Question

If \( u_1 \) and \( u_2 \) are the velocities of two bodies moving in the same direction before impact and \( v_1 \) and \( v_2 \) are the velocities of the same two bodies after impact, then the coefficient of restitution is

• A. \( \dfrac{v_1 - v_2}{u_1 - u_2} \)
• B. \( \dfrac{v_2 - v_1}{u_1 - u_2} \)
• C. \( \dfrac{u_1 - u_2}{v_1 - v_2} \)
• D. \( \dfrac{u_2 - u_1}{v_1 - v_2} \)

Hint:

Always subtract velocities in the direction of motion while finding relative velocities in impact problems.

Answer 1

The Correct Option is B

This is a definitional question about the coefficient of restitution rather than a numeric calculation, so here is the reasoning for each option.
By definition, the coefficient of restitution is the ratio of the relative velocity of separation (after impact) to the relative velocity of approach (before impact), and both must be measured consistently so that the ratio comes out non-negative for a physically ordinary collision.
Before impact, since both bodies move in the same direction with the first chasing the second, the approach speed is \( u_1-u_2 \) (positive); after impact, the separation speed, measured in the same consistent direction, is \( v_2-v_1 \).
Placing separation over approach in this consistent order gives the standard, always non-negative definition. \[ \boxed{e = \dfrac{v_2 - v_1}{u_1 - u_2}} \]

Answer 2

A further way to confirm the correct ordering is to test it against a known limiting case, a perfectly elastic head-on collision between two equal masses, where velocities are known to simply exchange.

1. Option \( \dfrac{v_1 - v_2}{u_1 - u_2} \): For equal masses in a perfectly elastic collision, the velocities swap completely, so after impact \( v_1 = u_2 \) and \( v_2 = u_1 \); substituting, this ratio becomes \( \dfrac{u_2-u_1}{u_1-u_2} = -1 \), but the coefficient of restitution for a perfectly elastic collision should equal \( +1 \), not \( -1 \), so this ordering gives the wrong sign.
2. Option \( \dfrac{v_2 - v_1}{u_1 - u_2} \): Using the same velocity-exchange substitution, \( v_2 = u_1 \) and \( v_1 = u_2 \), this ratio becomes \( \dfrac{u_1-u_2}{u_1-u_2} = 1 \), correctly reproducing \( e=1 \) for a perfectly elastic collision, exactly as physically expected.
3. Option \( \dfrac{u_1 - u_2}{v_1 - v_2} \): Substituting the same swap, this becomes \( \dfrac{u_1-u_2}{u_2-u_1} = -1 \), again the wrong sign for a perfectly elastic collision.
4. Option \( \dfrac{u_2 - u_1}{v_1 - v_2} \): Substituting, this becomes \( \dfrac{u_2-u_1}{u_2-u_1} = 1 \); while this happens to give the right numeric value in this special symmetric case, its structure reverses the before/after roles relative to the standard definition (approach speed placed with reversed sign convention in the numerator instead of denominator), so it does not generalise correctly to asymmetric, non-swapping collisions the way the standard definition does.

Testing each candidate against the well-known perfectly elastic collision case, where the coefficient of restitution must equal exactly \( 1 \), immediately singles out the structurally correct definition.

So the correct answer is \( \dfrac{v_2 - v_1}{u_1 - u_2} \).