- Download PDF
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.
2 Answers | Add Yours
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.
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
We’ve answered 323,933 questions. We can answer yours, too.Ask a question