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A 0.59 kg football is thrown with a velocity of 15 m/s to the right. A stationary...

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A 0.59 kg football is thrown with a velocity of 15 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.018 s.

What is the force exerted on the receiver? Answer in units of N.

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neela's profile pic

Posted (Answer #1)

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The acceleration required to  stop the ball with velocity of 15m /s in 0.018s  is  (velocity-0)/Time to stop it = 15/0.018m/s^2.

Therefore the force exerted by ball on the  catcher = ball's mass*ball's acceleration=0.59*(15/0.018) =491.6667N

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Posted (Answer #2)

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Given

The initial velocity of of ball = u = 15 m/s

Final velocity of ball = v = 0

Time taken to stop the ball = t = 0.018 s

Mass of the ball = 0.59 kg

Therefore acceleration = a = (u - v)/t = -15/0.018 m/s^2

and force f = m*a = 0.59 (-15/0.018) = 491.6666 N approximately

Answer: force exerted on the receiver is 491.6666 N.

 

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