Question

The speed of the trolley when the man jumps off the trolley is

• A. 2
• B. 4
• C. 8
• D. 16

Answer 1

Answer: (C)

8

Mass of trolley=$m=40kg$

Mass of the man=$M=80kg$

Since no external force acts in horizontal direction, the conservation of momentum along the direction holds.

Hence $(M+m)v_0=M{v}_{man}+m{v}_{trolley}$

$⟹(80+40)2=\left(80\right)(-1)+\left(40v\right)$

$⟹v=8m/s$