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A basketball player drops a 0.4-kg basketball vertically so that it is traveling at 5.7 m/s when it reaches the floor. The ball rebounds upward at a speed of 2.8 m/s. Determine (a) the magnitude and direction of the ball's change in momentum and (b) the average net force that the floor exerts on the ball if the collision lasts 0.09 s.

Respuesta :

Answer:

(a) p = 3.4 kg-m/s (b) 37.78 N.

Explanation:

Mass of a basketball, m = 0.4 kg

Initial velocity of the ball, u = -5.7 m/s (as it comes down so it is negative)

It rebounds upward at a speed of 2.8 m/s  (as it rebounds so positive)

(a) Change in momentum = final momentum - initial momentum

p = m(v-u)

p = 0.4 (2.8-(-5.7))

p = 3.4 kg-m/s

(b) Impulse = change in momentum

Ft = 3.4

We have, t = 0.09 s

[tex]F=\dfrac{3.4}{0.09}\\\\F=37.78\ N[/tex]

Hence, this is the required solution.

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