From a stationary boat, the mass of which is 80kg. a boy jumps to the shore. The weight of the boy is 40 kg

| June 25, 2021 | education

From a stationary boat, the mass of which is 80kg. a boy jumps to the shore. The weight of the boy is 40 kg., his speed during the jump is 2 m / s. What speed did the boat get?

ml = 80 kg.

mm = 40 kg.

Vm “= 2 m / s.

Vl “-?

Let us write down the law of conservation of momentum for the closed boy-boat system: ml * Vl + mm * Vm = ml * Vl “+ mm * Vm”, where ml, mm are the mass of the boat and the boy, respectively, Vm, Vm “- the speed of the boy before and after jump, Vl, Vl “- boat speed before and after the jump.

Since before the jump the boat and the boy were at rest Vm = Vm “= 0 m / s, then 0 = ml * Vl” + mm * Vm “.

ml * Vl “= – mm * Vm”.

Vl “= – mm * Vm” / ml.

The sign “-” means that the speed of the boat after the jump is directed to the opposite side of the boy’s jump.

Vl “= 40 kg * 2 m / s / 80 kg = 1 m / s.

Answer: the speed of the boat after the jump will become Vl “= 1 m / s.



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