A person weighing 70 kg runs at a speed of 6 m / s and jumps into a trolley weighing 140 kg standing on the rails.

| September 3, 2021 | education

A person weighing 70 kg runs at a speed of 6 m / s and jumps into a trolley weighing 140 kg standing on the rails. What are the initial modules of the impulse of a person and a cart?

mh = 70 kg.

Vch = 6 m / s.

mt = 140 kg.

Vt = 0 m / s.

V “-?

Let us write down the law of conservation of momentum for a closed system: man – trolley: mh * Vh + mt * Vt = mh * Vh “+ mt * Vt”, where mh is the mass of a person, Vh, Vch “is the person’s speed before and after the jump, mt – cart mass, Vт, Vт “- cart speed before and after the jump.

Since after the jump the person is on the cart, their speeds will be the same: Vch “= Vt” = V “.

The speed of the trolley before the jump is Vt = 0 m / s.

The impulse conservation law will take the form: mh * Vh = (mh + mt) * V “.

V “= mh * Vh / (mh + mt).

V “= 70 kg * 6 m / s / (70 kg + 140 kg) = 2 m / s.

Answer: after the jump, the cart with the person will move at a speed V “= 2 m / s.



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