Golf Swing Physics

1. The Collision

Guest article by Rod White  --  December 2008
This section is in two parts.

The first subsection, on the two-ball collision, provides a simplified explanation of what happens in the collision between the clubhead and ball.  

The second subsection, on the three-ball collision, considers what happens when two two-ball 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. 

Two-Ball 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 = M1V12
Momentum:   L = M1V1

After the Collision:
Kinetic energy:   T = M1U12 + M2U22 
Momentum:   L = M1U1 + M2U2

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, Tbefore=Tafter and Lbefore=Lafter. Since we know everything except the after-velocities U1 and U2 , 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
U1  =  V1 M1 - M2

M1 + M2
Velocity of second ball after the collision is
U2  =  V1 2 M1

M1 + M2
 =  V1 2

1 + M2 / M1
Note: You have probably seen the equation for ball speed after collision with the clubhead, cited by many references, as:
Vball  =  Vhead 1 + COR

1 + Mball / Mhead
Let's look at the two-ball 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 U2, 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 M1=M2, and the second ball takes on the original velocity V1.
  • When the first ball is much more massive than the second, the velocity of the second ball approaches 2V1.

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 M1M2

(M1 + M2)2

Three-Ball Collision

Now, let’s look in detail at the three-ball collision. It might look like a bit of a detour, but we will see shortly that two successive 2-ball 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 2-ball collision that occur within the 3-ball 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 M1 and M2, with the mass of M1 equal to 1 kg. As expected from our observations with the 2-ball 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 M2 and M3, with the mass of M3 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 three-ball 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:
U3  =   4 M1 M2

(M1 + M2)(M2 + M3)
V1
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.
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Last modified - Apr 13, 2014