A 100 g rubber ball is thrown down at 4.0 m/s from a height...

Question

A 100 g rubber ball is thrown down at 4.0 m/s from a height of 2.5 m. If the ball bounces to a height of 2.7 m, calculate the efficiency.

Answer: 81%

Answer 1

The efficiency probably refers to the retention of the mechanical energy (kinetic plus potential) of the ball as it bounces.  Assuming negligible air resistance, the ball should have constant mechanical energy as it travels to the ground, and constant mechanical energy as it rises again after the bounce.  The only point where it may lose energy, in this approximation, is during the bounce (since the force of the ground on the ball does not necessarily conserve mechanical energy).

 

So, all we need to do is take the ratio of the ball's mechanical energy after the bounce to its mechanical energy before the bounce.  Since the mechanical energy is constant before and constant after, we just need to look at one point before it bounces, and one point after.  We can use the points we are given information about:  the point where it is thrown (before bounce) and the highest point to which it returns (after bounce).

 

At the point where it is thrown, its mechanical energy is:

 

ME_o = KE_o + PE_o = (1/2)mv_o²+ mgh_o

 

At the point of its highest return (after bounce), its mechanical energy is:

 

ME_f = KE_f + PE_f = 0 + mgh_f  (the kinetic energy at its point of highest return is 0 because it has momentarily come to rest there)

 

So, the efficiency is:

 

eff = ME_f/ME_o = (mgh_f)/((1/2)mv_o²+mgh_o)

 

Note that we can cancel a factor of m out of this ratio, so, as long as we have the other information, the mass of the ball is irrelevant.  For the other terms, we have:

 

v_o = initial speed = 4.0 m/s

h_o = initial height = 2.5 m

h_f = final height = 2.7 m

 

With these values and the standard value for g, I obtained:

 

eff = 0.814 = 81.4%