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masraum masraum is online now
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Join Date: Oct 2001
Location: Central TX west of Houston
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I'll use caps for mass and Velocity, so the subscripts, i and f for initial and final and 1 and 2 for the two objects make more sense.

M1V1i + M2V2i = M1V1f + M2V2f
and
1/2 M1V1i^2 + 1/2 M2V2i^2 = 1/2 M1V1f^2 + 1/2 M2V2f^2

What it all comes down to is
V1i + V1f = V2i + V2f or
V1i - V2i = -(V1f - V2f)

excerpt from my college physics book concerning elastic collisions.
Physics for Scientists & Engineers 2nd ed, Serway
Quote:
Let us consider some special cases: If m1 = m2, then we see that v1f = v2i and v2f = v1i. That is, the particles exchange velocities if they have equal masses. This is what one observes in billiard ball collisions.

...

If m1 is very large compared with m2, we see from Equations 9.22 and 9.23 that v1f~=v1i and v2f~=v2i. That is, when a very heavy particle collides with a very light on initially at rest, the heavy particle continues its motion unaltered after the collision, while the light particle rebounds with a velocity equal to about twice the initial velocity of the heavy particle.
Eq 9.22 says that V1f equals the difference of the masses divided by the sum of the masses times the initial velocity of that particle.

so V1f = 5/15 * 10 or 3.333mph

Eq 9.23 says that V2f equals double the mass of particle 1 divided by the sum of the masses times particle 1's initial velocity.

so V2f = 20/15 * 10 or 13.333mph

Based on the math in the first two equations, I get different answers. I'm guessing that the difference between 10# and 5# is not enough to use the second set of equations.

Here are the results with the first set of equations. The one thing that seems odd to me is the final speed of the second particle. I'd expect it to be higher.
10*10 + 5*0 = 10*V1f + 5*V2f
.5*10*10^2 + .5*5*0^2 = .5*10*V1f^2 + .5*5*V2f^2

so
V1f = 10 - .5*V2f -- substituting that into the second equation we get

.5*10*100 + 0 = .5*10*(10 - .5*V2f)^2 + .5*5*V2f^2

when I solve, I get that V2f = 0 or 1.333.... since the weights aren't equal, it's not 0, then plugging 1.333 back into the first equation I get V1f = 9.3333

So part of that seems correct. The speed of the first ball hasn't changed by much since the second ball is lighter. So it only slows from 10 -->9.333..., but I'd expect the second ball to be going faster. So I'm not sure what, if anything I did wrong.
so, after, the 10 ball would be moving at 9.333 and the 5 would be going 1.333

Maybe I got the answers backwards. Maybe the masses are close enough that the first ball almost stops and the second ball is going almost as fast as the first.
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Old 11-29-2007, 10:56 PM
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