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anonymous
 one year ago
A 2.0 kg rubber ball falling at 0.85 m/s bounces from the floor and moves upward at 0.65 m/s. If the ball was in contact with the floor for 0.15 s:
what is the change in momentum of the ball?
what was the impulse acting on the ball?
what force did the floor exert on the ball?
anonymous
 one year ago
A 2.0 kg rubber ball falling at 0.85 m/s bounces from the floor and moves upward at 0.65 m/s. If the ball was in contact with the floor for 0.15 s: what is the change in momentum of the ball? what was the impulse acting on the ball? what force did the floor exert on the ball?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Consider the velocities as being in the same direction.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since the 2.0kg rubber ball had acquired a velocity of 0.65m/s [upward] from falling downward at 0.85m/s [2.0kg][0.85m/s][2.0kg][0.65m/s=1.7kgm/s+1.3kg/s=3.0mkg/s Therefore change in the momentum of the ball was 3.0mkg/s

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0To calculate the impulse, which is calculated by Impulse=Delta Momentum Impulse=3.0N/s Therefore the impulse on the ball was 3.0N/s. Note that impulse is equal to change in momentum.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The amount of force exerted by the floor to the ball, which resulted in the chance in momentum of 3.0kgm/s, is calculated using the formula F*t=Delta P (change in momentum) Therefore, F*(0.15s)=3.0kgm/s F=3.0kgm/s/0.15s=20N Therefore the force exerted by the floor to the ball is 20N.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Forward me any questions. )
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