The railcar moves along the rails. A person jumps off the trolley at a speed of 3.5 m / s in the opposite direction

The railcar moves along the rails. A person jumps off the trolley at a speed of 3.5 m / s in the opposite direction to the movement. The speed of the railcar, after a man jumped off it, became equal to 14 m / s. Find the initial speed of the trolley if you know that the mass of a person is 78 kg, and the mass of the trolley is 312 kg.

Vch “= 3.5 m / s.

Vd “= 14 m / s.

mh = 78 kg.

md = 312 kg.

V -?

To solve this problem, we will use the law of conservation of momentum in vector form for the system of a man-trolley: mh * V + md * V = mh * Vh “+ md * Vd”.

Where mh * V is the impulse of a person before jumping from the trolley, md * V is the impulse of the trolley before the jump, mh * Vh “, md * Vd” are the impulses of the person and the trolley, respectively, after the jump.

Since a person jumps in the opposite direction of the railcar movement, then for projections the law of conservation of momentum will take the form: mh * V + md * V = – mh * Vh “+ md * Vd”.

(mh + md) * V = md * Vd “- mh * Vh”.

V = (md * Vd “- mh * Vh”) / (mh + md).

V = (312 kg * 14 m / s – 78 kg * 3.5 m / s) / (78 kg + 312 kg) = 10.5 m / s.

Answer: before the jump, the trolley with a man was moving at a speed of V = 10.5 m / s.