L3 - Intro to impacts (part 2)

Question 1

The efficiency of an impact

The golfer is interested in knowing the energy transfer to the ball from the club at impact.

What is the balls energy after impact (inserting the v’ into the equation) ?

What is the club head’s energy ?

Answer 1

Question 2

The efficiency of an impact

The efficiency of the transfer is measured by the fraction of the original energy which is transferred.

Write the equation for efficiency

In the case of the golf drive ,approxiamtely, how efficiency is the collision ?

Answer 2

In the case of the golf drive –> efficiency = 0.43 . Therefore ,43% of the energy goes into the ball

Question 3

The efficiency of an impact

What does this plot suggest with regards achieving the most efficient collision?

If this was to be achieved? What would happen to the club head after the collision?

Answer 3

This graph suggests that the more similar the masses – the higher the efficiencies .i.e when M=m , e=1.

The most efficient collision when would be when M=m .Thus suggesting that club head should be same mass as ball for max collision efficiency.

However, If e=1 ,perfect elastic collision was possible , all energy would be transferred from club head to the ball and club head would be stationary after impact.

Question 4

The efficiency of an impact

Although better collision efficiency, a light weight golf club is not ideal.

Explain this statement using these equations

Answer 4

• If M = m at the maximum efficiency then:

v’ = V/2(1+e)

For an infinitely large mass M, then :

v’ = V (1+e)

• increase in mass of the clubhead from m to infinity has the following effect:

 decreases the denominator from 2 to 1,
 which would double the velocity of the ball, in spite of the inefficient energy transfer.

therefore a yielding a longer shot than lighter clubs

Question 5

The efficiency of an impact

Describe the relationship between mass of club and launch velocity.

What conflicting trends are highlighted ?

Answer 5

• Most efficient impact still M/m = 1 (but low mass therefore low momentum imparted ,thus, relatively slow launch velocity

o As Mass Increases , so does launch velocity, but to a lesser and lesser extent

o Mass affects V inbound! (club too heavy – player can’t swing as fast as a light club) M increase, thus, V decrease

• 2 conflicting trends
Ideal mass of club depends on how swing speed varies as M increases