Analysis of collisions from data

AI Thread Summary
The discussion focuses on analyzing collisions involving two balls with specified masses and initial speeds. The user calculated the momentum and kinetic energy for each mass using the formulas p = m*v and KE = 1/2mv². For mass A, the momentum is approximately 0.9511 kg m/s and kinetic energy is about 0.8179 J, while for mass B, the momentum is -1.5186 kg m/s and kinetic energy is approximately 2.8125 J. The total momentum of the system is -0.569 kg m/s, and the total kinetic energy is 3.6304 J. The calculations appear correct based on the provided data and equations.
xjcov
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Good afternoon! I would like to preface by saying, yes, this is for a project. I am only posting here to see if my method of solving is correct before I finish the project incorrectly.

Homework Statement


I chose two balls, mass A: .553 kg and mass B: .410 kg
I recorded their collision and plotted the points on Tracker: http://physlets.org/tracker/
Through the analysis of the horizontal component of mass A (before the collision), I have determined the speed is 1.72m/s. On the horizontal component of mass B (before the collision), the speed is -3.704m/s. I have to find the momentum (kg m/s) and kinetic energy (J) of each collision and of the total system. Is my attempt correct? If not, can someone point me in the right direction?

Homework Equations


momentum: p = m*v
kinetic energy: 1/2mv2

The Attempt at a Solution


mass A momentum: (.553)(1.72) = 0.9511 kg m/s
mass A kinetic energy: 1/2(.553)(1.722) = 0.8179 J

mass B momentum: (.410)(-3.704) = -1.5186 kg m/s
mass B kinetic energy: 1/2(.410)(-3.7042) = 2.8125 J

For the total system I just added the values.
momentum: .9511 + -1.5186 = -.569
kinetic energy: .8179 J + 2.8125 J = 3.6304 J
 
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That is fine so far.
 
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mfb said:
That is fine so far.
Awesome! Thanks for the help!
 
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