Golf ball hit by club (conservation problem)

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In summary, the conversation discusses the physics of a golf ball being hit by a club during a tee shot. The collision between the 300 g club head moving at 40 m/s and the 46 g ball results in the ball taking off with a speed of 70 m/s. The change in momentum of the ball is 3.22 Ns, the speed of the club head immediately after the collision is 29.3 m/s, and approximately 47% of the club's kinetic energy is transferred to the ball while the rest is lost to thermal energy. There is a discrepancy in the final kinetic energy, suggesting a possible error in the problem.
  • #1
XanziBar
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Homework Statement



In a typical tee shot, a golf ball is hit by the 300 g head of a club moving at a speed of 40 m/s. The collision with the ball happens so fast that the collision can be treated as the collision of a 300 g mass with a stationary ball—the shaft of the club and the golfer can be ignored. The 46 g ball takes off with a speed of 70 m/s.
A) What is the change in momentum of the ball?
B)What is the speed of the club head immediately after the collision?
C) What fraction of the club’s kinetic energy is transferred to the ball?
D) What fraction of the club’s kinetic energy is “lost” to thermal energy?

Homework Equations


Pi=Pf
KE=.5*m*v^2
P=mV

The Attempt at a Solution



For A I did Pf-Pi=.046*70-0=3.22 Ns

For B I did conservation of momentum: .3*40=.3*v+.046*70

v=(.3*40-.046*70)/.3=29.3 m/s

For C the initial KE of the club is .5*.3*40^2=240 J
the final KE of the ball is 0.5*0.046*70*70=112.7 J

112.7/240=.47

for D

KEi=240 J from before and the final is

0.5*0.3*29.3*29.3 + 0.5*0.046*70*70=241.5 J

which is where I am stuck because I am confused why the initial energy is less than the final energy. Help?
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  • #2
It's a round off error. The final speed of the club is 29.666... If you use this value, your final energy will be slightly less than the initial energy. It helps to solve the problem symbolically (if you can) and plug in at the very end. This way, problems like this will be avoided.
 
  • #3
Yeah I went back and used exact values before. And when I do the calculation

v=(.3*40-.046*70)/.3=29.2666 m/s=(439/15) m/s

Also, if I plug in your value of 29.666 I get 244.7 J in the end.

So I'm still stuck.
 
  • #4
XanziBar said:
Yeah I went back and used exact values before. And when I do the calculation

v=(.3*40-.046*70)/.3=29.2666 m/s=(439/15) m/s

Also, if I plug in your value of 29.666 I get 244.7 J in the end.

So I'm still stuck.
I did it in two different ways and I agree with you. The final kinetic energy is greater than the initial kinetic energy by about 1.2 J. Now what?

1. We are both wrong and we need a third opinion.
2. This is a poorly conceived problem.
3. The golfer cheated by attaching a firecracker on his club and added extra speed to the ball.

I vote for option 2. What do you think?
 
  • #5
You're right I think it is option 2. So I did the problem again "assuming" a completely elastic collision with a stationary target and if energy is conserved then the MOST the golf ball can go is 69.36 m/s. So I think whoever wrote this problem meant for the final speed of the ball to be less than 70 m/s. I think an elastic collision is when you get the most speed out of the target right?

Thanks for your help.
 

Related to Golf ball hit by club (conservation problem)

What is the conservation problem in golf ball and club collisions?

The conservation problem in golf ball and club collisions refers to the principle of conservation of energy and momentum. In this scenario, when a golf club hits a golf ball, they both experience a transfer of energy and momentum. The conservation problem arises when we try to understand how much energy and momentum is conserved in the collision.

What factors affect the conservation of energy and momentum in a golf ball and club collision?

The conservation of energy and momentum in a golf ball and club collision is affected by several factors. These include the speed and angle of the club head, the material and construction of the golf ball, and the surface on which the collision occurs. Additionally, factors such as air resistance and friction can also impact the conservation of energy and momentum.

How is the conservation of energy and momentum calculated in a golf ball and club collision?

The conservation of energy and momentum in a golf ball and club collision can be calculated using the equations for kinetic energy and momentum. The initial kinetic energy and momentum of the golf club are equal to the final kinetic energy and momentum of the golf ball after the collision. By solving these equations, we can determine the conservation of energy and momentum in the collision.

Why is the conservation of energy and momentum important in golf ball and club collisions?

The conservation of energy and momentum is important in golf ball and club collisions because it helps us understand and predict the behavior of the golf ball after being struck by the club. It also allows us to make adjustments to our swing and equipment to optimize the transfer of energy and momentum, resulting in better shots.

What are some real-world applications of the conservation of energy and momentum in golf?

The conservation of energy and momentum in golf has several real-world applications. For example, understanding these principles can help golf equipment manufacturers design clubs and balls that optimize energy transfer, leading to better performance on the course. It also allows golfers to analyze their swings and make adjustments to improve their shots. Additionally, the conservation of energy and momentum is used in golf course design to create challenging and strategic holes for players.

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