A trolley of mass $200\; kg$ moves with a uniform speed of $36\; km / h$ on a frictionless track. A child of mass $20\; kg$ runs on the trolley from one end to the other ( $10\; m$ away) with a speed of $4 \;m s ^{-1}$ relative to the trolley in a direction opposite to the its motion, and Jumps out of the trolley. What is the final speed of the trolley ? How much has the trolley moved from the time the child begins to run?
A trolley of mass $200\; kg$ moves with a uniform speed of $36\; km / h$ on a frictionless track. A child of mass $20\; kg$ runs on the trolley from one end to the other ( $10\; m$ away) with a speed of $4 \;m s ^{-1}$ relative to the trolley in a direction opposite to the its motion, and Jumps out of the trolley. What is the final speed of the trolley ? How much has the trolley moved from the time the child begins to run?
Mass of the trolley, $M=200 kg$
Speed of the trolley, $v=36 km / h =10 m / s$
Mass of the boy, $m=20 kg$
Initial momentum of the system of the boy and the trolley
$=(M+m) v$
$=(200+20) \times 10$
$=2200 kg m / s$
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 $=M v^{\prime}+m\left(v^{\prime}-4\right)$
$=200 v^{\prime}+20 v^{\prime}-80$
$=220 v^{\prime}-80$
As per the law of conservation of momentum:
Initial momentum $=$ Final momentum
$2200=220 v^{\prime}-80$
$\therefore v^{\prime}=\frac{2280}{220}=10.36 m / s$
Length of the trolley, $l=10 m$
Speed of the boy, $v^{\prime \prime}=4 m / s$
Time taken by the boy to run, $t=\frac{10}{4}=2.5 s$
Distance moved by the trolley $=v^{\prime \prime} \times t=10.36 \times 2.5=25.9 m$
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