Golf club 🏌️and ball ⛳ impulse problem

[Clueless_2]

Homework Statement

A golfer strikes a 50 gram golf ball with a 225 gram clubhead travelling at 30 m/s. After impact the clubhead is travelling at 18 m/s. (iii) What is the initial velocity of the golf ball?

Relevant Equations

F dt = dp (p = momentum, d = change / difference)
p = mv
total p initial = total p final

I have done the first two parts of this question.

(i) Find the clubhead's change of momentum. ( dp = p_f - p_i = (0.225*18 - 0.225*30) = -2.7 kg ms^-1 )
(ii) Find the impulse given to the golf ball. (F dt = dp = 2.7 Ns from part (i) with just a change in the units. Because of the conservation of momentum, the reduction of the clubhead's momentum comes with an increase in the ball's momentum)
(iii) What is the initial velocity of the golf ball?

At first appearances, there seems to be two unknowns and only one equation. The velocity of the golf-ball, initial and final, are not specified explicitly. The only equation is the total initial momentum equals the final momentum. So I made two different assumptions and checked the logic and the answers.

Assumption 1: The golf ball is initially stationary. But this seems to makes this question trivial, since no work was needed to get to the answer. The answers (no worked solutions) begged to differ as well, stating the initial velocity was 54 ms^-1.

Assumption 2: The golf ball's final velocity is the same as the clubhead, since the question did not say they came apart after the impact. Then ...

(initial p of clubhead) + (initial p of ball) = (final p of clubhead) + (final p of ball)
0.225*30 + 0.05*v_i,b = 0.225*18+0.05*18 (where v_i,b is the initial speed of the golf ball)
6.75+ 0.05*v_i,b = 4.05+0.9
0.05*v_i,b = 4.95-6.75
v_i,b = -1.8 / 0.05
v_i,b = -36 m/s

So the initial velocity is -36 m/s, which means the golf ball was initially going in the opposite direction the clubhead was travelling. This seems unlikely, and once again it doesn't match the back-of-the-book answer of 54 ms^-1.

So i am veritably stuck ... for now.

lingering thoughts: I am wondering whether there is another equation that I missed? Because then I can solve for the two unknowns without making assumptions about the initial and final velocity of the golf ball.

Credits: Andrew Baylis & Gary McPhee et. al. 1994, Physics Study Guide Units 3 & 4, Longman Chesire Pty Ltd

 

[Nugatory]

By “initial velocity” they mean the speed of the ball immediately after it the collision with the club is over and the ball and club separate, and that the ball is at rest before it is hit - this is consistent with the 54 m/sec answer.

That is, the thing that you are calling v_i,b is zero and you are being asked to calculate the post-collision velocity.

 

[Steve4Physics]

Hi @Clueless_2. Welcome to PF.

I agree with what @Nugatory said. The wording of the question is poor. If it helps, I believe the intended question is this:

A golfer strikes a stationary 50 gram golf ball with a 225 gram clubhead.
At the start of the impact the clubhead is travelling at 30 m/s.
At the end of the impact the clubhead is travelling at 18 m/s (in the same direction).
(iii) What is the velocity of the golf ball at the end of the impact?
________

Note. The clubhead’s velocity is 18m/s at the end of the impact. That does not mean the ball’s velocity is also 18m/s.

Imagine you are a (miniaturised!) observer attached to the clubhead. What you would see is:
a) the ball coming towards the clubhead;
b) an impact which lasts some short time;
c) the ball bouncing off the clubhead (i.e. moving away from you).

In step c), even though you are moving at 18m/s, the ball is moving faster.

Have another go, bearing all that in mind!

 

[Lnewqban]

Welcome, @Clueless_2 !

Momentum has two factor that define how much of it a rigid body has.

Initially, the ball had mass but zero velocity respect to the ground; therefore, zero momentum.

Some of the momentum that the club initially had was transferred to the ball; hence, since its mass remained the same, its velocity had to be reduced.

Finally after impact and transfer of momentum, the ball, which mass did not change either, had some momentum to be measured by its increased...

