How to isolate $v_m$?

Question

Note: I am not asking anything pertaining to the physics of this question; only the mathematics. The physics is just given as a context to the problem for those interested, as opposed to simply saying "simplify this equation".

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?

Here is a picture of accompany this description:

I have attempted to solve this problem, but am stuck with the following equation:

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

What I ask is to help isolate $v_m$. I attempted to using the quadratic equation, but that resulted in a very challenging equation that I am unable to simplify. I feel as though there may be an easier way to solve for $v_m$.

Any help is appreciated.

Answer 1

\begin{array}{ccc} \sqrt{\frac{2gh(m + M) - mv_m^2}{M}} &=& \frac{(m + M)\sqrt{2gh} - mv_m}{M} \\ \\ \frac{2gh(m + M) - mv_m^2}{M} &=& \frac{((m + M)\sqrt{2gh} - mv_m)^2}{M^2} \\ \\ (2gh(m + M) - mv_m^2)M &=& ((m + M)\sqrt{2gh} - mv_m)^2 \\ \\ (2gh(m + M) - mv_m^2)M &=& 2gh(m + M)^2 - 2\sqrt{2gh}(m + M)mv_m + m^2v_m^2 \\ \\ 2gh(m + M)M - mMv_m^2 &=& 2gh(m + M)^2 - 2\sqrt{2gh}(m + M)mv_m + m^2v_m^2 \\ \\ 2gh(m + M)M &=& 2gh(m + M)^2 - 2\sqrt{2gh}(m + M)mv_m + (m^2+mM)v_m^2 \\ \\ 2gh(m + M)M &=& 2gh(m + M)^2 - 2\sqrt{2gh}(m + M)mv_m + m(m+M)v_m^2 \\ \\ 2ghM &=& 2gh(m + M) - 2\sqrt{2gh}mv_m + mv_m^2 \\ \\ 0 &=& 2ghm - 2\sqrt{2gh}mv_m + mv_m^2 \\ \\ 0 &=& m(2gh - 2\sqrt{2gh}v_m + v_m^2) \\ \\ 0 &=& m(\sqrt{2gh} - v_m)^2 \\ \\ \end{array} Hence $v_m = \sqrt{2gh}$.

Answer 2

I think if you just square both sides you can get an equation in terms of $v_{m}^2$ and $v_m$. Should be easy to solve then

Answer 3

This equation simplifies to

$$(v_m-\sqrt{2 g h})^2=0$$

You can immediately see that the solution here satisfies the original equation. All you need to do is square both sides - this really isn't a beast. I get

$$2 g h M (m+M) - mMv_m^2 = (m+M)^2 2 g h + m^2 v_m^2 - 2 m (m+M) \sqrt{2 g h} v$$

Rearrange to get a quadratic in $v$:

$$m(m+M) v_m^2-2 m (m+M) \sqrt{2 g h} v_m + (m+M)^2 -2m (m+M) \sqrt{2 g h}=0$$

Much cancellation ensues.