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(In the following figure). 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. 

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:
\[0 = - mu + (M + m)V\]
\[ \Rightarrow V = \left( \frac{mu}{M + m} \right), \text{ towards right }\]
When the man on the right jumps, his velocity w.r.t. the car is u.
\[0 = mu - MV'\]
\[ \Rightarrow V' = \frac{mu}{M}\]
(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)
\[V' - V = \frac{mu}{M} - \frac{mu}{(M + m)}\]
\[ = \frac{mMu + m^2 u - Mmu}{M(M + m)}\]
\[ \Rightarrow V' - V = \frac{m^2 u}{M(M + m)}\]