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html#c1

 

[kuruman]

I think @Clueless_2 understands momentum conservation in collisions as evidenced by the analysis under Assumptions 1 and 2 in post #1. OP's problem was interpreting "initial velocity" which was introduced rather sloppily by the author of the problem. It is "initial" in the context of the ball's parabolic trajectory after the collision is complete. I am sure OP will have no difficulty getting the 54 m/s answer once that point is clarified.

 

[Clueless_2]

Thanks Nugatory!

I have bolded the edits based on what I think you wrote.

(prior p of clubhead) + (prior p of ball) = (following p of clubhead) + (following p of ball)
0.225*30 + 0.05*0 = 0.225*18+0.05*v_f,b (where v_f,b is the following speed of the golf ball)
6.75 = 4.05+0.05*v_f,b
2.7 = 0.05*v_f,b
v_f,b = 54 m/s

The golf ball’s initial velocity after the collision with the clubhead is 54 m/s.

Thanks for essentially rewriting the question for better clarity!

 

[Clueless_2]

Hi @Clueless_2. Welcome to PF.

I agree with what @Nugatory said. The wording of the question is poor. If it helps, I believe the intended question is this:

A golfer strikes a stationary 50 gram golf ball with a 225 gram clubhead.
At the start of the impact the clubhead is travelling at 30 m/s.
At the end of the impact the clubhead is travelling at 18 m/s (in the same direction).
(iii) What is the velocity of the golf ball at the end of the impact?
________

Note. The clubhead’s velocity is 18m/s at the end of the impact. That does not mean the ball’s velocity is also 18m/s.

Imagine you are a (miniaturised!) observer attached to the clubhead. What you would see is:
a) the ball coming towards the clubhead;
b) an impact which lasts some short time;
c) the ball bouncing off the clubhead (i.e. moving away from you).

In step c), even though you are moving at 18m/s, the ball is moving faster.

Have another go, bearing all that in mind!

Thanks @Steve4Physics ! Thanks for breaking the question down and rewriting it for me ! I see that the ball would end up with lower velocity to the clubhead's 18 m/s, especially after following through with Nugatory's suggestion and getting the "correct" outcome.

I do have another question. You laid the problem's scenario out in an a), b), and c) order. In step b), is the ball moving at 18 m/s like the clubhead? I suspect there may be subtleties with the contact aspect, but - if I may - let's assume - hyper unrealistically - that they are two blocks coming in contact with each other.

 

[Clueless_2]

Welcome, @Clueless_2 !

Momentum has two factor that define how much of it a rigid body has.

Initially, the ball had mass but zero velocity respect to the ground; therefore, zero momentum.

Some of the momentum that the club initially had was transferred to the ball; hence, since its mass remained the same, its velocity had to be reduced.

Finally after impact and transfer of momentum, the ball, which mass did not change either, had some momentum to be measured by its increased...

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html#c1

Thanks for the welcome and providing quite a detailed description! Helped me to reinforce in my mind what is going on in this problem.

 

[Clueless_2]

I think @Clueless_2 understands momentum conservation in collisions as evidenced by the analysis under Assumptions 1 and 2 in post #1. OP's problem was interpreting "initial velocity" which was introduced rather sloppily by the author of the problem. It is "initial" in the context of the ball's parabolic trajectory after the collision is complete. I am sure OP will have no difficulty getting the 54 m/s answer once that point is clarified.

@kuruman Thanks for the reassurance! And it turned out as you expected to my delight :). Your quote "Doing physics without math is like cooking food without heat." made me grin ear to ear and chuckle.

 

[erobz]

I do have another question. You laid the problem's scenario out in an a), b), and c) order. In step b), is the ball moving at 18 m/s like the clubhead? I suspect there may be subtleties with the contact aspect, but - if I may - let's assume - hyper unrealistically - that they are two blocks coming in contact with each other.

Search "Golf Impacts - Slow Motion"

You can watch some videos of the impact and think about what is happening to the club and ball during the brief moments of contact. There is a miniscule gap in the analysis. The idea is we look at just before contact and just after contact to infer what happened on average in terms of forces, accelerations, etc... What's precisely happening during the contact is more involved if you watch closely.

