A ball of mass m = 0.1 kg, falling without initial velocity from a height of H = 5 m, hit a solid horizontal plate.

A ball of mass m = 0.1 kg, falling without initial velocity from a height of H = 5 m, hit a solid horizontal plate. The modulus of the ball’s impulse change as a result of the impact is P = 1.8 kg × m / s. What part of the mechanical energy n lost the ball on impact?

m = 0.1 kg.

g = 10 m / s2.

H = 5 m.

ΔP = 1.8 kg * m / s.

n -?

n = E2 * 100% / E1, where E1, E2 are the energy of the ball before and after impact.

The energy of the ball before hitting E1 is expressed by the formula: E1 = m * g * H.

E1 = 0.1 kg * 10 m / s2 * 5 m = 5 J.

E2 = m * V2 ^ 2/2.

Let us express the modulus of change in the momentum of the ball ΔP upon impact: ΔP = m * V2 + m * V1.

V2 = (ΔP – m * V1) / m.

m * g * H = m * V1 ^ 2/2.

V1 = √ (2 * g * H).

V1 = √ (2 * 10 m / s2 5 m) = 10 m / s.

V2 = (1.8 kg * m / s – 0.1 kg * 10 m / s) / 0.1 kg = 8 m / s.

E2 = 0.1 kg * (8 m / s) ^ 2/2 = 3.2 J.

n = 3.2 J * 100% / 5 J = 64%.

Answer: upon impact, the ball lost 36% of its mechanical energy.