Two balls with masses of 200g and 600g are moving towards each other at the same speed -2m / s. Determine the speed of the balls after inelastic interaction.
m1 = 200 g = 0.2 kg.
m2 = 600 g = 0.6 kg.
V1 = V2 = 2 m / s.
In the case of an inelastic impact of the balls, the law of conservation of momentum is valid in vector form: m1 * V1 + m2 * V2 = m1 * V1 “+ m2 * V2”.
Since before the collision the balls were moving towards each other, and after non-elastic interaction the balls will move with the same speed V1 “= V2” = V, then the law of conservation of momentum will take the form: – m1 * V1 + m2 * V2 = (m1 + m2 ) * V.
V = (m2 * V2 – m1 * V1) / (m1 + m2).
V = (0.6 kg * 2 m / s – 0.2 kg * 2 m / s) / (0.2 kg + 0.6 kg) = 1 m / s.
Answer: after an inelastic impact, the balls will move at a speed of V = 1 m / s in the direction of a ball with a mass of m2 = 0.6 kg.
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