No conservation of momentum with bouncing ball?

nhmllr

So I was thinking about the conservation of momentum. If you throw a handball at a wall, the wall will provide an equal normal force, thus sending the handball back at the same velocity (in a perfect scenario). The ball has a momentum vector, the wall never moves, and thus only has a zero-amplitude vector. But in this closed system, the net momentum vector changes! How?

russ_watters

The ball is not perfectly elastic: some of the spring energy in the bounce is converted to heat.

nhmllr

So what would happen if the ball was perfectly elastic? It would still bounce off, right?

russ_watters

It would bounce back with an equal momentum to what it started with.

AceInfinity

Sound is also a form of energy in which the initial energy of the ball gets converted to, so it loses energy there. In a perfectly elastic system the momentum is conserved completely and none is wasted. meaning p = p' (momentum before = momentum after)

nhmllr

But the vector is in a completely different direction!

Pengwuino

the wall does move, except that it's connected to the ground (aka Earth) which means it only appears to not move. If you really could isolate the ball and wall/earth system, the momentum would be conserved. Of course, look at the masses you're talking about and you can understand why the wall seems to not move.

nhmllr

Ah- I see. little mass x big velocity = huge mass x tiny velocity. So I suppose in deep space the wall would actually start moving back, and the conservation would be more obvious.

Thanks

sophiecentaur

Don't forget that when you threw the ball, the conservation law also applied and the Earth rotated backwards a tiny bit.
The amount it moves forwards would be twice that value for a perfectly elastic collision and an equal value for a totally inelastic collision. All the Mv's add up to zero in every case.