Two men, each of mass m, stand on the edge of a stationary buggy of mass M. Assuming the friction to be negligible, find the velocity of the buggy after both men jump off with the same horizontal velocity u relative to the buggy: (1) simultaneously, (2) one after the other. In what case will the velocity of the buggy be greater and how many times?

Explanation

(i) Let `vecv_1` be the velocity of the buggy after both man jump off simultaneously. For the closed system (two men +buggy), from the conservation of linear momentum,
`Mvecv_1+2m(vecu+vecv_1)=0`
or, `vecv_1=(-2mvecu)/(M+2m)` (1)
(ii) Let `vecv` be the velocity of buggy with man, when one man jump off the buggy. For the closed system (buggy with one man + other man) from the conservation of linear momentum:
`0=(M+m)vecv+m(vecu+vecv)` (2)
Let `vecv_2` be the sought velocity of the buggy when the second man jump off the buggy, then from conservation of linear momentum of the system (buggy + one man):
`(M+m)vecv=Mvecv_2+m(vecu+vecv_2)` (3)
Solving equations (2) and (3) we get
`vecv_2=(m(2M+3m)vecu)/((M+m)(M+2m))` (4)
From (1) and (4)
`v_2/v_1=1+(m)/(2(M+m))gt1`
Hence `v_2gtv_1`