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[SOLVED] frictionless railcar

dwsmith

Well-known member
Feb 1, 2012
1,673
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?
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
It is from that book but I figured out what they did before you posted; hence, the post was marked solved before then.