# MV or 1/2MV^2

1. Jun 27, 2008

### mtworkowski@o

My experience is that because a lighter bowling ball can be thrown faster, the force that the ball has is altered. Should I use MV or 1/2MV^2?

2. Jun 27, 2008

### Defennder

Neither. Force is not described by 1/2mv^2 or mv.

3. Jun 27, 2008

### nicksauce

Can you maybe clarify your question? Neither MV nor 1/2MV^2 represents a force in any way.

4. Jun 27, 2008

### razored

MV = P which is Momentum.
Ke = 1/2 mv^2 which is Kinetic Energy which can be exchanged for Potential energy.

Force is typically defined as , F=ma.

5. Jun 27, 2008

### mtworkowski@o

Yes. In the context of trying to deliver energy to the pins, I'm thinking M and V are trade offs as far as energy is concerned. A heavier ball at slower speed or a somewhat lighter ball at higher V.

6. Jun 27, 2008

### pmb_phy

The correct definition of force is F = dp/dt.

Pete

7. Jun 28, 2008

### xArcherx

Just though I would throw in the more accurate formula for momentum (or so it seems)...

p = ɣmv
ɣ = (1-v^2/c^2)^(-1/2)

8. Jun 28, 2008

### mtworkowski@o

I'm getting the dp/dt thing from calc 1 97 years ago. thank you Pete. But I missed the thing from xArcherX. Sorry. I'm rusty as hell. Break it down, as they say, please. thanks.

Last edited: Jun 28, 2008
9. Jun 28, 2008

### Gear300

xArcherx's perspective is essentially the same as what you've been learning, but in general form. In classical physics, we mostly deal with what we can experience, which is one of the things that makes modern physics different (it tells us we've only been seeing part of the picture, and that phenomenon start acting differently over different scales).

If you've noticed, there is a c in the equations, which is the speed of light. Compared to v (the velocity of the bowling ball), its probably much bigger, so the formula for momentum can more or less be regarded as p = mv for this situation with the bowling ball.

10. Jun 28, 2008

Staff Emeritus
Why would you do that? It adds a totally unnecessary complication for no good reason other than to make you look smart. (And I'm afraid adding totally unnecessary complications doesn't make one look very smart)

A bowling ball might travel at 20 mph. That means we are talking about a correction in the 15th decimal place. Not only is this totally unmeasurable, it's tiny compared to the correction to the mass of the bowling ball as it travels and picks up and drops off various bits of this and that. So not only is it an unmeasurable correction, it's not even the biggest unmeasurable correction.

11. Jun 28, 2008

### mtworkowski@o

But when looking at a trade off between speed and weight, we use MV and not 1/2MV^2?

12. Jun 28, 2008

### cristo

Staff Emeritus
What are you actually trying to calculate? It would be a lot easier if we had a well-defined problem.

Note that the two things you have formulae for are momentum and kinetic energy, respectively.

13. Jun 28, 2008

### mtworkowski@o

which is better, the 16lb ball or a 15. More detail, if I or anybody can throw a 16 lb ball at the pins and deliver a certain amount of energy(let's say thats as fast as he can comfortably throw the ball) than would a lighter ball ( assuming he can throw it a wee bit faster) deliver more energy. I don't know if its momentum, kinetic energy or if either one will work.

Last edited: Jun 28, 2008
14. Jun 28, 2008

### D H

Staff Emeritus
Firstly, it is both conservation of momentum and conservation of energy that dictate what happens. I good starting point is to assume perfectly elastic collisions. Together, conservation of momentum and conservation of energy tell what happens.

Secondly, your physiology rather than physics is the determining factor here. You are missing a key factor, which is throwing a ball accurately. Suppose that you can throw a 16 lb ball fast enough to get more pin action than you can get with a 15 lb ball. That does not necessarily mean you should prefer the heavier ball. If the heavier ball makes you less accurate you might want to use the lighter ball.

15. Jun 28, 2008

### mtworkowski@o

Thanks DH..........I did better with the 15. But this is more of an academic question I guess. Compare the energy delivery of both balls when an equal force is used in each case. Maybe that says it better. Sorry.

16. Jun 28, 2008

### xArcherx

Yeah it is pretty much unnecessary at speeds that are not relativistic. Although, personally, I like to work with as close to exacts as I can. I also like to try and look smart , even if it does backfire on me (something that happens way too often):tongue:

17. Jun 29, 2008

### mtworkowski@o

to all. you know my question is perfectly clear. If you can't answer it than i'm out. you can just debate it for the next thousand years, which i'm sure you will. By The idea of using relativistic physics on a bowling bowl. Ha Ha Hal. What a joke.

18. Jun 29, 2008

### gamesguru

If we assume that you want to accelerate each bowling ball to the same speed, because you only have one swing to do it and want the same final speed, then you can assume,
$$F\propto m$$.
Here's why.
$$\frac{1}{2}mv^2=W=\int_0^d ma dx=ma\int_0^d dx=mad$$
So...
$$\frac{1}{2}mv^2=Fd$$
and finally,
$$F=\frac{1}{2d}mv^2$$
with d and v constants,
$$F\propto m$$.

19. Jun 29, 2008

### Janus

Staff Emeritus
Assuming that you use the same length swing, and are able to exert the same force throughout the swing, then the energy delivered to the ball is found by

$$E= fd$$

IOW, the ball will deliver the same energy regardless of its mass.

20. Jun 30, 2008

### mtworkowski@o

So are you saying it's a one to one trade off between mass and velocity? So I don't use 1/2MV^2. That was my original question, I think.

21. Jun 30, 2008

### mtworkowski@o

You know, I'm reading your remarks again and I think adding swing length is not a good thing to do. I'm iterested in the energy that can be delivered to the pins. Now that is either kinetic energy or momentum or both that we're talking about. In one formula V is squared in the other it's not. I though if you trade mass for V and V is squared, than you have an advantage in using a lighter ball and throwing it faster with the same force applied from your end. I hope that makes the question clearer. Sorry.

22. Jun 30, 2008

### mtworkowski@o

PS. E=fd..............? I remember W=fd. I'm a bit confused by this E thingy.

23. Jun 30, 2008

### Janus

Staff Emeritus
Swing length determines how much energy you impart to the ball, ergo how much energy the ball has to impart to the pin.

So, assuming that you use the same swing to roll the lighter ball as you do the heavier ball, and assuming that you can apply the same amount force throughout the swing, the smaller ball will have a greater velocity when you release it, But it will have the same energy as the heavier ball would.

For instance, you can roll a 12lb ball faster than a 16 lb ball, but not $1 \frac{1}{3}$ times faster but only $\sqrt{1 \frac{1}{3}}$ times faster. As far as the energy of the ball goes, the gain of velocities effect is matched by the reduced masses effect.

BTW E stands for energy, which is the same as work, which the W stands for.
IOW E=W

24. Jun 30, 2008

### mtworkowski@o

Could anyone tell me if MV is the formula or is it 1/2MV^2. Read my original question. This is a joke. Now I know whyu the Japanese are ahead of us. Answer my freakin' question please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

25. Jun 30, 2008

### D H

Staff Emeritus
You have to use both momentum and energy, just like I said in post #14.