A trolley of mass M is at rest over a smooth horizontal surface as shown in figure. Two boys each of mass 'm' are standing over the trolley. They jump from the trolley (towards right) with relative velocity `v_r` [relative to velocity of trolley just after jumping]
(a) together
(b) one after the other.
Find velocity of trolley in both cases.

Answer

A, B

Explanation

(a) Let velocity of trolley just after jumping is `v_1` (towards left). Relative velocity of boys towards right is `v_r`. Therefore, there absolute velocity is `v_r-v_1`, towards right.

Net force on the system in horizontal direction is zero. Hence, linear momentum of the system in horizontal direction is zero.
or `p_i=p_f`
`implies 0=2m(v_r-v_1)-Mv_1`
`impliesv_1=(2mv_r)/(2m+M)`
(b) Let `v_1` be the velocity of trolley after the first boy jumps. Then,

`impliesp_i=p_f`
`0=m(v_r-v_1)-(M+m)v_1`
`impliesv_1=(mv_r)/(M+2m)` ...(i)
Now, the second boy jumps from the moving trolley and let `v_2` be the velocity of trolley after the second boy also jumps. Then,

`p_i=p_f`
`implies-(M+m)v_1=m(v_r-v_2)-Mv_2`
`:. v_2=(mv_r+(M+m)v_1)/((M+m))`
`=((m)/(M+m))v_r+v_1`
Substituting the value of `v_1` from Eq. (i), we have
`v_2=mv_r[(1)/(M+m)+(1)/(M+2m)]`