Question

max. 5 ms t (ms)

A 300 g ball is dropped from a height of 2.2 m , bounces on a hard floor, and rebounds to a height of 1.7 m . The figure shows the impulse received from the floor. (Figure 1)What maximum force does the floor exert on the ball? Express your answer using two significant figures.

Answer 1

(1). speed of the ball before the bounce :
=> V₁= - √(2 g h₁)
=> V₁= - √(2 x 9.8 x 2.2) = - 6.57 m/s

(2). the ball speed after rebound :
=> V₂= √(2 g h₂)
=> V₂= √(2 x 9.8 x 1.7) = 5.772 m/s

(3). impulse = change in momentum
=> I = Δp
=> I = m (V₂- V₁)
=> I = 0.3 (5.772 - (-6.57))
=> I =3.70 kg m/s

(4). area of triangle = │impulse│
=> (½ x 5x10^-3 x Fm) = 3.7
=> 2.5 x 10^-3 x Fm = 3.7
=> Fm = 1480 N
(Fm = maximum force)

Newton ↔ Kg.m/s^2

Fm = 1480 Kg.m/s^2