The ball that comes back

Problem

The ball that comes back. You throw a ball horizontally against a wall, the ball bounces elastically and comes back towards you. Discuss the change in momentum of the ball and the impulse transferred from the wall to the ball (and vice versa).

Solution

Take the outward direction (towards the wall) as positive. The ball of mass $m$ arrives with velocity $+v$ and, since the collision is elastic, leaves with the same speed in the opposite direction, $-v$.

Change in momentum of the ball: $\Delta p = p_\text{fin} - p_\text{in} = m(-v) - m(+v) = -2mv$ It is not $0$: the sign flips, so the ball undergoes a large change, equal in magnitude to twice the initial momentum.

By the impulse-momentum theorem, the impulse the wall exerts on the ball is exactly this change: $J_\text{wall} = \Delta p = -2mv$ By Newton’s third law, the ball exerts an equal and opposite impulse on the wall: $J_\text{ball} = +2mv$ The wall (and the Earth to which it is anchored) has enormously greater mass, so its velocity change $\Delta V = J_\text{ball}/M_\text{wall}$ is negligible: it stays practically still. Result: $\ev{\Delta p_\text{ball} = -2mv,\quad J_\text{wall\to ball} = -2mv,\quad J_\text{ball\to wall} = +2mv}$

Links

Topics: Momentum and collisions Concepts: Impulse · Momentum · Newton’s third law · Elastic collision Skills: Symbolic setup