# Conservation of energy when ball rebounds [duplicate]

Consider a ball that strikes a stationary box and has momentum $$mv$$. If the ball rebounds with the same speed it had before the collision, its momentum is $$-mv$$. Therefore, the box must have a momentum of $$2mv$$ in order for momentum to be conserved. However this would mean the kinetic energy after the collision is more than the kinetic energy before the collision. Where would this energy come from, or is the situation I described impossible?

• Have you made some assumptions? Nov 14, 2019 at 15:13

The energy could come from potential energy somewhere. For example, if the ball has a compressed spring attached to it that is set to release upon impact against the box, them the potential energy could be converted into kinetic energy of the system, and this is where the increase would come from.

You are correct that if there are no mechanisms like this at play though, then the scenario is impossible exactly as you describe it, as you cannot get kinetic energy from nowhere.

However, typically we do describe balls striking walls as elastic collisions where the ball comes in with momentum $$mv$$ and leaves with momentum $$-mv$$, but in this we assume a negligible change in velocity to the wall. Therefore, we have a collision where kinetic energy remains constant, i.e an elastic collision.

More formally, the wall (or massive box) gains a momentum $$2mv$$, but since the wall is so massive with mass $$M\gg m$$, the wall's velocity after the collision is $$V=\frac{2mv}{M}\ll v$$, so its kinetic energy $$\frac12MV^2\ll\frac12mv^2$$ due to the squared velocity term. So, your scenario is essentially possible without additional energy if the box is much more massive than the ball.

The reason this occurs is that a change in kinetic energy requires work to be done, which depends on the distance over which a force is applied. If the box is massive enough then it will barely move due to the collision, and hence the work done on the box will be negligible. This gives you the negligible discrepancy in kinetic energy if you choose to ignore the kinetic energy of the massive box.

In your question the kinetic energy of the ball is the same before and after collision as the ball has the same speed.
However two implicit assumptions have been made: that the box has an infinite mass and the collision is elastic.

If this is not the case then the rebound speed of the ball is slightly less than $$v$$ and the box will be moving and so have some kinetic energy as well.

The initial kinetic energy of the ball will equal the final kinetic energy of the ball plus the final kinetic energy of the box (plus mechanical energy lost due the collision being inelastic).