A 0.251 kg ball strikes a smooth level floor with a velocity of 5.37 m/s, 55.0 d

Question

A 0.251 kg ball strikes a smooth level floor with a velocity of 5.37 m/s, 55.0 degrees below the horizontal. It rebounds off the floor with an unknown velocity and reaches a maximum rebound height of 0.735 m. If the duration of the collision with the floor is 53.5 milliseconds, then what was the average (magnitude of the) normal force (in Newtons) the ground applied to the ball? Hint: Use the impulse-momentum theorem. Assume g = 10.0 m/s2, and don't neglect the weight during your analysis. Also respect momentum as a vector: pay attention to your signs!

Explanation / Answer

normal force = rate of change of momentum

= m(v1-v2)/dt

v2 = sqrt(2*g*h) (calculated from the rebound height)

= sqrt(2*10*.735)

= 3.834 m/s

since direction of v2 is reversed after bouncing back thus v2= -3.834 m/s

v1= 5.37 m/s

F = 0.251*[5.37-(-3.834)]/(53.5*10^-3)

= 43.18 N

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