From a boat weighing 200 kg, moving at a speed of 2 m / s, a boy weighing 50 kg dives horizontally

From a boat weighing 200 kg, moving at a speed of 2 m / s, a boy weighing 50 kg dives horizontally from the stern at a speed of 4 m / s. What will be the speed of the boat after the boy’s jump?

To solve this problem, it is necessary to use the law of conservation of momentum. The sum of impulses before interaction is equal to the sum of impulses after interaction:

(mm + ml) u0 = ml + mm.

Hence the speed of the boat after the boy dives is equal to:

ul = ((mm + ml) u0 – mmum) / ml.

We substitute all known quantities in the derived formula:

ul = ((50 + 200) * 2 – 50 * 4) / 200 = (250 * 2 – 50 * 4) / 200 = (500 – 200) / 200 = 300/200 = 3/2 = 1.5 m / s.

Answer: 1.5 m / s.