A person weighing 70 kg, running at a speed of 7.2 km / h, catches up with a cart weighing 50 kg

A person weighing 70 kg, running at a speed of 7.2 km / h, catches up with a cart weighing 50 kg, moving along the same straight line at a speed of 2 m / s, and jumps onto it. With what speed and in what direction will the cart with the person move? Friction is neglected.

Given:

m1 = 70 kilograms – the mass of a person;

v1 = 7.2 km / h = 2 m / s – the speed of a person’s movement;

m2 = 50 kilograms – the mass of the cart;

v2 = 2 m / s – trolley speed.

It is required to determine v (m / s) – the speed and direction of movement of the cart, after a person jumps on it.

According to the condition of the problem, the person and the cart initially move in the same direction.

Then, according to the law of conservation of momentum:

m1 * v1 + m2 * v2 = (m1 + m2) * v;

v = (m1 * v1 + m2 * v2) / (m1 + m2);

v = (70 * 2 + 50 * 2) / (70 + 50) = (140 + 100) / 120 = 240/120 = 24/12 = 2 m / s.

Answer: a cart with a person will move at a speed of 2 m / s in the same direction as before interaction.