A ball of mass 180 g is released from rest at a height of1.50 m above the floor

Question

A ball of mass 180 g is released from rest at a height of1.50 m above the floor and it rebounds straight up to a heightof 0.700 m. A. Determine the magnitude of the the ball's change inmomentum due to its contact with the floor. B. If the contact time with the floor was0.0850 s, what was the average force the floor exerted on theball? A ball of mass 180 g is released from rest at a height of1.50 m above the floor and it rebounds straight up to a heightof 0.700 m. A. Determine the magnitude of the the ball's change inmomentum due to its contact with the floor. B. If the contact time with the floor was0.0850 s, what was the average force the floor exerted on theball? B. If the contact time with the floor was0.0850 s, what was the average force the floor exerted on theball?

Explanation / Answer

Given : .                Mass (m) = 180 g = 0.18 kg .                Initial velocity (v1) = 2gh1 = (2 * 9.8*1.50)   = 5.42   m /s .                Final velocity (v2)   = 2gh2 = ( 2 *9.8*0.70)   = 3.70 m / s .                As it rebounds back ,     v2   =   - 3.70   m/s . (a)    Change in momemtum is : .        P    =   m(v2    - v1)   = 0.18 ( - 3.70 - 5.42 ) .                                              =   - 1.641    kg - m /s .               Magnitudeis :   P   = 1.641 kg - m/s . (b)     Time (t) = 0.0850 s . Froce is : .               F      =   P /t    = 1.641 / 0.0850  =   19.31   N . Hope this helps u!

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