Let’s Play Pool!

Pool balls at KMC in the Big Island.

NOTE:  To keep things simple, I neglected air resistance and friction in my discussion.

When Jordan and I were playing pool at Senior Camp this year, she said to me, “You should use this for your next physics blog! The collisions between the pool balls are nearly elastic.” At the time, I had no idea what an elastic collision was so I responded to her with a simple “Yeah, that’s a great idea!” Now, with a solid background in momentum and collisions, I would have whipped out my Casio Fx-CG10 as soon as Jordan reminded me about elastic collisions and I would have calculated the final velocities of the balls right there on the spot!

In pool, as Jordan pointed out, the collisions between the balls are nearly elastic. That is, both momentum AND kinetic energy are conserved. In pool, you start with two balls: your attack ball and your target ball. Once you set the attack ball into motion, one of your balls is moving while the other remains stationary. Soon enough, your attack ball hits your target ball and both balls scatter in different directions at different velocities! How exciting, right?!

A picture of me getting excited about elastic collisions at Senior Camp while Jordan watches me fail to shoot the pool ball in the pocket.

To find the new magnitude and direction of the balls’ velocities, we would need their angles of separation, which we could then use to componentize both balls in to their respective x- and y-velocity components. Assuming we have all of the necessary information and assuming that we are able to find the resultant vector of each ball’s velocities, we can observe conservation of momentum and kinetic energy in elastic collisions through the following relationships. Because both balls have the same mass and because ball 2 is initially stationary, the conservation of momentum equation (M1V1i + M2V2i = M1V1f + M2V2f) simplifies to V1i = V1f + V2f. Similarly, the conservation of kinetic energy equation (1/2M1V1^2i = 1/2M1V1^2f + 1/2M2V2^2f) simplifies to V1^2i = V1^2f + V2^2f.

Through the above relationships, we can see that both momentum and kinetic energy are conserved in pool (and all other examples of elastic collisions for that matter). Undoubtedly, Senior Camp would have been a much better experience if I had already known how to calculate the velocities of the pool balls. Oh well, at least I know what I’m going to do the next time I play pool.

I Can Be More Powerful Than You!

A photo of me sitting at the top of the Koko Head stairs hike.

NOTE: To keep things simple, I neglected air resistance and friction in all of my calculations and explanations.

Looking up from the bottom of the Koko Head stairs.

My dad and I often hike the Koko Head stairs during free time on the weekends. For the past few weekends, I have been able to complete the 805 meter long hike in about 15 minutes (V=0.89m/s) while my dad has been able to complete it in about 20 minutes (V=0.67m/s). At the top of the hike, our total work done can be calculated by W=KE+PE. Given our velocities, masses of 59kg and 82kg for me and my dad, respectively, and that the height of the mountain is 368m, our total work at the top is as follows:

My Work:

W=(0.5)(59kg)(0.89m/s)^2 + (59kg)(9.8m/s^2)(368m)

W=212800.967 J

My Dad’s Work:

W=(0.5)(82kg)(0.67m/s)^2 + (82kg)(9.8m/s^2)(368m)

W=295743.205 J

How embarrassing… My 55 year old dad does more work up a mountain than I do! Really though, this isn’t surprising. Since my dad weighs significantly more than I do, his work is going to be greater than mine at the top of the mountain because his PE will always be greater. KE, in this case, will not affect total work by much at all.

Surely, because my dad does the most work up the mountain, he has the most power too, right? Not necessarily. In this case, however, his power is actually greater than mine. Given our times of 15 minutes and 20 minutes for me and my dad, respectively, we can calculate power as P=Work/Time.

A view from the top of the Koko Head stairs.

My Power:

P=212800.967 J / 900s

P=236.45 W

My Dad’s Power:

P=295743.205 J / 1200s

P=246.45 W

As a cross country runner, I don’t know if I’m going to be able to let my dad beat me in terms of power for much longer. After all, I am only 10 watts behind him! If I were able to cut down my time the next few times I hike Koko Head stairs, I will probably be able to soon beat him in terms of power. Although his total work will be significantly larger than mine, I can, with a fast enough time, become more powerful than my dad. Who knew that hiking up Koko Head could become so technical and competitive?! Ah, the life of an athletic physics nerd.