Question

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?

Answer 1

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) × 10

= 2200 kg 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

= 220v' – 80

As per the law of conservation of momentum: 

Initial momentum = Final momentum

2200 = 220v' – 80

∴ v' = 2280 / 220 = 10.36 m/s

Length of the trolley, l = 10 m

Speed of the boy, v'' = 4 m/s

Time taken by the boy to run, t = 10/4 = 2.5 s

∴ Distance moved by the trolley = v'' × t= 10.36 × 2.5 = 25.9 m

Answer 2

Let there be an observer travelling parallel to the trolley with the same speed. He will observe the initial momentum of the trolley of mass M and child of mass m as zero. When the child jumps in opposite direction, he will observe the increase in the velocity of the trolley by Δv.

Let u be the velocity of the child. He will observe child landing at velocity (u – Δu) Therefore, initial momentum = 0

Final momentum = MΔ v – m (u – Δv)

Hence, MΔ v – m (u – Δv) = 0

Whence Δv =mu/ M + m

Putting values Δv =4 x 20/ 20 + 220 = ms^-1

∴ Final speed of trolley is 10.36 ms^-1.

The child take 2.5 s to run on the trolley.

Therefore, the trolley moves a distance = 2.5 x 10.36 m = 25.9 m.