A ball with a mass of 2 kg, moving at a speed of 4 m / s, collides inelastically with a ball with a mass of 1 kg moving in the same direction, the speed of which is 2 m / s. Find the kinetic energy of the balls after impact.
m1 = 2 kg.
V2 = 4 m / s.
m2 = 1 kg.
V2 = 2 m / s.
The kinetic energy of the balls Ek is determined by the formula: Ek = (m1 + m2) * V2 / 2, where V is the velocity of the balls after inelastic collision.
Since the balls interact only with each other during impact, this system can be considered closed.
For a closed-loop system, the momentum conservation law is valid: the sum of the impulses of a closed-loop system does not change.
m1 * V1 + m2 * V2 = (m1 + m2) * V – vector.
Since before the collision the balls were moving in one direction, then after the collision they will move in the same direction with the speed V = (m1 * V1 + m2 * V2) / (m1 + m2).
V = (2 kg * 4 m / s + 1 kg * 2 m / s) / (2 kg + 1 kg) = 3.3 m / s.
Ek = (2 kg + 1 kg) * (3.3 m / s) 2/2 = 16.3 J.
Answer: after the collision, the kinetic energy of the balls will be Ek = 16.3 J.
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