# Free Fall Lab ```PH 105
Names:
Date:
Coefficient of Restitution Lab
The “coefficient of restitution”  is a measure of how elastic a collision is. The coefficient of restitution is
defined as the ratio of the speed of separation after a collision to the speed of approach before the
collision:

vafter
v before
For a collision with perfect conservation of mechanical
energy (an “elastic collision”),  = 1. The coefficient of
restitution is a convenient way of specifying how much
kinetic energy in a collision is transferred to internal
energy.
In order to illustrate the utility of the coefficient of
restitution, consider an object of mass m at rest dropped
from a height h0, as shown in the figure on the right.
Neglecting air resistance (and rotational motion),
conservation of mechanical energy allows one to
determine the velocity v0 at which the object reaches the
ground:
Ei  K i  Ui  0  mgh0 and
t0
t1
t2
Ef  K f  Uf  21 mv 02  0
Ei  Ef  v0  2gh0
The coefficient of restitution defined above gives us the velocity after the ball rebounds once as:
v1  v0   2gh0
Applying conservation of mechanical energy again, the maximum height reached after rebound can easily
be calculated:
h1 
2gh0    2h
v12
2
0
2g
2g

which yields
vafter
h1

v before
h0
One can repeat this analysis, and show that the height after the second rebound is h2 =  4h0, and after n
bounces hn =  2nh0 (vn= nv0). What is curious is that hn does not reach zero at any finite n, i.e., the ball
bounces an infinite number of times. (This is a version of Zeno’s paradox if you like.)
We can similarly calculate the time required for each bounce and rebound cycle. For the initial drop from a
height h, the time to reach the ground is t 0 
2h 0
g
, as usual, and each subsequent rebound and drop is
calculated in the same way (counting twice for the round trip). For example, the time required for the first
two bounces is:
t1  2
2h1
2h0
 2
 2t 0
g
g
t2  2
2h2
2h0
 2 2
 2 2 t 0
g
g
Continuing the analysis, we can sum the time for each of the infinite number of bounces, noting that
tn = 2 nt0:




ttotal  t0  t1  t 2    t 0 1  2  2 2  2 3    2t 0   n   t0
n  0 
 1 
 1  
ttotal  2t0 
  t0  t0 

 1  
 1  


1/2
PH 105
Names:
Date:
Perhaps surprisingly at first, the time it takes to bounce an infinite number of times is a convergent
geometric series, since  &lt;1. We see that the total time before the ball stops bouncing is proportional to
the square root of the original height times a factor that involves , a result we can use to determine .
Procedure:
You will need a meter stick, a ball to bounce, and a stopwatch. 1
Part 1:
From the equations above, we note that the ball’s height after n bounces is hn   h0 , or, the ratio of the
2n
height after n bounces to the original release height will be
hn
  2n .
h0
Drop your ball vertically from a given height (about 1m is good), next to an upright meter stick. Measure

the rebound height after a certain number of bounces (say 3). Drop your ball again from the same height,
and measure the rebound height after a different number of bounces. Do this for several numbers of
bounces (n).

hn 
When you are finished, plot log 10  versus n (e.g., in Excel). From the equation above, the slope of
h0 
this plot should be 2log10  , from which you can determine .
Part 2:
Just as in part 1, drop your ball next to the meter stick from a known height h0. Using a stopwatch,
 it takes the ball to stop bouncing (e.g., when you can no longer hear separate
measure how long
bounces).
Repeat
this for a number of different initial heights and record the time for each h0.

h0 .
2 1  

.
g 1  
Now plot the time required to stop bouncing (in seconds) versus the square root of the initial height
Rearranging the expression for the total time above, this should give a straight line of slope
g
1
2
Some algebra yields the coefficient of restitution:  
.
g
(slope)
1
2
Part 3:

(slope)

Repeat either Part 1 or Part 2 for a different type of ball. Should the value of  change? Alternatively,
perform the simulation in Interactive Physics.

Finish and Turn in:



Bonus:


Turn in your plots from Parts 1 &amp; 2, and the values for  you determined in each case.
Do the results agree?