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

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. Why didn’t the ball rise to its original height? How much mechanical energy has been converted into heat? How many degrees is the balloon warmed up? (0.15J; 0.006 ° C)

Given: m (mass) = 50 g (0.05 kg); h1 (starting height) = 1.5 m; h2 (lifting height) = 1.2 m; we do not take into account the heating of the plate; g (acceleration due to gravity) ≈ 10 m / s2; Cc (specific heat capacity of steel) = 500 J / (kg * K).

1) The ball did not rise to its original height, since part of its mechanical energy turned into heat as a result of the impact.

2) Heat energy: ΔQ = ΔEp = m * g * (h1 – h2) = 0.05 * 10 * (1.5 – 1.2) = 0.15 J.

3) Ball heating: Δt = ΔQ / (Cc * m) = 0.15 / (500 * 0.05) = 0.006 ºС.