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 (figure 9-E10). 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

Solution:

There will be two cases.

CASE - 1

Let the velocity of the railroad car with respect to earth is V after jump of left man.

0 = - mu + (M + m) VV = mu / (M + m) towards right.

CASE - 2

When man on right jumps , the velocity of it with respect to car is u.

0 = mu - Mv'v' = mu/Mv' is the change in velocity of platform when platform is taken as reference and we assumed car to be at rest.

Therefore , net velocity towards left i.e. velocity of car with respect to earth = mv/M - mv/(M + m)= (mMu + m²v - Mmu) / M(M + m)= m²v / M(M + m)

Hope it helps!