# My new bowling balls

1. May 12, 2006

### haxmya

I think this is a fairly basic question, but I have been out of physics since high school. Let's assume that I have two identical bowling balls, except that one is 15 lbs and the other is 16 lbs. If we avoid discussion about ball rotation and friction on the lane, how much additional speed does the 15 lbs ball need to match the energy (?) of the 16 pounder.

Possibly more at the heart of my confusion, is whether or not I'm dealing with Force which equals mass times acceleration if I remember right OR transmission of energy. Force seemed right to me to begin with, until I realized that the ball itself doesn't have positive acceleration at the time. So it seems that it may be how much energy the ball has to impart?

Do you guys think I'm on the right track there? Do you have any clarification on the two?

Thanks Folks! I sure do love bowling

2. May 12, 2006

### Staff: Mentor

First cut at an answer, $$KE = \frac{1}{2}m v^2$$

The ball has a lot of rotational energy also, but the initial impact on the pins will mostly involve the straight-line KE being transferred to the pins, I would think.

3. May 12, 2006

### haxmya

Thanks Berkeman. So kinetic energy is the key. I'll have to try and do some calculations and see how that turns out.

4. May 13, 2006

### matthew baird

I did some for you....using berkeman's formula,^ see above ^

KE=1/2(6.804kg)(8.9408m/s)^2
KE=271.944

6.804kg=15lbs 8.9408m/s=20mph

Now using that same KE found from throwing a 15lb bowling ball that reaches a velocity of 20mph, we will see at what velocity the 16 pounder will reach the same KE

271.944=1/2(7.257kg)(V)^2
135.972=7.257kg(V)^2
V^2=18.735
V=4.33m/s

7.257kg=16lbs

Okay, if this is the correct formula to use there appears to be a huge difference between the kinetic energy levels of each ball. The 16lb ball has the same energy that the 15lb ball has at HALF the speed!
Is this right? Looks fishy to me haha

Last edited: May 13, 2006
5. May 13, 2006

### Kurdt

Staff Emeritus
It would actually be 8.657m/s as you have divided by two initially instead of multiplying the kinetic energy by two. Just a simple algebraic error there. If you take the ratio of the two speed they really are negligible. ratio ofthe speed of the 15ib to 16lb is ~1.03 so really theres not too much difference. In laymans terms you need to bowl the 15 pounder 0.03 times faster than the 16 pounder to achieve the same translational kinetic energy. Hope this helps.

6. May 13, 2006

### KingNothing

I see it as a simple problem of momentum.

Momentum = mass * velocity

In other words, the 15lb ball will have to go 16/15 (sixteen fifteenths) the speed of the 16lb ball.

7. May 15, 2006

### BobG

Yes, the momentum is the key. Both momentum and energy will conserved, but some of the energy will have a different form other than kinetic energy - for example, some of the energy will create sound.

The difference in momentum will determine which ball has more effect on the movement of the pins.

Of course, the momentum the ball has when released depends primarily on how high the backswing is. If a person can't bring a 16 pound ball up as high on the backswing as they can a 14 pound ball or 12 pound ball, the loss of velocity on the release will more than negate the extra mass (actually, I think having enough wrist strength to hold the ball at the proper angle through the whole swing is usually more of an issue than being able to bring the ball back far enough on the backswing).

8. May 15, 2006

### matthew baird

Hahaha I always make those simple algebraic mistakes..like missing one + or - sign on a calculus test question!! I knew the answer looked funky..I read over my work again and realized what I had done, at that time I was unable to find this thread again to fix the error, until now.

9. May 16, 2006

### haxmya

Great discussion guys. Thanks all!

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