A soft tennis ball is dropped onto a hard floor from a height of 1.85 m and rebo

Question

A soft tennis ball is dropped onto a hard floor from a height of 1.85 m and rebounds to a height of 1.33 m. (Assume that the positive direction is upward.) (a) Calculate its velocity just before it strikes the floor. m/s (b) Calculate its velocity just after it leaves the floor on its way back up. m/s (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms. m/s2 (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid? m

Explanation / Answer

a) The ball is drooped from a height 1.85 m .The potential energy it had before falling from that height
PE = mgh
Where m is the mass of the ball
At the floor this potential energy is converted into the kinetic energy of the ball
KE = (1/2) m v2
m gh = (1/2) mv2
v = sqrt (2gh)
v = sqrt (2 x 9.8 x 1.85)
v = 6.021 m/s
v = - 6.021 m/s (downward direction is negative )
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b) Again
v = sqrt (2 g h)
Here h = 1.33 m
v = sqrt (2 x 9.8 x 1.33 )
v = 5.10 m/s
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c) Acceleration = Change in velocity / time
Change in velocity = Final - initial = 5.10 m/s - (- 6.021 m/s) = 11.121 m/s
a = 11.121 m /s / 3.50 x 10-3 s
a = 3177.4 m /s2
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d) The work done in the process of compression and the expansion is
W = F . 2 dx
dx is the compression (a factor of two is there because we take the complete cycle of compression and expansion)
This is equal to the change in KE of the process
Change in KE = (1/2) m (vf2 - vi2)
m a.2dx =   (1/2) m (vf2 - vi2)
dx =   (vf2 - vi2) / 4 a
dx = 5.102 - 6.0212 / 4 x  3177.4
dx = 8.059 x 10-4 m

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