Two bodies weighing 0.4 kg and 0.6 kg moved towards each other and stopped

Two bodies weighing 0.4 kg and 0.6 kg moved towards each other and stopped after the impact. what is the speed of the second body if the first was moving at a speed of 3m / s.

m1 = 0.4 kg.

m2 = 0.6 kg.

V1 = 3 m / s.

V = 0 m / s.

V2 -?

Since the balls interact only with each other during impact, they can be considered a closed system for which the law of conservation of momentum is valid.

m1 * V1 + m2 * V2 = m1 * V1 “+ m2 * V2” – vector law of momentum conservation.

m1 * V1, m1 * V1 “impulses of the first ball before collision, m2 * V2, m2 * V2” – impulses of the second ball before and after impact.

Since before the collision the balls were moving towards each other and after the collision stopped V1 “= V2” = 0, then the law of conservation of momentum will take the form: m1 * V1 – m2 * V2 = 0.

m1 * V1 = m2 * V2.

V2 = m1 * V1 / m2.

V2 = 0.4 kg * 3 m / s / 0.6 kg = 2 m / s.

Answer: before the collision, the second ball had a velocity V2 = 2 m / s.