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.
(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
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