A steel ball weighing 50 g falls from a height of 1.5 m onto a stone slab and, bouncing off it

A steel ball weighing 50 g falls from a height of 1.5 m onto a stone slab and, bouncing off it, rises to a height of 1.2 m. How many degrees the ball will heat up in this case (neglect energy losses).

m = 50 g = 0.05 kg.

h1 = 1.5 m.

h2 = 1.2 m.

g = 9.8 m / s ^ 2.

C = 500 J / kg * ° C.

Δt -?

According to the law of conservation of energy, the change in the potential energy of the ball goes into its internal energy and it heats up.

ΔEp = Q.

ΔЕп = m * g * h1 – m * g * h2 = m * g * (h1- h2).

The amount of heat Q that goes to heat the ball is expressed by the formula: Q = C * m * Δt.

m * g * (h1- h2) = C * m * Δt.

Δt = m * g * (h1- h2) / C * m.

Δt = 0.05 kg * 9.8 m / s ^ 2 * (1.5 m – 1.2 m) / 500 J / kg * ° C * 0.05 kg = 0.006 ° C.

Answer: The temperature of the ball will increase by Δt = 0.006 ° C.