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3 years ago

7

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.

1 answer:

Alik [6]3 years ago

4 0

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

$F=\dfrac{3.4}{0.09}\\\\F=37.78\ N$

Hence, this is the required solution.

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The force exerted between two point charges $q_1$ and $q_2$ separated a distance d is given by Coulomb's formula

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Applying Coulomb's formula for $q_1$ is

$\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}$

Now for $q_2$

$\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}$

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$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{k\ q_1\ q_3}{d_1^2}=\frac{k\ q_2\ q_3}{d_2^2}$

Simplfying and solving for $q_1$

$\displaystyle q_1=\frac{q_2\ d_1^2}{d_2^2}$

$\displaystyle q_1=\frac{3.10^{-9}\ 0.02^2}{0.04^2}$

$\boxed{\displaystyle q_1=+0.375\ {10}^{-9}}$

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