# How to derive the velocity in the double ball drop problem?

The double ball drop problem is as follows:

A ball of mass $m$ is placed on top of a ball of mass $M$ (where $m < M$), and the balls are dropped simultaneously from some height $h$. When the balls hit the floor, the ball on top will shoot upwards. What is the velocity of this ball at the moment it shoots upwards?

And so I ask; is there any easy derivation of this problem, without the use of calculus?

I have attempted to solve it used the conservation of momentum, kinetic energy, and mechanical energy, but I seem unable to solve it. My attempt is as follows:

First, we can solve for the velocity of the larger ball using the conservation of momentum:

$$p_\text{before} = p_\text{after}$$

$$(m + M)\sqrt{2gh} = mv_m + Mv_M$$

$$v_M = \frac{(m + M)\sqrt{2gh} - mv_m}{M}$$

Then, using the conservation of kinetic energy:

$$\frac{1}{2}(m + M)\sqrt{2gh}^2 = \frac{1}{2}mv_m^2 + \frac{1}{2}Mv_M^2$$

$$v_M = \sqrt{\frac{2gh(m + M) - mv_m^2}{M}}$$

And then you can equate the two to get:

$$\sqrt{\frac{2gh(m + M) - mv_m^2}{M}} = \frac{(m + M)\sqrt{2gh} - mv_m}{M}$$

However, this, when solved, gives $v_m = \sqrt{2gh}$, which does not align with what I have been taught (saying that the velocity should be about three times this).

What am I doing wrong in this derivation? The math is correct, I'm sure. It's somewhere in the physics itself, a concept that I'm missing.

Any help is appreciated.

-
@Qmechanic This is not a homework question. – Xenon Jun 7 '13 at 16:24
Hi Sim. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. – Qmechanic Jun 7 '13 at 16:25
@Qmechanic My apologies, I suppose a homework tag is appropriate then. – Xenon Jun 7 '13 at 16:26

When the large ball strikes the ground, its momentum switches from $p\rightarrow -p$. Now you can consider the process as a two ball collision. The small ball traveling down with momentum $m\sqrt{2gh}$ and the large ball traveling up with momentum $M\sqrt{2gh}$. Use this as you momentum conservation condition: $$(M-m)\sqrt{2gh} = mv_m + Mv_M$$