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.
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