A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a fr

1.	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 the trolley's 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» Initial total momentum = p_i = (m₁ + m₂)u₁ = (20 + 200)(36 x 5/18) = 220 x 10 = 2200 kg ms^-1 Final total momentum = p_f = m₁v₁ + m₂v₂ = 20(v₂ - 4) + 200 v₂ [(By using v₁ = v₂ - 4)] p_f = (200 v₂ - 80) Using the law conservation of total momentum, i.e., p_i = p_f We have 2200 = (220 v₂ - 80) v₂ = 2280/220 = 10.38 ms^-1 = 25.9 m

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 the trolley's 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»

Initial total momentum = p_i = (m₁ + m₂)u₁

= (20 + 200)(36 x 5/18)

= 220 x 10 = 2200 kg ms^-1

Final total momentum = p_f = m₁v₁ + m₂v₂

= 20(v₂ - 4) + 200 v₂ [(By using v₁ = v₂ - 4)]

p_f = (200 v₂ - 80)

Using the law conservation of total momentum, i.e., p_i = p_f

We have 2200 = (220 v₂ - 80)

v₂ = 2280/220 = 10.38 ms^-1 = 25.9 m