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

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

Vm “= 2 m / s.

ml = 80 kg.

mm = 40 kg.

Vl “-?

When a boy jumps from a boat, they can be considered a closed system for which the law of conservation of momentum is valid.

Since the boy was in the boat before the jump, the law of conservation of momentum will have the form: (mm + ml) * V = mm * Vm “+ ml * Vl”, where (mm + ml) * V is the impulse of the boy with the boat before the jump , mm * Vm “- the impulse of the boy when jumping, ml * Vl” – the impulse of the boat after the jump.

Since the boy was in a stationary boat before the shot, their speed before the jump was V = 0.

mm * Vm “+ ml * Vl” = 0.

mm * Vm “= – ml * Vl”.

The “-” sign indicates that after the jump, the boy and the boat will move in opposite directions.

Vl “= mm * Vm” / ml.

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

Answer: the boat will move in the opposite direction of the boy’s jump at a speed Vl “= 1 m / s.