Two plasticine balls weighing 200g and 300g are suspended on identical threads 50 cm long.

Two plasticine balls weighing 200g and 300g are suspended on identical threads 50 cm long. The balls are in contact. The first ball was rejected 90 degrees and released. How high should the balls rise after inelastic collision?

m1 = 200 g = 0.2 kg.

m2 = 300 g = 0.3 kg.

h1 = 50 cm = 0.5 m.

g = 9.8 m / s2.

h2 -?

Inelastic is a blow in which, after interaction, the balls move together with the same speed. For an inelastic impact, the law of conservation of total mechanical energy is valid: E0 = E.

The initial total mechanical energy E0 is expressed by the formula: E0 = m1 * g * h1, where m1 is the mass of the first ball, g is the acceleration of gravity, h1 is the height of the first ball.

Let us express the total mechanical energy of the balls E, after an inelastic impact at the maximum deflection height: E = (m1 + m2) * g * h2.

m1 * g * h1 = (m1 + m2) * g * h2.

h2 = m1 * g * h1 / (m1 + m2) * g = m1 * h1 / (m1 + m2).

h2 = 0.2 kg * 0.5 m / (0.2 kg + 0.3 kg) = 0.2 m.

Answer: with an inelastic impact, the balls will rise to a height of h2 = 0.2 m.