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Snow Day Blog 4


On this beautiful snow day, (well, a few days after the storm), I decided to take my dog outside to play his favorite game... frisbee!! But first, I had to convince him to help me with my physics blog. To start, I instructed my dog, Teddy, to sit and stay in one spot in the yard. He was VERY eager to get his frisbee, so I had to work fast. I was kneeling a few feet away from him with the frisbee, stationary in my hand, extended outward. This way, the video could capture Teddy running to grab his frisbee (at initial velocity 0). Using Logger Pro, I was able to solve for Teddy's initial velocity, and with these variables I was able to solve for the final velocity once Teddy and the frisbee were moving as one object. This scenario is known as a perfectly inelastic collision because the two objects (dog and frisbee) are eventually moving together as one.




Known Variables:
m1 = dog =22.73kg
 *converting 50lb to kg: 1kg = 2.2lb*
m2 = frisbee = .1814kg 
v1 = 2.65m/s

Now, using these known variables, m1, m2, v1, I used the formula: m1v1 + m2v2 = (m1+m2)vf, to solve for the final velocity of the system (dog + frisbee together).

The answer, 14.82m/s. Seems very fast for a dog running with his frisbee, no? This outcome could have been this extreme for a few reasons. First of all, the video stopped recording too soon once the dog grabbed the frisbee from my hand. This could have resulted in a faulty reading from Logger Pro, due to the absence of a few extra data points.

To find the amount of kinetic energy lost in the equation, I used the conservation of kinetic energy formula: 1/2(m)(v)^2 - KE = 1/2(m)(v)^2

1/2(22.73)(2.65)^2 - KE = 1/2(22.73 + .1814)(14.82)^2
79.811 - KE = 2516.042
KE lost in collision = 2436.231 J

After torturing my dog enough with physics homework, he finally got to enjoy some frisbee time :)



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