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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?

diagram of bouncing balls

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.

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@Qmechanic This is not a homework question. –  Sim 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. –  Sim Jun 7 '13 at 16:26
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1 Answer

up vote 4 down vote accepted

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

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