On a frictionless track, a trolley moves with a speed of 36 km/h with a mass of 200 Kg. A child whose mass is 20 kg runs on the trolley with a speed of 4 m s1 from one end to other which is 20 m. The speed is relative to the trolley in the direction opposite to its motion. Find the final speed of the trolley and the distance the trolley moved from the time the child began to run.
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)×10 =2200kg m/s Let v′ be the final velocity of the trolley with respect to the ground. Final velocity of the boy with respect to the ground =v′−4 Final momentum =Mv′+m(v′−4) =200v′+20v′−80 From the conservation of linear momentum: Initial momentum= Final momentum 2200=220v′−80 ∴ v′=2280/220=10.36m/s Length of the trolley, l=10m Speed 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
