Golf Swing Physics
1. The Collision
Guest article by Rod White

December 2008
This section is in two parts.
The first subsection, on the twoball
collision, provides a simplified explanation of what
happens in the collision between the clubhead and ball.
The second subsection, on the threeball
collision, considers what happens when two twoball
collisions are combined. This may seem like a bit of a detour, but hang
in there; it leads to an explanation of why the club head is so much
heavier than the ball. More importantly, it captures the essential
features of the compromise required to maximize the overall energy
transfer in the golf stroke.

TwoBall
Collision
Let us look at a moving ball colliding with another ball that is
initially stationary; our aim is to get as much kinetic energy as
possible into the second ball.
In an efficient collision the first ball stops. This is the criterion
for 100% efficiency. Any movement of the first ball after the collision
represents kinetic energy that has not been transferred to the second
ball.
Case
1: small ball collides with large ball
This slide shows an animation of the collision. After the collision,
the large ball moves slowly, and the little ball has recoiled and is
moving slowly backwards. As evident from the movement of the
small ball after the collision, not all of the kinetic energy of the
small ball has been transferred to the large ball. The collision is not
100% efficient.
Additional note: In these animations it is assumed that the
collisions are ideal, there is no energy lost through compression of
the balls (i.e., the coefficient of restitution = 1.0)
Math note
To a physicist, this page illustrates perfectly the laws of
conservation of energy and momentum. Let's calculate the
energy and momentum for each ball, before and after the collision.
Before the
collision:
Kinetic energy:

T
= ½ M_{1}V_{1}^{2} 
Momentum: 
L = M_{1}V_{1} 
After
the
Collision:
Kinetic energy: 
T
= ½ M_{1}U_{1}^{2}
+ ½ M_{2}U_{2}^{2} 
Momentum: 
L = M_{1}U_{1}
+ M_{2}U_{2} 
Where V
is the velocity before collision, U
is the velocity after the
collision, and M
is the mass of the balls.
The conservation laws say the the "before" should be equal to the
"after", for both kinetic energy and momentum. So we now have two
equations, T_{before}=T_{after}
and L_{before}=L_{after}.
Since we know everything except the aftervelocities U_{1} and
U_{2} ,
we should be able to solve the two equations for the two unknowns 
and we will below.

Case
2: large ball collides with small ball
This
time the animation has the larger ball colliding with the small ball.
Again the collision is not 100% efficient, and afterwards, the large
ball continues to move forward: it retains some of the kinetic energy.
Note
that the little ball moves faster after the collision than the big ball
did before the collision. We will come back to this point in a few
slides.

Math note
Above we saw how conservation of energy and momentum give us two
equations in two unknowns. So it is now
just algebra to calculate the velocities after the collision in terms
of
the initial velocity of ball 1 and the masses of the two balls. If we
solve the equations, we get...
Velocity of first ball after the collision is
U_{1}
= V_{1} 
M_{1}
 M_{2
}
M_{1}
+ M_{2} 
Velocity of second ball after the collision is
U_{2}
= V_{1} 
2
M_{1}_{
}
M_{1}
+ M_{2} 
=
V_{1} 
2_{
}
1 +
M_{2} /_{ }M_{1} 
Note:
You have probably seen the equation for ball speed after collision with
the clubhead, cited by many references, as:
V_{ball}
= V_{head} 
1
+ COR
1
+ M_{ball} / M_{head} 
Let's look at the twoball collision as a golf shot, with
ball 1 as the clubhead and ball 2 as the golf ball. Our model
so far has no energy loss, so the CoR is 1.0. Given this value of CoR
and assignment of masses, the familiar
equation for ball speed is exactly
the same as the equation for U_{2},
the speed after collision of ball 2. Now we know where
the equation for ball speed comes from.

Case
3: two equal sized balls in collision
This time the two balls have the same mass, and the colliding ball
comes to a dead stop with the collision. The collision is now 100%
efficient.
We can also make a couple of other observations
about
the collision  and you can even see where they
came from if you followed the math notes:
 The condition for the first ball to stop is M_{1}=M_{2},
and the second ball takes on the original velocity V_{1}.
 When the first ball is much more massive than the
second, the velocity of
the second ball approaches 2V_{1}.

This graph plots the
collision efficiency versus the ratio of the two masses. The efficiency
is the fraction
of the total kinetic energy transferred to the second ball.
The graph
shows the collision follows the Goldilock’s principle: the colliding
ball can
be too big, too small, or just right. Just right is when the two masses
are
equal.
So why does the typical golf driver have a mass of 200 g and
the ball a
mass of
46 g, surely the collision would be more efficient if the two masses
were the
same?
We will answer that question shortly.

Math note
The equation of this curve is
Efficiency
= 
4
M_{1}M_{2}
(M_{1}
+ M_{2})^{2} 

ThreeBall
Collision
Now, let’s look in
detail at the threeball collision. It might look like a bit of a
detour, but
we will see shortly that two successive 2ball collisions illustrates
an important
compromise that occurs in the golf swing.
Remember, when we ran
the animation of the large ball colliding with the small ball, how the
small
ball moved faster than the large ball? Although it’s not easy
to tell,
the small ball now moves faster than before. The intermediate ball
makes the transfer
of energy from the large ball to the small ball more efficient.
This
graph plots the efficiency for the two separate 2ball collision that
occur within the 3ball collision; first M1 with M2, and then
M2 with M3. In the graph I have assumed that the big ball has
a mass of 1 kg, and the small ball has a mass of 50 g. The mass of the
intermediate ball can take any value as indicated on the horizontal
axis of the graph.
The right hand dotted curve shows the efficiency of the collision
between M_{1}
and M_{2},
with the mass of M_{1}
equal to 1 kg. As expected from our observations with the 2ball
collisions, the energy transfer is most efficient when the intermediate
mass is also 1 kg.
The left hand dotted curve shows the efficiency of the collision
between M_{2}
and M_{3},
with the mass of M_{3}
equal to 50 g. It is most efficient when the intermediate mass is
50 g.
After the first
collision, the intermediate mass carries a given percentage of the
energy, and
then this fraction of the energy is reduced by a further percentage in
the
second collision. The efficiency of the overall threeball collision is
therefore
found by multiplying the efficiency the two collisions, as given by the
two
dotted curves. The result is the solid red curve. Clearly, the best
overall
efficiency is obtained when the intermediate mass is in between the
other two.
If we read the mass off the graph we can see it is a little greater
than 200 g.
Math note
The velocity of the third ball is given by:
U_{3}
= 
4 M_{1}
M_{2}
(M_{1}
+ M_{2})(M_{2}
+ M_{3}) 
V_{1} 
Differentiation with respect to M2, and setting the result to zero
gives the condition for the maximum energy transfer:
In other words, the optimum intermediate mass is the geometric mean of
the other two.
To improve the efficiency even further we could introduce a fourth
ball,
or an infinite sequence of balls, with the masses graded between the
large
and small masses. Now the efficiency approaches 100%. Practical
examples
of this type of exponentially graded coupling include the bullwhip,
horn loudspeakers, and graded electromagnetic absorbers used in stealth
technology.

Last modified  Apr 13, 2014
