A racquet ball with mass m = 0.256 kg is moving toward

Ответ:

shealene7

22.01.2022

Physics

Explanation:

Given that,

Mass of the, m = 0.256 kg

Initial peed of the ball, v = 11.8 m/s

The ball is in contact with the wall for t = 0.066 s

(1) The magnitude of the initial momentum of the racquet ball is given by :

p = mv

$p=0.256\ kg\times 11.8\ m/s\\\\p=3.02\ kg-m/s$

(2) It is projected at angle of 29 degrees. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal.

The change in momentum of the racquet ball is given by :

$\Delta p = m(v\cos \theta - (-v\cos \theta))\\\\\Delta p = m(v\cos \theta - (-v\cos \theta))\\\\\Delta p = 2mv\cos \theta\\\\\Delta p = 2\times 0.256 \times 11.8\times \cos (29)\\\\\Delta p=5.28\ kg-m/s$

(3) Impulse is equal to the change in momentum.

$\Delta p=Ft\\\\F=\dfrac{\Delta p}{t}\\\\F=\dfrac{5.28}{0.066 }\\\\F=80\ N$

Hence, this is the required solution.

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