MHB Find Velocity of Recoiling Railcar with Conservation of Momentum

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The discussion focuses on using conservation of momentum to determine the velocity of a recoiling railroad flatcar when two hobos jump off simultaneously. The derived formula for the car's velocity is v = (2m_h / (m_fc + 2m_h))u, where m_h is the mass of each hobo and m_fc is the mass of the flatcar. The momentum equation is set up as m_fc v = 2m_h (u - v), leading to the final solution after rearranging terms. The original question is confirmed to be from the book "Classical Mechanics" by John Taylor. The thread concludes with the problem being marked as solved.
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Two hobos, each of mass $m_{\text{h}}$, are standing at one end of a stationary railroad flatcar with frictionless wheels and mass $m_{\text{fc}}$.
Either hobo can run to the other end of the flatcar and jump off with the same speed $u$ (relative to the car).

Use conservation of momentum to find the speed of the recoiling car if the two men run and jump off simultaneously.

Let $v$ be the velocity of the recoiling car.
Then
\begin{alignat*}{3}
m_{\text{fc}}v & = & 2m_{\text{h}}(u - v)\\
v & = & \frac{2m_{\text{h}}}{m_{\text{fc}}}(u - v)
\end{alignat*}

The solution is $v = \frac{2m_{\text{h}}}{2m_{\text{h}}+m_{\text{fc}}}u$.
How did they get that?
 
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Well, expanding the product gets us

$$m_{fc} v = 2m_h (u-v) = 2m_h u - 2m_h v,$$

and from that

$$m_{fc} v + 2m_h v = (m_{fc} + 2m_h) v = 2m_h u.$$

Finally,

$$v = \frac{2m_h}{m_{fc} + 2m_h} u.$$

If you don't mind me asking, is this question from the book Classical Mechanics by John Taylor? :D

Cheers.
 
It is from that book but I figured out what they did before you posted; hence, the post was marked solved before then.
 
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