If the collision of the ball with the cushion takes 0.4 s, what is the force the
ID: 1372089 • Letter: I
Question
If the collision of the ball with the cushion takes 0.4 s, what is the force the cushion exerts on the ball? Show your calculations. Not sure how to set up this problem for a lab. The ball's mass is 0.16kg.
37641606932295 24476591877545 C.)6 5 3 7, 1 7 8 9 8 8 8 8 8 8 1 1 1 0 0 0 0 0 0 0 0 0 0 0 m1-10-0-0-0-0-00-00-0-O 9 yl-4 6 9 5 631 9 8 7 0 8 95 45 3 412 9 9 8676 55 88876544444444 C) 07029629823495 9 4 82110794097 0114715727539 n/) 1233221832811 06287455135467 8526225111444 4 171581581470 54 112223334445 000-O-O xm (n 00-0-0-0-0-Q-0-0-0-0-0-0-0 07307307307307 6306306306306 6306306306306 ) 6122344566788 ,0 0 0 0 0 0 0 0 0 0 0 0 0 12345678901234 89 10-11-1Explanation / Answer
According to given data collision has occured at 0.2667 s. At this time , ball has maximum y coordinate. After this ball has smaller y coordinates, which shows that ball hit the cushion at 0.2667 s.
Also at this time ball changes direction of its vertical velocity from positive to negative.
To find force exerted on ball, we can use
Force = Rate of change of momentum
Along y direction
Force = ( change in momentum)/time = (final momentum - initial momentum)/ time
final momentum at 0.2667 s = mvf = 0.16 x (-0.161) = -0.026 kgm/s
initial momentum at 0.2000 s = mvi = 0.16 x 0.774 = 0.124 kgm/s
Force Fy = (-0.026- 0.124)/( 0.2667-0.2000) = -2.25 N
Along x direction
Force = ( change in momentum)/time = (final momentum - initial momentum)/ time
final momentum at 0.2667 s = mvf = 0.16 x (-0.616) = -0.099 kgm/s
initial momentum at 0.2000 s = mvi = 0.16 x (-0.745) =- 0.119kgm/s
Force = Fx = (-0.099- 0.119)/( 0.2667-0.2000) = -3.27 N
Net force = ( Fx^2 + Fy^2)^0.5 = 3.97 N
Direction = arctan ( Fy/Fx) = 34.5 degree counterclockwise with respect to - x axis
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