A ball weighing 1 kg, rolling without sliding, hits the wall, rolls back from it. The speed of the ball before impact is 10 cm / s

A ball weighing 1 kg, rolling without sliding, hits the wall, rolls back from it. The speed of the ball before impact is 10 cm / s, after impact is 8 cm / s. Find the amount of mechanical energy transferred to the internal energy upon impact.

Given:

m = 1 kilogram is the mass of the ball;

v1 = 10 cm / s = 0.1 m / s – the speed of the ball before hitting the wall;

v2 = 8 cm / s = 0.08 m / s is the speed of the ball after hitting the wall.

It is required to determine W (Joule) – the amount of energy transferred to the internal energy after the ball hit the wall.

Before hitting the wall, the ball had kinetic energy E1, after impact – E2. Then, according to the law of conservation of energy:

E1 = E2 + W

W = E1 – E2 = m * v1 ^ 2/2 – m * v2 ^ 2/2 = m / 2 * (v1 ^ 2 – v2 ^ 2) =

= 1/2 * (0.1 ^ 2 – 0.08 ^ 2) = 0.5 * (0.01 – 0.0064) = 0.5 * 0.0036 = 0.0018 Joules.

Answer: upon impact, 0.0018 Joules = 1.8 mJ passed into the internal energy.




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