A boy weighing 50 kg dives in a horizontal direction from a boat weighing 200 kg. What will be the speed

A boy weighing 50 kg dives in a horizontal direction from a boat weighing 200 kg. What will be the speed of the boat after the jump, if before the jump its speed was equal to 1 m / s? In the first case, the jump was made in the direction of the boat’s movement at a speed of 2 m / s, in the second case – against the movement at the same speed.

This task is to preserve the momentum (momentum).

m1 = 200 kg, m2 = 50 kg, v0 = 1 m / s, 1) vm = 2 m / s, 2) vm = -2m / s.

v l =?

The solution is in two versions.

1) Let’s compose the equation of conservation of momentum before and after the boy’s jump.

(m1 + m2) * v0 = m1 * vl + m2 + vm;

(200 + 50) kg * 1m / s = 200kg * vl + 50kg * 2m / s;

250 kg * 1m / s = 200 kg * vl + 100 kg * m / s, vl = 150/200 = 3/4 (m / s) = 0.75 m / s.

2) (m1 + m2) * v0 = m1 * vl – m2 * vm,

vl = (200 + 50) kg * 1 m / s + 50kg * 2 m / s = 200 kg * vl, vl = 350/200 = 1.75 m / s.