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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 ?

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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

Posted by

Devendra Khairwa

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