A very small 51 g steel ball is released from rest and falls vertically onto a s

Question

A very small 51 g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for 0.50 ms. The ball rebounds elastically and returns to its original height. The total time interval for a round trip is 3.00 s. What is the magnitude of the average force exerted on the ball by the plate during contact with the plate?

I know the answer is 3000 N, but Im not sure of the equations to use to there. please include ALL steps and equations and where the numbers are coming from. thank you

Explanation / Answer

time interval for a round trip t = 3 seconds

time for downward journey is t1 = 3/2 = 1.5 sec

velocity of ball before hitting the steel plate is u = gt= 9.8*1.5 = 14.7 m/s

if the ball rebounds elastically then it rebounds with samevelocity then v = -14.7 m/s

change in velocity is u-(-v) = 14.7 +14.7 = 29.4 m/s

change in momemtum of the ball is = mass *change invelocity

mass of ball = 0.051 kg

then change in momemtum of the ball is = 0.051 *29.4 = 1.4994kg m/s

time of contact of the ball t ' = 0.5 ms =0.5*10-3s

then average force exerted on the ball during contact with theplate

                            = change in momemtum of the ball /time of contact of theball

                            =  1.4994 /0.5*10-3 = 2998 N or nearly 3000N

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