Super bounciness...conservation?

Question

A large superball and a small superball are dropped from the same initial height H. Both balls bounce backup to a height h, which is just a bit less than H. If, however, the small superball is placed immediately above the large superball and both are dropped to the ground simultaneously from height H, the small superball rebounds to a height h' > H! Explain this in terms of Conservation Laws.

Is it because the larger ball is providing the smaller one with extra "bounce"? I'm not exactly sure...thanks =)

Answer 1

'Large' and 'small' probably refer to the masses of the balls, rather than the sizes. Assume one ball (with mass m1) is initially directly above the other (mass m2), and both are dropped at the same time from rest (with a slight vertical space between the two).

The lower ball hits first, bounces up and collides with the upper ball near the ground. The two have about the same speed v = sqrt(2gH) going into this collision because both balls have fallen from approximately the same height. If the two masses were equal, then both would simply reverse direction, the top ball going upward at speed v and the bottom ball going down (for the second time) and immediately up at the same speed v --> both balls would continue up at the same speeds to the original height H.

Now suppose the masses are slightly different with m2 > m1. Then the initial and final velocities (taking the upward direction as positive) for the collision are respectively:

v1 = -v
v2 = v

and:

v1 = v + x1
v2 = -v + x2

where x1,x2 << v. Energy and momentum conservation require that:

(m2-m1)v = m1(v + x1) + m2(-v + x2)
(m1+m2)v^2 = m1(v + x1)^2 + m2(-v + x2)^2
= m1(v^2 + 2vx1) + m2(v^2 - 2vx2) (Approximately)
-->
2(m2-m1)v = m1x1 + m2x2
0 = 2m1vx1 - 2m2vx2
-->
x1 = (m2/m1 - 1)v
x2 = (1 - m1/m2)v

So the top (lighter) ball ends up with a higher speed (m2/m1)v and the lower (heavier) ball ends up with the lower speed (m1/m2)v. This is reflected by the one bouncing higher than H and the other lower than H.

There can't be more than one collision since the top ball comes out of the first (and only) collision with a higher speed than the bottom ball so the latter can never catch up to it. If the mass ratio were reversed (heavier ball on top) the situation would be much more complicated since there could be multiple collisions.

Answer 2

It depends upon the volume of the balls - the smaller ball is a certain ratio volume of the big ball but the surfaces of the balls are is NOT the same ratio - therefore the conservation of energy would be "wound up" more tightly in the smaller ball. The smaller ball releases more energy at the instant of impact than the larger ball. The difference in height could be sufficient for these results - what happens when all your measurements are based on the centre point of the balls?

(Of course, all of this assumes that other factors are eliminated - air resistance, gravitational variations, etc.)

Answer 3

If the small ball is "above" the large one, then I guess they are not dropped from the *same* height H.

Maybe you need to re-read the statement of this problem.