Physics Laboratory Experiment 15
Coefficient of Restitution
Object:
materials.
To experimentally determine the coefficient of restitution of several
Apparatus: Meterstick; elastic bands; tennis balls; ping-pong ball; super ball; and
other assorted balls.
Theory:
Consider a head-on, elastic collision between two bodies of mass m1 and
m2 traveling with initial velocities of v1 and v2 respectively.
v1>v2
m1
m1
m2
v1
v2
Before Collision
u2>u1
m2
u1
Collision
u2
After Collision
Diagram 15-1
Immediately after collision the mass have velocities u1 and u2 respectively.
We define the coefficient of restitution, e, as the ratio of the relative rate of
separation after collision to the relative rate of approach before collision, or:
e=
relative rate of separation
relative rate of approach
or:
e=
u2 - u1
v1 - v 2
(1)
Consider a sphere made of some material dropped from rest onto the floor. A
collision takes place between the sphere and the floor (Earth) and the sphere naturally
rebounds to some new height, y’. We will call y the original distance fallen and y’ the
rebound height. Since the mass of the Earth is so much greater than the sphere, its relative
velocity with respect to the sphere is zero before and after the collision.
Sphere is a
falling body
with initial
velocity = 0.
Sphere is at
maximum height
velocity = 0.
y
y
y’
v1
e=
u2 - u1 0 - u1
=
v1 - v 2 0 - v 2
To find the impact velocity we write:
u12 = 0 - 2g(-y)
and then
u1 = 2gy
To find the rebound velocity:
0 = u1 - 2gy'
2
u1 = 2gy'
2
and then
u1 = 2gy'
Now
e=
-u1 u1
2gy'
= =
-v 2 v 2
2gy
and simplifying,
e=
y'
y
(2)
By dropping a sphere through a given distance y and measuring the rebound height we
can determine the coefficient of restitution for the sphere and the surface contacted.
Procedures:
Part 1:
1. We will first consider collisions between a set of balls and the floor. One member of
the group will hold a meterstick vertically, another will release a ball from the 100.0
cm mark, and another will observe the rebound height. Be as consistent as possible.
Drop the ball from the same point each time. The member of the group observing
the rebound height must be in a position to eyeball the rebound height. Mark the
rebound height with an elastic band. Measure the rebound height from the bottom of
the ball. Make three (3) trials with each different ball.
2. In each case, using equation (2), calculate the coefficient of restitution for the floor
and balls in question using the average values of y and y’.
3. Next transfer the meterstick to the tabletop and repeat procedures 1 and 2 finding the
coefficient of restitution between the tabletop and the balls used.
4. You are to experimentally measure the coefficient of restitution for four (4) different
balls using procedures 1, 2, and 3.
5. Pick your most elastic ball. Experimentally, measure the 2nd, 3rd, 4th, and if possible,
the 5th rebound heights. In each case mark the rebound heights with elastic bands.
6. Use equation (2) and calculate the rebound heights of you most elastic ball. How do
they compare?
7. For the elastic sphere of Procedures 5 and 6, plot a graph of rebound height vs.
rebound number.
Part 2:
In Mathematics and Physics we commonly deal sequences. A sequence of numbers is a
set of numbers arranged in a definite order, so that there is a first number, a second
member, etc., of the set. The numbers of a sequence are called its terms. The Dan Brown
novel and the motion picture that followed, The Da Vinci Code, made the Fibonacci
Sequence famous. Leonardo Fibonacci (1170 to 1250) was the first, in 1202, to write the
sequence 0, 1, 1, 2, 3, 5, 8, 13, 21, etc. Add 0 and 1 and you get the next term, 1. Add 1
and 1 and you get the next term, 2. Add 1 and 2 and the next term is 3, and so on.
There are many different types of special types of sequences called progressions.
Progressions differ by the rules as to how the terms increase or decrease. The one we are
the most interested in is called a geometric progression.
A geometrical progression is a sequence of numbers each of which, after the first,
may be obtained my multiplying the proceeding number of the sequence by a certain
fixed number called the common ratio. We will exclude from consideration as trivial all
sequences that contain the term zero.
To determine whether a given sequence is a geometric progression, divide each
number after the first by the preceding number. If the quotients are equal, the numbers
constitute a geometric progression; if any two of the quotients are unequal, the sequence
is not a geometric progression.
Example: Consider the sequence 2, 6, 18, 54, 162. Show that it is a geometric
progression.
Starting with the second term, divide each one by the preceding term:
6 18 54 162
= =
=
=3
2 6 18 54
The common ratio is 3 and the sequence is a geometric progression.
A geometric progression has four elements and they are:
1. the first term, a;
2. the common ratio, r ;
3. the number of terms, n;
4. the nth or last term, l.
In terms of a and r the numbers of a geometric progression are successively
a, ar, ar 2, ar 3,
.
In each of these terms the exponent of r is one less than the number of the term. For the
nth or last term we therefore have
l = ar n-1
(3)
List the sequence of rebound heights from Procedure 5 in Part 1. Is this sequence a
geometric progression? What is the common ratio? How does this compare to the
coefficient of restitution? How are equations (2) and (3) related?