A boy of mass 40 kg jumps with a horizontal velocity of 5 ms-1 on to a stationary cart with frictionless wheels. The mass of the cart is 3 kg. What is his velocity as the cart starts moving? Assume that there is no external unbalanced force working in horizontal direction.
Suppose the velocity of the boy on the cart as the cart starts moving is v.
The total momenta of the boy and cart before the interaction
= 40 kg x 5 ms-1 + 3 kg x 0 ms-1 = 200 kg-ms-1
Also, the total momenta after the interaction
= (40 + 3) kg x v ms-1 = 43 v kg-ms-1
As per the law of conservation of momentum, the total momentum is conserved during the interaction. In other words,
Thus, the boy on cart would move with a velocity of 4.65 ms’1 in the direction in which the boy jumped on to the cart.