anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Consider the velocities as being in the same direction.
anonymous
  • anonymous
Since 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
  • anonymous
To 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.

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anonymous
  • anonymous
The 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
  • anonymous
Forward me any questions. )

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