Two lead bars m1 = 100g and m2 = 200g are moving towards each other at speeds v1 = 4m / s and v2 = 5m / s.

Two lead bars m1 = 100g and m2 = 200g are moving towards each other at speeds v1 = 4m / s and v2 = 5m / s. What kinetic energy will the balls 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.

Eк -?

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.

The masses of the balls m after an absolutely inelastic impact will be the sum: m = m1 + m2.

m = 0.1 kg + 0.2 kg = 0.3 kg.

– m1 * V1 + m2 * V2 = (m1 + m2) * V.

The speed of the balls V after impact is expressed by the formula: V = (m2 * V2 – m1 * V1) / (m1 + m2).

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

Eк = 0.3 kg * (2 m / s) 2/2 = 0.6 J.

Answer: the balls after impact will have kinetic energy Eк = 0.6 J.