# A railway car moving at a speed of 0.5 m / s collides with a platform moving towards the opposite

**A railway car moving at a speed of 0.5 m / s collides with a platform moving towards the opposite at a speed of 0.1 m / s, after which they move together with a certain speed. Determine this speed if the mass of the car is 20 tons, and the mass of the platform is 8 tons …**

Vv = 0.5 m / s.

Vp = 0.1 m / s.

mw = 20 t = 20,000 kg.

mp = 8 t = 8000 kg.

V -?

Let us write the law of conservation of momentum for a closed system wagon-platform in vector form: mw * Vw + mp * Vp = mw * V + mp * V, where mw * Vw, mw * V are the impulses of the car before and after the collision, mp * Vp, mп * V – platform impulses before and after the collision.

Since the car and the platform are moving towards each other, then for projections the law of conservation of momentum will take the form: mw * Vw – mp * Vp = mw * V + mp * V.

The speed of the car with the platform after the collision will be determined by the formula: V = (mw * Vw – mp * Vp) / (mw + mp).

V = (20,000 kg * 0.5 m / s – 8,000 kg * 0.1 m / s) / (20,000 kg + 8,000 kg) = 0.33 m / s.

Answer: the wagon with the platform after the collision will move in the direction of the wagon movement at a speed of V = 0.33 m / s.