Calculate Average Force by finding Impulse over Impact Time

In summary: You got it. That's exactly right. And it's definitely a lot of force, which is why you need to be careful on the golf course. :)In summary, the conversation discussed the process of calculating the average force exerted by a club on a golf ball during impact, as well as estimating the distance over which the club and ball are in contact during the swing. The conversation delved into the conversion of units and the correct use of initial and final velocities in the equations. It also touched upon the concept of action and reaction forces in Newton's third law and how it applies to this scenario.
  • #1
PhysicsPhil
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[SOLVED] Calculate Average Force by finding Impulse over Impact Time

Homework Statement



A movie depicting a 55-g golf ball hit with a club shows that the impulse time between the club and the ball is 0.2 ms, and the initial speed of the ball is 136 mph.

a) What is the average force exerted by the club on the ball during impact?
b) Estimate the distance over which the club and the ball are in contact during the swing.

Homework Equations



The Attempt at a Solution



First converted everything to SI units...
Mass of ball: 0.055kg
Impact time: 2x10^-4 seconds
Initial velocity of golf ball: 60.797 m/s

(a)
Fave = change in momentum (impulse) / impact time.
= Pfinal - Pinitial / impact time = mVf - mVi / impact time = m(Vf - Vi)/impact time

= [tex] (0.055kg)(0m/s - 60.797m/s) / 2x10^-4 s = -16719.175 N [/tex]

I have a feeling this is wrong because the Force should not be negative. Where I am getting tripped up in is what the final velocity of the ball should be ... I just assumed it was 0 m/s (when the ball would eventually come to rest) but I have a few doubts about whether this is what I am to do.

(b) Fave = M * Impact speed / 2 * change in collision distance / impact speed =

M * Impact speed^2 / 2*change in collision distance

[tex](0.055kg)(60.797m/s)^2/ 2deltah = -16719.175N[/tex]

delta h = -0.0060797m ...
 
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  • #2
I've thought about my question more and now I'm thinking maybe I misinterpreted one of the calculations. Perhaps change in momentum can't equate to m(Vf - Vi) ...
 
  • #3
= (0.055kg)(0m/s - 60.797m/s) / 2x10^-4 s = -16719.175 N

Well, you have your Vi and Vf mixed up. The ball is initially at rest, so Vi=0. Vf = 60.797 m/s because that's the velocity the ball is going after the force has acted on it. Change that up and you will get a positive number.
 
  • #4
Throughout the equation you were (correctly) calculating from the point of view of the ball (using the ball's mass, velocities, etc.). The question asks what the average force exerted by the club onto the ball is. Re-read Newton's laws for a hint.
 
  • #5
hage567 said:
Well, you have your Vi and Vf mixed up. The ball is initially at rest, so Vi=0. Vf = 60.797 m/s because that's the velocity the ball is going after the force has acted on it. Change that up and you will get a positive number.

Thanks. That makes sense and I feel silly for not realizing it. I think what threw me off was how the question worded "the initial speed of the ball is 136mph" - so I immediately went Vi = 136mph ...

Now I realize that its final velocity (in regards to velocity post-impact) is 136mph (60.797 m/s) and that its initial velocity is infact 0 m/s since it is as at rest.
 
  • #6
simon1987 said:
Throughout the equation you were (correctly) calculating from the point of view of the ball (using the ball's mass, velocities, etc.). The question asks what the average force exerted by the club onto the ball is. Re-read Newton's laws for a hint.

I found the three laws of Newton on a neat website (and am going to print them out and post it on my wall!).

My guess is you are referencing the third law: III. For every action there is an equal and opposite reaction.

But this doesn't seem to make conceptual sense to me. Yes by the conservation of momentum if you add up the momentum of the club and the ball it will equal 0. The reason the ball goes much farther is because it has to have a much larger velocity to compensate for its smaller mass relative to the club. But does this relation really hold when you examine forces? The club applies 16719.175 Newtons of Force on the ball, (and my assumption from action-reaction pairs is that that the ball, via your hint, is that the ball will exert a force of -16719.175N on the club ...) but this is a large amount of force, and while I'm not a golfer, it seems like this is an excessive amount of force for the golfer to withstand...

EDIT: I read this statement which sort of made it clear to me: The average force on an object during a collision is the change in the objects momentum divided by the collision time. The change of the club's momentum divided by its collision time would give me a negative force. But if I look at the change of momentum (positive) for the ball and divide it by the impact time ; then I would get a positive force (which I now do!)
 
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  • #7
PhysicsPhil said:
EDIT: I read this statement which sort of made it clear to me: The average force on an object during a collision is the change in the objects momentum divided by the collision time. The change of the club's momentum divided by its collision time would give me a negative force. But if I look at the change of momentum (positive) for the ball and divide it by the impact time ; then I would get a positive force (which I now do!)

Bingo!
 

1. How do you calculate average force by finding impulse over impact time?

To calculate average force, you need to first find the impulse, which is the change in momentum. This can be calculated by multiplying the mass of the object by its change in velocity. Then, divide the impulse by the impact time, which is the amount of time the force is applied. This will give you the average force exerted during the impact.

2. What is impulse and how is it related to average force?

Impulse is the change in momentum of an object. It is equal to the average force applied multiplied by the time it is applied. This means that the greater the impulse, the greater the change in momentum and the greater the average force.

3. Can you provide an example of calculating average force by finding impulse over impact time?

Sure, let's say a baseball with a mass of 0.15 kg is thrown at a velocity of 25 m/s and is caught by a catcher in 0.1 seconds. The change in momentum would be 0.15 kg * (25 m/s - 0 m/s) = 3.75 kgm/s. Dividing this by the impact time of 0.1 seconds, we get an average force of 37.5 N.

4. Is average force affected by the duration of the impact time?

Yes, the average force is directly proportional to the impact time. This means that the longer the impact time, the smaller the average force will be. Conversely, a shorter impact time will result in a greater average force.

5. How is calculating average force by finding impulse over impact time useful in science?

This concept is useful in studying the effects of collisions and impacts, such as in sports or car accidents. It can also be used in engineering to design materials and structures that can withstand certain amounts of force during impacts. Additionally, it is a fundamental principle in the study of mechanics and motion.

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