Two trolleys weighing 1 kg and 2 kg are moving towards each other at a speed of 0.4 and 0.1 m / s.

Two trolleys weighing 1 kg and 2 kg are moving towards each other at a speed of 0.4 and 0.1 m / s. Determine the speed of the bogies after a non-elastic collision.

m1 = 1 kg.

m2 = 2 kg.

V1 = 0.4 m / s.

V2 = 0.1 m / s.

V “-?

Impact is called inelastic when, after collision, the bodies stick to each other and move at the same speed.

Let’s write the law of conservation of momentum in vector form: m1 * V1 + m2 * V2 = (m1 + m2) * V “.

Since the carts were moving towards each other before the collision, the conservation law will take the form: m1 * V1 – m2 * V2 = (m1 + m2) * V “.

V “= (m1 * V1 – m2 * V2) / (m1 + m2).

V “= (1 kg * 0.4 m / s – 2 kg * 0.1 m / s)) / (1 kg + 2 kg) = 0.07 m / s.

Answer: the bogies after the collision will move with the speed V “= 0.07 m / s in the direction of the first bogie.