# From a boat moving towards the shore at a speed of 0.5 m / s, a man jumped to the shore at a speed of 2 m / s

**From a boat moving towards the shore at a speed of 0.5 m / s, a man jumped to the shore at a speed of 2 m / s relative to the shore. At what speed will the boat move after a man’s jump, if the mass of a man is 80 kg, and the mass of the boat is 120 kg?**

Given:

m1 = 120 kilograms – the mass of the boat;

m2 = 80 kilograms – the mass of a person;

v = 0.5 m / s – boat speed before a person jump;

v2 = 2 m / s – the speed of the person after the jump.

It is required to determine v1 (m / s) – the speed of the boat after the jump.

We will take the initial movement of the boat as a positive direction. Also, we believe that after the jump, the boat will start moving in the opposite direction. Then, according to the law of conservation of momentum (momentum):

(m1 + m2) * v = m2 * v2 – m1 * v1;

m1 * v1 = m2 * v2 – (m1 + m2) * v;

v1 = (m2 * v2 – (m1 + m2) * v) / m1 = (80 * 2 – (80 + 120) * 0.5) / 120 =

= (160 – 200 * 0.5) / 120 = (160 – 100) / 120 = 60/120 = 0.5 m / s.

Answer: the speed of the boat after a person jump will be equal to 0.5 m / s.