From a boat weighing 200 kg, moving at a speed of 1 m / s, a boy weighing 50 kg dives, moving in a horizontal

From a boat weighing 200 kg, moving at a speed of 1 m / s, a boy weighing 50 kg dives, moving in a horizontal direction. What will be the speed of the boat after the boy’s jump, if he jumps from karma with a speed of 4m / s

ml = 200 kg.

Vl = 1 m / s.

mm = 50 kg.

Vm “= 4 m / s.

Vl “-?

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

Since the boy jumps in the opposite direction of the boat’s movement, then in projections the law of conservation of momentum will look like: ml * Vl + mm * Vl = ml * Vl “- mm * Vm”.

Vl “= (ml * Vl + mm * Vl + mm * Vm”) / ml.

Vl “= (200 kg * 1 m / s + 50 kg * 1 m / s + 50 kg * 4 m / s) / 200 kg = 2.25 m / s.

Answer: the speed of the boat after the jump will become Vl “= 2.25 m / s.