- #1
ph123
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Let two particles collide. Particle 1 has initial velocity v, directed to the right, and particle 2 is initially stationary. Now assume that the mass of particle 1 is 2m, while the mass of particle 2 remains m. If the collision is elastic, what are the final velocities v1 and v2 of particles 1 and 2?
Express answer in terms of the initial velocity, v.
My conservation of energy equation for the two particles is (I already canceled out the masses):
v1^1 + 0.5v2^2 = v^2
My conservation of momentum equation is (masses already cancelled):
2v1 + v2 = 2v
Thus,
v2 = 2v-2v1
v1 = (2v-v2)/2
I tried plugging these values for v1 and v2 into the conservation of energy formula to try and isolate v for each variable. For v1, I ran into difficulty because of the middle term of the binomial equation, -8vv1.
v1^1 + 0.5v2^2 = v^2
v1^2 + 0.5[(2v-2v1)(2v-2v1)] = v^2
v1^2 + 0.5[4v^2 - 8vv1 + 4v1^2] = v^2
-v^2 = v1^2 - 4vv1 + 2v1^2
3v1^2 - 4vv1 = -v^2
From here I spent like 10 lines trying to rearrange terms to separate v from that middle term (now -4vv1), but am unable to do so. Any ideas? Thanks!
Express answer in terms of the initial velocity, v.
My conservation of energy equation for the two particles is (I already canceled out the masses):
v1^1 + 0.5v2^2 = v^2
My conservation of momentum equation is (masses already cancelled):
2v1 + v2 = 2v
Thus,
v2 = 2v-2v1
v1 = (2v-v2)/2
I tried plugging these values for v1 and v2 into the conservation of energy formula to try and isolate v for each variable. For v1, I ran into difficulty because of the middle term of the binomial equation, -8vv1.
v1^1 + 0.5v2^2 = v^2
v1^2 + 0.5[(2v-2v1)(2v-2v1)] = v^2
v1^2 + 0.5[4v^2 - 8vv1 + 4v1^2] = v^2
-v^2 = v1^2 - 4vv1 + 2v1^2
3v1^2 - 4vv1 = -v^2
From here I spent like 10 lines trying to rearrange terms to separate v from that middle term (now -4vv1), but am unable to do so. Any ideas? Thanks!