Velocities in m/s, heights in m; e has no unit.
Quick example
A ball dropped from 4 m rebounds to 2.25 m: \( e = \sqrt{2.25/4} = 0.75 \). The next bounce reaches \( 0.75^2 \times 2.25 = 1.27 \) m.
Common mistakes
- Writing \( e = v/u \) without signs for two moving bodies, which can make e negative or greater than 1.
- Using total velocities instead of components along the line of impact in oblique impact.
- Writing \( e = h_2/h_1 \) instead of its square root.
How MU asks it
Theory 3 marks: 'Define the coefficient of restitution'. Numerical 4 to 6 marks: bouncing ball (find e or total distance travelled), or as one of the two equations in collision problems. Always write the signed form with a stated positive direction.