Question

A $(trolley+child)$ of total mass $200kg$ is moving with 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 the speed of $10m{s}^{-1}$ 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?

Answer 1

Mass of the trolley, $M=200kg$

Speed of the trolley, $v=36km/h=10m/s$

Mass of the boy, $m=20kg$

Initial momentum of the system of the boy and the trolley

$=(M+m)v$

$=(200+20)\times 10$

$=2200kgm/s$

Let $v^{\prime }$ be the final velocity of the trolley with respect to the ground.

Final velocity of the boy with respect to the ground $=v^{\prime }-4$

Final momentum $=M{v}^{\prime }+m(v^{\prime }-4)$

$=200v^{\prime }+20v^{\prime }-80$

From the conservation of linear momentum:

Initial momentum$=$ Final momentum

$2200=220v^{\prime }-80$

∴ $v^{\prime }=2280/220=10.36m/s$

Length of the trolley, $l=10m$

Speed of the boy, $v^{\prime \prime }=4m/s$

Time taken by the boy to run, $t=10/4=2.5s$

∴ Distance moved by the trolley $=v^{\prime \prime }\times t$= $10.36\times 2.5=25.9m$