Originally posted by devin
Intelligent Stuff
o.O
....
I know!
*
MajesticLight shuffles through papers to get the aforementioned physics lab report that is now complete.
Ok, here we go:
When two objects collide, they produce equal and opposite forces on each other (F1 on 2 = -F2 on 1, henceforth known as F1 and F2) as is stated by Newton’s third law of motion. Using Newton’s second law, F = ma, and from the definition of acceleration, a = dv/dt, it follows that F = m*dv/dt. So, F1 = m1*dv1/dt and -m2*dv2/dt = -F2. Multiplying through by dt and taking the integral of the first two quantities yields S F1*dt = S m1*dv1. Likewise, the second pair of quantities yields S -F2*dt = S -m2*dv2. Multiplying both sides of F1 = -F2 on 1 by dt and then integrating yields S F1*dt = S -F2. After making appropriate substitutions, the penultimate equation becomes m1v1 (evaluated between v1i and v1f) = m2v2 (evaluated between v2i and v2f). Once the quantities are evaluated and the terms sorted, the final equation becomes m1v1i + m2v2i = m1v1f + m2v2f. This equation is used for the conservation of momentum.
Darn integral signs didn't show up... hence the capital S instead.
The limit of integration on forces is time (from ti to tf). Since at t = ti v = vi and t = tf v = vf, substitutions can be made on the limits of integration of mass*change in velocity. The limit of integration on mass*velocity is from vi to vf.
Couldn't find my book for this one. The scratch sheet I derived it from had lots of, well, scratches. As in mistakes that had been scribbled out. But I managed to do it

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