# A man running at a speed of 7 m / s, catches up with a cart moving at a speed of 2 m / s, and jumps on it.

**A man running at a speed of 7 m / s, catches up with a cart moving at a speed of 2 m / s, and jumps on it. at what speed the cart will move after this, the mass of the person and the cart are, respectively, 70 and 30 kg**

Given:

v1 = 7 m / s (meters per second) – the speed of a person catching up with the cart;

v2 = 2 m / s – trolley speed;

m1 = 70 kilograms – the mass of a person;

m2 = 30 kilograms – the mass of the cart.

It is required to determine v (m / s) – with what speed the cart will move after a person catches up with it and jumps.

According to the condition of the problem, initially the carts and the person moved in the same direction. Then, according to the law of conservation of momentum (momentum), we obtain:

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

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

v = (70 * 7 + 30 * 2) / (70 + 30) = (490 + 60) / 100 = 550/100 = 5.5 m / s.

Answer: after interaction, the speed of the cart will be 5.5 m / s.