Two lead balls with masses m1 = 100 g and m2 = 200 g are moving towards each other with velocities

Two lead balls with masses m1 = 100 g and m2 = 200 g are moving towards each other with velocities v1 = 4 m / s and v2 = 5 m / s. What kinetic energy will the first ball have after their absolutely inelastic collision?

m1 = 100 g = 0.1 kg.

m2 = 200 g = 0.2 kg.

V1 = 4 m / s.

V2 = 5 m / s.

Ek1 “-?

The kinetic energy of the body Ek is determined by the formula: Ek = m * V2 / 2, where m is the mass of the body, V is the speed of its movement.

Ek1 “= m1 * V1” 2/2, where m is the mass of the first ball, V1 “is the speed of the first ball after collision.

Let’s write down the law of conservation of momentum for projections on the axis directed towards the motion of the ball with mass m1: m1 * V1 – m2 * V2 = m1 * V1 “+ m2 * V2”.

Since the impact is inelastic, the velocities of the balls after impact will be the same: V1 “= V2”.

m1 * V1 – m2 * V2 = (m1 + m2) * V1 “.

V1 “= (m1 * V1 – m2 * V2) / (m1 + m2).

V1 “= (0.1 kg * 4 m / s + 0.2 kg * 5 m / s) / (0.1 kg + 0.2 kg) = – 2 m / s.

The “-” sign indicates that after the impact, the balls will move in the direction of the movement of the second ball before the impact.

Ek1 “= 0.1 kg * (2 m / s) 2/2 = 0.2 J.

Answer: after impact, the kinetic energy of the first ball will be Ek1 “= 0.2 J.