Jumping off a Train

Recently, I came across this cool problem (which is from Kleppner and Kolenkow, as well as INPhO 2014):

Two men, each with mass \(m\), stand on a railway flatcar of mass \(M\) initially at rest. They jump off one end of the flatcar with velocity \(u\) relative to the car. The car rolls in the opposite direction without friction. Find the final velocities of the flatcar if they jump off at the same time, and if they jump off one at a time. Generalize to the case of \(N ≫1\) men, with a total mass of \(m_\mathrm{tot}\).

I highly suggest you try solving it before looking at my solution.

My Solution

Let’s start with the easy part: finding the final velocity if they jump off at the same time. In this case, we have \(p_0=0\) and \(p_f=m_\mathrm{tot}(v-u)+Mv\), where \(v\) is the final velocity of the flatcar. By conservation of momentum, \(p_0=p_f\) and we have that

\[v=\dfrac{m_\mathrm{tot}u}{M+m_\mathrm{tot}}\]

In the second case, by a similar logic, we have that after the first man jumps, the flatcar moves at

\[v_1=\dfrac{mu}{M+2m}\]

We can find the velocity after the second man jumps by transforming into the frame moving with the flatcar at a velocity \(v_1\). In this frame, the second man exerts a further impulse onto the train of magnitude \(v_2=\dfrac{mu}{M+m}\). Therefore, the final speed of the flatcar is then

\[v=v_1+v_2=mu\left(\dfrac{1}{M+2m}+\dfrac{1}{M+m}\right)\]

Extending it to \(N\) people, we then have

\[v=\dfrac{m_\mathrm{tot}u}{N}\sum_{i=1}^N\dfrac{1}{M+(i/N)m_\mathrm{tot}}\]

Now comes the cool part. For a large enough \(N\), we can turn this into an integral with \(x=i/N\) and \(\mathrm{d}x=\mathrm{d}i/N\). Substituting it in, we get

\[v \approx m_\mathrm{tot}u\int_0^1 \dfrac{\mathrm{d}x}{M+xm}=\ln\left(\dfrac{M+m_\mathrm{tot}}{M}\right)u\]

Which turns out to be almost exactly the rocket equation! Definitely a cool problem, and the first one where I’ve seen the idea of turning a series into an integral.

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