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

| July 3, 2021 | education

From a boat weighing 200 kg, moving at a speed of 1 m / s, a boy weighing 50 kg jumps in a horizontal direction at a speed of 7 m / s. What is the speed of the boat after the boy’s jump if he jumps from the stern to the side opposite to the movement of the boat?

ml = 200 kg.

Vl = 1 m / s.

mm = 50 kg.

Vm “= 7 m / s.

Vl “-?

Let’s write the law of conservation of momentum in vector form: (ml + mm) * Vl = ml * Vl “+ mm * Vm”.

Let’s draw the OX coordinate axis in the direction of the boat’s movement before the boy’s jump.

1) When the boy jumps from the stern, the law of conservation of momentum for projections will take the form: (ml + mm) * Vl = ml * Vl “- mm * Vm”.

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

Vl “= ((200 kg + 50 kg) * 1 m / s + 50 kg * 7 m / s) / 200 kg = 3 m / s.

2) When the boy jumps from the bow of the boat, the law of conservation of momentum for projections will take the form: (ml + mm) * Vl = ml * Vl “+ mm * Vm”.

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

Vl “= ((200 kg + 50 kg) * 1 m / s – 50 kg * 7 m / s) / 200 kg = – 0.5 m / s.

The “-” sign indicates that after jumping from the bow of the boat, it will begin to move in the opposite direction of the initial movement.



One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.