# Elastic materials affecting impact

1. Jun 7, 2010

### Ketman

Remember those executive toys that were popular in the '80s? You have a steel ball on the end of a thin rod hitting another steel ball that is stationary. As near as we can see, the target ball gets all the momentum of the moving ball, and the moving ball stops dead - close to 100% transfer. But then you have a golfer hitting a ball that is not rigid like a steel ball, but is beefed up by elastic material, rubber or something, to give it a bit of spring. The thing is, if it helps the golf ball get more momentum, how is it that it can't help the steel ball? I mean, you can't get more than 100%, can you?

2. Jun 7, 2010

### blitz.km

Momentum is conserved only when the external force is zero.
the momentum increases when the golfer hits the ball.
hope i am right!

3. Jun 7, 2010

### Ketman

And what if the golfer withdraws the force before impact, leaving only his momentum?

4. Jun 8, 2010

### Ketman

No one has a solution?

5. Jun 8, 2010

### proculation

Like blitz.fm said, the momentum (energy) is equal from the 3rd law of Newton if there is no external forces.

If you release the force from your hands 0.0001 second before the hit, you will apply the conservation of momentum equation m1v1 = m2v2.

That means your golf rod will go back, stay still or reduce the speed with that same formula.

6. Jun 9, 2010

### Ketman

That equation might make sense where one object transfers all its momentum to another. But when that isn't the case (as it wouldn't be with the golf club and the ball), there are four velocities involved, not two - the initial and final velocities of the club, and initial and final velocities of the ball. However those cannot be worked out using conservation of momentum.

Both the answers I've had make no reference to the elasticity of the bodies. Newton's experimental law involving bodies of different elasticities gives us something called a "coefficient of restitution". My question relates to that.

7. Jun 16, 2010

### Ketman

All right, I'm drawing a blank here, so let me try it another way. If the golf ball is the same mass as a normal golf ball, but made of a very hard material, such as steel or ivory, instead of the springy compound we are familiar with, would it fly off the tee with a greater velocity or lesser?

8. Jun 16, 2010

### proculation

You were right.

With a totally elastic material, now you can use m1v1 = m2v2

9. Jun 16, 2010

### Ketman

Nope. There cannot be a 100% transfer of momentum unless the club comes to a complete halt after impact. So that equation doesn't apply. The question comes down to whether and how the composition of the golf ball increases or decreases the percentage of momentum transferred. But whatever the answer, the amount transferred cannot be 100%.

10. Jun 16, 2010

### dulrich

I think the correct answer has two pieces.

On the one hand the more elastic the ball, the farther it will fly upon impact. But, remember, the steel ball is more elastic than the golf ball. The golf ball suffers deformation and so loses kinetic energy due to heat. This is an inelastic collision. The steel ball will collide more elastically (less energy lost) than the golf ball.

On the other hand, this is not a collision -- your hands are providing an external force that continues to push through the entire collision (i.e., one will "follow-through" the swing). In this case, the momentum transfered to the ball is proportional to the impulse delivered by the club. Impulse is force multiplied by the duration the force is applied. Normally, the force from the club is delivered during the full time period of the impact (with proper "follow-through"). This works the opposite direction of the first consideration: extending the time of impact will increase the momentum transferred.

I suspect that on the scale from perfectly elastic (steel) to completely inelastic (mud ball), that the golf ball is pretty elastic, so the second consideration is more important.

11. Jun 16, 2010

### Sithdarth

What you are looking at is nearly an elastic collision so you need both of these equations:

$$m_{1}\ast v_{1i} + m_{2}\ast v_{2i} = m_{1}\ast v_{1f} + m_{2}\ast v_{2f}$$
$$\frac{1}{2}m_{1}\ast v_{1i}^{2} + \frac{1}{2}m_{2}\ast v_{2i}^{2} = \frac{1}{2}m_{1}\ast v_{1f}^{2} + \frac{1}{2}m_{2}\ast v_{2f}^{2}$$

Of course that is only an approximation. What you are looking at has to do with collisions that aren't perfectly elastic so we need the coefficient of restitution of the objects in the collision. For simplicities sake I'm going to assume the club head is perfectly elastic so the two equations you need are:

$$m_{1}\ast v_{1i} + m_{2}\ast v_{2i} = m_{1}\ast v_{1f} + m_{2}\ast v_{2f}$$

$$C_{R} = \frac{v_{2f}-v_{1f}}{v_{1i}-v_{2i}}$$ (making sure to use only magnitudes in this second equation and ignoring directions)

Using the two of those and the coefficients of restitution, or CR, for a golf ball and say a steel ball bearing of roughly the same mass as a golf ball you can work out which one would be faster off the head of the club. Of course it requires that the club be moving without an applied force when it contacts the balls. From what I can find steel ball bearings have a coefficient of restitution of about .597 and golf balls have one of about .858. Which means with equal masses for the balls and equal masses and speeds for the club heads hitting the balls the golf ball will leave the club head faster.

12. Jun 16, 2010

### dulrich

Interesting. Can you provide a link for your coefficients of restitution? I assumed the steel ball would have a higher COR than the golf ball.

Although, as I think about whether a golf ball or a ball bearing would bounce higher from a drop from equal heights you are probably right...

