Power and Efficiency without mass?

In summary, the textbook solution is poor due to incorrect use of units, premature plugging in of numbers, and lack of need to find an energy.
  • #1
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Homework Statement
Possible error in text (has happened before.)
A basketball loses 35% of its kinetic energy when it hits the floor. Initially dropped from a height of 1.2m. How much time will it be in the air between 3rd and 4th bounces?
Relevant Equations
Ek = Eg = mgh = m(9.81)(1.2)

s = v2t-1/2at^2
Really perplexed me as the answer provided does not solve for mass ... seemingly wants to use mass as a unit of height? Or perhaps I'm off base ... really just curious if this is even correct or possible?

Thanks in advance
 

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  • #2
Sounds okay. Is the answer going to be different depending on the mass of the b-ball ?
 
  • #4
It is correct. The solution shows that the answer is independent of the mass. Look at it this way. If you drop a mass from height ##h_0##, its mechanical energy before the bounce is ##E_0=mgh_0##. If it loses 35% of that during the bounce, it will have ##KE_1=0.65mgh_0## left in the form of kinetic energy.

Question: To what maximum height ##h_1## will the ball rise?
Answer: $$mgh_1=KE_1=0.65mgh_0\implies h_1=0.65h_0.$$Note that the answer is independent of the mass. If you drop two unequal masses, they will hit the floor at the same time; if in addition they lose the same fraction of energy after the bounce, they will rise to the same maximum height.
 
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  • #5
The textbook solution is poor in several ways.

1. It plugs in numbers without units. E.g. we read "##mgh=11.8m##" instead of "##mgh=m\cdot 11.8m^2/s^2##". That may be why you thought it was turning mass into a distance.

2. It plugs in numbers too soon. By leaving g as g it would have cancelled out in the first part, as @kuruman shows in post #4. That makes for less work and greater accuracy.

3. It was never necessary to find an energy. Since after three bounces it will only have ##0.65^3## of its initial energy, we know it will rise to a height of ##1.2\cdot 0.65^3=0.33##m. After that, it is just a matter of finding how long it takes to fall to the ground from that height and doubling it.
 

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