From a stationary boat weighing 50 kg, a boy weighing 40 kg jumped onto the shore at a speed of 1 m / s

From a stationary boat weighing 50 kg, a boy weighing 40 kg jumped onto the shore at a speed of 1 m / s relative to the shore, directed horizontally. What speed did the boat gain relative to the shore? Ignore the water resistance.

1. Let’s write down the law of conservation of momentum.

m1 * v1 + m2 * v2 = m1 * v1 ‘+ m2 * v2’.

2. Since the bodies were at rest in the initial state, we take v1 and v2 equal to zero.

m1 * v1 ‘+ m2 * v2’ = 0.

3. Find the value of the boat speed after interaction.

v2 ‘= m1 * v1’ / m2 = 40 kg * 1 m / s / 50 kg = 0.8 m / s.

Answer: 0.8 m / s.