13. Jun 16, 2010

### Sithdarth

http://hypertextbook.com/facts/2006/restitution.shtml" [Broken] I figured with an almost one meter drop the forces involved would be well below the forces in a golf swing. Obviously the more force you apply to an object the closer you get to its elastic limit. Going beyond the elastic limit of a material substantially decreases the coefficient of restitution. A steel ball bearing might be more elastic than a golf ball before you reach the steel's elastic limit but the elastic limit of the golf ball is going to be much higher. Therefore in the realm of collisions with the force of say a golf club a golf ball should have a higher CR then a steel ball because it hasn't yet reached its elastic limit.

Last edited by a moderator: May 4, 2017
14. Jun 16, 2010

### Ketman

Thanks to dulrich and sithdarth. This is more like it. I think the source of my problem lay in the assumption I had made that steel had a very high coefficient of restitution, and that was why it was the chosen material for those aforementioned executive toys. It certainly seems like a nearly 100% transfer of momentum to the naked eye. And I'm surprised to find now that its COR is only 0.597. That is very low! But if a golf ball has a higher COR than a hard material like steel, that explains why they use it. But does it also mean that if they used the same materials in the executive toy it would be better?

Edit: Our posts have crossed. This new info about steel's COR changes things. I suppose the elastic limit you're talking about is the point where physical damage happens - i.e. the steel gets dented?

Last edited: Jun 16, 2010
15. Jun 16, 2010

### dulrich

Yeah, I think that's right. In fact, now that you mention it, I have wondered why those toys that I've played came to rest so quickly. If it were perfectly elastic, the toy should just bounce and bounce and bounce. But actually you get about a couple dozen bounces or so before it stops. I just never stopped to really think about it.

In fact, if someone made one of those executive toys from golf balls, it might sell too!

16. Jun 16, 2010

### Sithdarth

Yup and that generalization holds for all materials. The CR of any material will naturally change a little bit depending on the forces involved. By far the greatest change occurs when you pass the elastic limit. After that any deformation is essentially permanent. The steel works well for the toy because it operates below or very close to the elastic limit of the steel. A full speed golf swing even with the driving force removed at the last second should be well above the elastic limit of the steel ball. You'd end up with a dented ball and not nearly as much velocity as a golf ball which wouldn't dent. When choosing a material you generally have to take into account both its elasticity and its elastic limit.

17. Jun 17, 2010

### Ketman

Without wishing to blow more smoke over the question, wouldn't it be true to say that determining the velocities after a collision between two bodies of different materials would depend upon the lower COR of the two? And that in measuring CORs by experiment you'd need a surface with a COR close to unity? I ask that because I'm doubtful about the validity of those experimental results where they used a concrete surface to bounce balls of different materials. When it turns out that a wooden ball bounces higher than a steel one, I have to doubt the reliability of the surface. I'm not sure concrete itself has a very high COR, and possibly not a very high elastic limit. Where deformation of the surface is a possibility, I'd expect denser materials to have a more drastic effect in killing the bounce, and maybe that's why steel scored so low in the trials. What do you think?

18. Jun 17, 2010

### Sithdarth

If you really want to get down to it the elastic limit is also known as http://en.wikipedia.org/wiki/Yield_(engineering)" [Broken]and is measured in terms of pressure. For concrete its about 68 MPa (megapascals), for steel ball bearings it's probably about 500 MPa and based just on what golf balls are generally made of I wouldn't expect it to be more than about 20 MPa (probably less). The key thing here is that they are pressures. When a steel ball bearing hits something or is hit by something it doesn't deform very much. Which means in the case of a spherical piece of steel the area of contact stays quite small. This means a relatively small force can produce a relatively large pressure and push it past it's elastic limit. On the other hand a golf ball deforms quite a bit distributing the force of the impact over a much larger area lowering the pressure created from that force allowing it to stay below it's elastic limit.

There are a few possible conclusions. If the cement didn't deform when they dropped the steel ball bearing on it than neither object went past their elastic limits and therefore the COR of steel really has to be lower than the COR of a golf ball in the elastic regime. Since the COR of the concrete would have been constant over the whole experiment. If just the cement plastically deformed then it would be hard to say but if the steel ball bearing produced 68 MPa of pressure with such a relatively tame collision then chances are the golf club is going to push it past its elastic limit. If both the cement and the steel ball bearing plastically deformed then the steel is going to behave similarly when struck with the much greater force of the club and its COR is still going to be much lower than the golf ball. Mainly because we know the deformation of the golf ball allows it to stay below its elastic limit when hit by a club otherwise every golf ball would only be usable for a couple of good drives.

Of course the other side of all of this is that plastic is lighter and cheaper than steel. To make a steel ball as light as a golf ball it'd have to be very small or somewhat hollow. A hollow steel ball would plasticly deform much easier than a solid one and would be basically useless. A solid steel ball would be rather small to have the same weight as a golf ball. Add to that the fact that golf balls can withstand much larger impacts by deforming and reducing the pressure applied to them and it is clear why golf balls are made the way they are. Comparing materials is rarely as simple as looking at a single number.

Last edited by a moderator: May 4, 2017