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.

Why Can’t I Get Down?!

A photo of the Area 51 ride at the 50th State Fair.

As Jordan and I were walking through the 50th State Fair trying to find a ride to go on this past summer, she turned to me and said, “We should go on that Area 51 ride! It is completely physics based and it’s supposed to be really fun.” Curious as to how the ride worked, I asked Jordan for an explanation. Not having taken AP Physics B at the time, however, most of her explanation went over my head. Now that I have been enlightened by the beauty of circular motion, however, I can easily explain the phenomenon of the Area 51 ride.

Before the ride begins, everyone stands up against a circular wall with seemingly no safety measures like seat-belts or straps in place. At first a bit worried about how safe the ride was, Jordan assured me that the laws of physics behind circular motion would be the only safety measures needed to keep us safe.

A photo of me and Jordan inside the Area 51 ride. She is trying to assure me that, because of centripetal force, I am not going to die on the ride.

The ride starts to spin at a slow angular velocity but then increases its angular velocity as time goes on. Given that the distance of any person from the center of rotation to the wall (the radius) is constant, using the relationship v=ωr we see that the linear velocity increases as angular velocity increases. Furthermore, given that friction is directed upwards and weight is directed downward, we find that our resultant, centripetal force (Fc=mv^2/r) is directed inward toward the center of rotation. As the centripetal force, also the normal force, increases with an increase in linear velocity, friction increases as well because f=(Fn)(μs). Eventually, the ride is going so fast that friction is greater than weight directed downward, which makes the rider feel as if they are stuck to the wall unable to easily move or get down.

Not having any understanding of circular motion at the time, I was confused as to why I was stuck on the middle of a spinning wall unable to get down. Now, with a thorough understanding of circular motion, the fact that I can be stuck to a spinning wall is not surprising, it’s just physics! How phun!

I Swear, I Wasn’t Meaning To Speed!

A photo taken from my car while driving down the H1 freeway.

After dropping Jordan off at her house at the end of our day together, I returned home via the H1 freeway. On my return trip home, I was actively thinking and trying to decide what I was going to write my physics blog about this week. All of the sudden, because physics is all around us, an idea finally hit me (with quite a bit of force too!).

In an attempt to maintain a speed of exactly 50mph (the speed limit), I put my car on cruise control to ensure that I wouldn’t speed. The cruise control feature worked great while I was traveling on flat ground because it found and maintained a force that caused an acceleration of 0m/s^2, that is, a force that caused net force to be 0. On flat ground, my car maintained a constant velocity of 50mph the entire time. Once I reached the point of the freeway where my car was going down an incline, however, the cruise control feature failed me, as cruise control cannot control breaks.

While the car traveled down the incline, I asked myself, “Why, if the car is exerting the same force as it was before, does it now accelerate?” With simple knowledge of force and inclined planes, this question is easy to answer. Yes, the car exerted the same force down the incline as it exerted before the incline, but it now had the horizontal force of mgx (mgSintheta) pushing it forward as well. The horizontal force of mgx, in this case, caused net force to be greater than zero because the force of the car + mgx was greater than the force of friction that opposed the car’s motion. With net force greater than zero, the car must accelerate because Fnet = ma. Using this equation, we find that, with a constant mass, an increase in net force will cause an increase in acceleration because the two variables are directly related.

Analyzing the effects of inclined planes on net force and on my car’s motion during the drive home, I found, made the lonely car ride home much more phun.

P.S. Jordan says “hi” 🙂

How Does That Golfball Do It?!

One afternoon this past summer, Justin Park, Shane Hayakawa, Matt Callahan and I decided to go golfing at the Ala Wai Golf Course. As we made our way through the 18-hole course, I would carefully watch my golfball whiz through the air after each drive and would wonder why the ball formed a parabolic path in the air every time that I hit it. After each drive, the ball would continue to increase its vertical position in the air until it reached a peak height, when it would then fall back down to the ground in the same amount of time that it took the ball to reach its highest point. At the same time, it seemed as though the horizontal velocity of the ball was constant. Having already taken AP Physics B, Shane, Matt and Justin already knew the physics behind the golfball phenomenon, meanwhile I on the other hand was missing out on this knowledge. I now, however, educated in the topic of motion, can explain the reasoning behind the golfball’s motion using projectile motion.

A photo of me after pitching my golfball onto the green.

In golf, the golfer hits the ball with a certain initial velocity at a certain angle. Both the magnitude of the velocity and the angle that the golfer hits the ball are very important as they determine the initial x and y velocities, which can then, with time, help to determine how high and how far the golfball will go. The parabolic motion of a golfball is explained by all of these factors in that the ball must reach a peak height in its trajectory before gravity brings the ball back down to the ground. The initial y-velocity constantly decreases because of gravity while the horizontal velocity remains constant because there are no horizontal forces affecting the golfball.

It is interesting to note that the average professional golfer can hit a golfball at around a velocity of 100mph, wow! Knowing this, we can use vectors to find that, if the golfer strikes the ball at a 40 degree angle, the initial (and final because there are no horizontal forces acting on velocity) horizontal velocity is 100mphCos40 = 76.6mph and the initial vertical velocity is 100mphSin40 = 64.3mph. Now those are some fast traveling balls!

Golf, though often touted as a tedious and sometimes boring sport, is actually quite fun when you use physics to analyze the ball’s motion as you play.

Me and Justin riding on a golf-cart after our round of golf.

Cross Country – Run the Displacement!

This past weekend the boys and girls `Iolani Cross Country teams ran their second race of the season. This weekend’s race, the Kaiser Invitational, was held at Kaiser High School. Kaiser’s course consists of multiple loops around their campus and their fields and is either two miles for Intermediates or three miles for JV and Varsity runners. Although the total distance ran between an Intermediate runner and a JV/Varsity runner differs by about a mile, the total displacement is exactly the same: ~400 meters (the finish line is about 400 meters from the start line).

A photo of me and Ty Moriwaki posing at the starting line.

Upon completing the race, I thought to myself: I wonder what my average velocity and average acceleration was over the course of this race. Well, that’s easy! Your average velocity should simply be your distance divided by time, right? Wrong! In fact, average velocity is only your displacement divided by your time. Knowing this, I can now calculate my average velocity and average acceleration. Given a displacement of ~400 meters and a race time of 20:25 (1,225 seconds) for 3 miles, Vav = 400m / 1,225s: 0.327 m/s. Now that I have my average velocity, I can calculate my average acceleration to be average velocity divided by time, aav = 0.327m/s / 1,225s: 0.000267 m/s^2. Given that the average Intermediate runner can run 3 miles in 13:00 (780 seconds), it may seem puzzling that an Intermediate’s average velocity and average acceleration are higher (0.513 m/s and 0.000658 m/s^2, respectively) than those of a Varsity runner’s, but that’s just simple physics.

A photo of me running toward the finish line at the end of the Kaiser Invitational Varsity race.