A girl of mass `40kg` jumps with a horizontal velocity of `5m//s` onto a stationary cart with frictionless wheels. The mass of the cart is `3kg`. What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction.

Explanation

Let v be the velocity of the girl on the cart as the cart starts moving. The total momenta of the girl and cart before the interaction
`=40 kg xx 5 ms^(-1)+3 kg xx 0 m s^(-1)`
`=200 kg ms^(-1)`.
Total momenta after the interaction
`=(40+3) kg xx v ms^(-1)`
`=43 v kg ms^(-1)`. According to the law of consevation of momentum, the total momentum is conserved during the interaction. That is
`43v=200`
` implies v=200//43=+4.65 ms^(-1)`.
The girl on cart would move with a velocity of `4.65 ms^(-1)` in the direction in which the girl jumped (fig. 9.18).