Question

Two persons each of mass m are standing at the two extremes of a railroad ear 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.

Answer 1

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.

$v_i=v_f$

$0=-mu+(M+m)V$

$V=\frac{mu}{(M+m)}$ towards right.

CASE - 2

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

$0=mu-M{v}^{\prime }$

$v^{\prime }=\frac{mu}{M}$

v' 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

$V_{net}=\frac{mv}{M}-\frac{mv}{(M+m)}$

$=\frac{(mMu+m^{2}v-Mmu)}{M(M+m)}$

$=\frac{m^{2}v}{M(M+m)}$