A 280 g ball is dropped from a height of 1.6 m , bounces on a hard floor, and re

Question

A 280 g ball is dropped from a height of 1.6 m , bounces on a hard floor, and rebounds to a height of 1.1 m . (Figure 1)  shows the magntiude of the normal force exerted on the ball while it is contact with the floor. If the ball is on contact with the floor for a time t = 2 ms , what maximum force, Fmax, does the floor exert on the ball?  Hint: Given the small amount of time and the small mass of the ball, the impulse imparted the ball's weight is negligible. Express your answer using three significant figures and include the appropriate units.

Explanation / Answer

Impulse given = mv-mu

= 0.280*(sqrt(2gh1) - - sqrt(2gh2))

= 0.280* (sqrt (2*9.8*1.6) + sqrt (2*9.8*1.1))

= 2.868 Ns

Also impulse = 0.5* Fmax*delta t =2.868

Fmax = 2*2.868/0.002

= 2.87*10^3 N answer

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