Question

A (trolley+child) of total mass $200kg$ is moving at a uniform speed of $36km/h$ on a frictionless track. The child of mass $20kg$ starts running on the trolley from one end to the other ($10m$ away) with a speed of $10{ms}^{-1}$ relative to the trolley in the direction of the trolley's motion and jumps out of the trolley with the same relative velocity. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run and just before jump?

Answer 1

Using momentum conservation

$(M+m)v=M{v}^1+mu(200+20)\left(\frac{36}{3.6}\right)m/s=200v^1+20\times (10+v^1)220\times 10=200v^1+200+20v^1(20+200)v^1=2200-200=2000v^1=\frac{2000}{220}=9.09m/s$

$v^1=$ Velocity of trolley that is remains unchanged after the boy jump.

Now time taken by boy to cover$10m$ with respect to trolley.

$t=\frac{10}{10}=1sec$

Now distance cover by trolley in $1sec$ is

$D=v^{1}t=9.09\times 1=9.09m$