A vertically falling ball weighing 200 g hit the floor at a speed of 5 m / s and jumped

A vertically falling ball weighing 200 g hit the floor at a speed of 5 m / s and jumped to a height of 8 cm. Find the change in the ball’s momentum upon impact?

m = 200 g = 0.2 kg.

g = 10 m / s2.

V1 = 5 m / s.

h2 = 8 cm = 0.08 m.

Δp -?

The momentum of the ball p is a vector physical quantity equal to the product of the mass of the ball m by the speed of its motion V: p = m * V.

Δp = p2 – p1 = m * V2 – m * V1 = m * (V2 – V1).

m * g * h2 = m * V22 / 2 – the law of conservation of total mechanical energy.

V2 = √ (2 * g * h2).

V2 = √ (2 * 10 m / s2 * 0.08 m) = 1.26 m / s.

Since V2 and V are directed in opposite directions, then Δp = m * (V2 – (-V1)) = 0.2 kg * (1.26 m / s + 5 m / s) = 1.252 kg * m / s.

Answer: the change in the momentum of the ball upon impact was Δp = 1.252 kg * m / s.