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Vanyuwa [196]
3 years ago
11

A basketball has a mass of 567 g. Heading straight downward, in the direction, it hits the floor with a speed of 2 m/s and rebou

nds straight up with nearly the same speed. What was the momentum change ?
Physics
1 answer:
Vsevolod [243]3 years ago
5 0

Answer:

\Delta p=2.27\frac{kg\cdot m}{s}

Explanation:

The momentum change is defined as:

\Delta p=p_f-p_i\\\Delta p=mv_f-mv_i\\\Delta p=m(v_f-v_i)(1)

Taking the downward motion as negative and the upward motion as positive, we have:

v_f=2\frac{m}{s}(2)\\v_i=-2\frac{m}{s}(3)

Replacing (2) and (3) in (1):

\Delta p=567*10^{-3}kg(2\frac{m}{s}-(-2\frac{m}{s}))\\\Delta p=2.27\frac{kg\cdot m}{s}

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Answer:

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b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

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