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Question

# A (trolley+child) of total mass 200 kg is moving with a uniform speed of 36 km/h on a frictionless track. The child of mass 20 kg starts running on the trolley from one end to the other (10 m away) with the 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?

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Solution

## Mass of the trolley, M=200kgSpeed of the trolley, v=36km/h=10m/sMass of the boy, m=20kgInitial momentum of the system of the boy and the trolley =(M+m)v =(200+20)×10 =2200kg m/sLet v′ be the final velocity of the trolley with respect to the ground.Final velocity of the boy with respect to the ground =v′−4Final momentum =Mv′+m(v′−4) =200v′+20v′−80From the conservation of linear momentum:Initial momentum= Final momentum 2200=220v′−80 ∴ v′=2280/220=10.36m/sLength of the trolley, l=10mSpeed of the boy, v′′=4m/s Time taken by the boy to run, t=10/4=2.5s∴ Distance moved by the trolley =v′′×t= 10.36×2.5=25.9m

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