Question

Two persons each of mass m are standing at the two extremes of a railroad car of mass M resting on a smooth track. The person on left jumps to the left with a horizontal speed u with respect to the state of the car before the jump. Thereafter, the other person jumps to the right, again with the same horizontal speed u with respect to the state of the car before his jump. Find the velocity of the car after both the persons have jumped off.
Figure

Answer 1

It is given that:
Mass of each persons = m
Mass of railroad car = M

Let the velocity of the railroad w.r.t. earth, when the man on the left jumps off be V.

By the law of conservation of momentum:

$$\begin{gathered} 0=-mu+(M+m)V \\ \Rightarrow V=\left(\frac{mu}{M+m}\right),\mathrm{towards}\mathrm{right} \end{gathered}$$

When the man on the right jumps, his velocity w.r.t. the car is u.

$$\begin{gathered} 0=mu-MV\text{'} \\ \Rightarrow V\text{'}=\frac{mu}{M} \end{gathered}$$

(V is the change in velocity of the platform when the platform itself is taken as reference, assuming the car to be at rest.)

∴ Net velocity towards left, (i.e. the velocity of the car)

$$\begin{gathered} V\text{'}-V=\frac{mu}{M}-\frac{mu}{(M+m)} \\ =\frac{mMu+m^{2}u-Mmu}{M(M+m)} \\ \Rightarrow V\text{'}-V=\frac{m^{2}u}{M(M+m)} \end{gathered}$$