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3 years ago

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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 ?

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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From the question we are told that

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$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

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The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

$h_c = \frac{v^2}{g}$

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