 

[haruspex]

@kuruman Thanks for the reassurance! And it turned out as you expected to my delight :). Your quote "Doing physics without math is like cooking food without heat." made me grin ear to ear and chuckle.

Ceviche anyone?

 

[haruspex]

In step b), is the ball moving at 18 m/s like the clubhead?

There will be a fair bit of elasticity. In the first phase of contact, the two mass centres are still coming closer together, the clubhead slowing, the ball accelerating, and the compression of the ball increasing.
At maximum compression they are moving at the same speed, which you can calculate using momentum conservation.
Thereafter, the mass centres move apart, the clubhead still slowing, the ball still accelerating, and the compression decreasing.
When the compression returns to zero the bodies separate.

 

[Steve4Physics]

Thanks @Steve4Physics ! Thanks for breaking the question down and rewriting it for me ! I see that the ball would end up with lower velocity to the clubhead's 18 m/s,

Do you really mean 'lower'?

By the way, for part 2 of the original question (in Post #1), you found the impulse applied to the (initially stationary) golf ball was 2.7Ns.

Can you use that value to find the golf ball’s final velocity? (I'd guess that's the method the author of the question intended.)

 

[kuruman]

Ceviche anyone?

Exactly my point. Although it can be done, one has to consider the other side of the coin. Imagine a diet consisting entirely of ceviche, gazpacho, steak tartare, etc.

 

[haruspex]

Exactly my point. Although it can be done, one has to consider the other side of the coin. Imagine a diet consisting entirely of ceviche, gazpacho, steak tartare, etc.

I would not consider gazpacho or steak tartare cooked. Ceviche, in a sense, is.

 

[kuruman]

I would not consider gazpacho or steak tartare cooked. Ceviche, in a sense, is.

In a sense. I would call that cured. I think I will stop posting about cooking on this thread and send you a PM.

 

[Clueless_2]

In a sense. I would call that cured. I think I will stop posting about cooking on this thread and send you a PM.

Was reading this in the middle of my lunch. Will go back to my lunch before responding to all my fantastic respondees!

 

[Clueless_2]

Do you really mean 'lower'?

By the way, for part 2 of the original question (in Post #1), you found the impulse applied to the (initially stationary) golf ball was 2.7Ns.

Can you use that value to find the golf ball’s final velocity? (I'd guess that's the method the author of the question intended.)

@Steve4Physics No, I didn't mean lower! Thanks for catching that.

You say "I found" but it seemed a trivial step to go from the change in momentum to the impulse, since they are clearly related in the equation (impulse = Fdt = dp).

Here's a shot at finding the value with the impulse as is (which shares much with post #6 in this thread).

F dt = dp
2.7 = p_f - p_i
2.7 = m_b*v_f - m_b*v_i = (0.05 kg)*v_f - 0 (I have rewritten what the problem calls "initial velocity" (v_i) with "final velocity" (v_f) for clarity : that is, the ball before being hit has an initial velocity of zero () , and the ball's velocity after being hit is the final velocity)
v_f = 2.7/0.05 = 54 ms^-1

Makes for a shorter answer, as well as shows the continuity in the problem. Cheers!

 

[Clueless_2]

There will be a fair bit of elasticity. In the first phase of contact, the two mass centres are still coming closer together, the clubhead slowing, the ball accelerating, and the compression of the ball increasing.
At maximum compression they are moving at the same speed, which you can calculate using momentum conservation.
Thereafter, the mass centres move apart, the clubhead still slowing, the ball still accelerating, and the compression decreasing.
When the compression returns to zero the bodies separate.

After watching the video recommended by @erobz, I could how your words complement the pictures. The golf ball compressing into a weird shape and decompressing into another shape is a sight to see!

 

[hutchphd]

In a sense. I would call that cured. I think I will stop posting about cooking on this thread and send you a PM.

Denatured perhaps. A bad term in a physics discussion I fear.