Q28  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 $4 ms ^{-1}$ relative to the trolley in a direction opposite to 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 ?

D Devendra Khairwa

The initial momentum of the system (boy + trolley) is given as :

$=\ (m\ +\ M)V$

$=\ (200\ +\ 20)10\ =\ 2200\ Kg\ m/s$

Now assume v' is the final velocity of the trolley with respect to the ground.

Then the final momentum will be :

$=\ Mv'\ +\ m(v'\ -\ 4)\ =\ 220v'\ -\ 80$

Conserving momentum :

$220v'\ -\ 80\ =\ 2200$

or                                                      $v'\ =\ 10.36\ m/s$

The time taken by the boy is :

$=\ \frac{10}{4}\ =\ 2.5\ s$

Hence the distance moved by the trolley is :

$=\ 10.36\times 2.5 \ =\ 25.9\ m$

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