A 0.65 kg ball is dropped from 2.5 m. Every time it bounces, it reached 3/4 of original height. Find :
a. the kinetic energy when it hit the floors for the first time
b. lost in mechanical energy at first bounce
c. lost in mechanical energy at second bounce
d. number of bounce so that the ball is at rest
Ep = mgh
Un = ar^(n-1)
The Attempt at a Solution
a. Ek = Ep = mgh = 0.65 * 9.8 * 2.5 = 15.925 J
b. lost in mechanical energy = ∆ Ep = mg*∆h = 0.65 * 9.8 * (2.5 - 2.5 * 3/4) = 3.98125 J
c. lost in mechanical energy at second bounce (with respect to initial mechanical energy, when h = 2.5 m) = 0.65 * 9.8 * (2.5 - 2.5 * (3/4)^2) = 6.97 J
d. the balll will be at rest if the height = 0. The height of the ball bounced off is a geometric sequence with a = 2.5 and r = 3/4. Using the formula : Un = ar^(n-1), it can be shown that h will never be zero, so the ball will never be at rest.
Do I get it right?