A girl of mass 40 kg jumps with a horizontal velocity of 5 m s-1 onto a stationary cart with frictionless wheels. The mass of the cart is 3 kg. What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction.

Solution:
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 × 5 m s–1 + 3 kg × 0 m s–1
= 200 kg m s–1
.
Total momenta after the interaction
= (40 + 3) kg × v m s–1
= 43 v kg m s–1
.
According to the law of conservation of momentum, the total momentum is conserved during the interaction.
That is, 43 v = 200
⇒ v = 200/43 = + 4.65 m s–1
.
The girl on cart would move with a velocity of 4.65 m s–1 in the direction in which the girl jumped

The girl jumps onto the cart.

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