Elastic collision, conceptual

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 cancelled 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!

 

Well, there are two different terms in the quadratic. How can you solve for two terms?

 

You lost me. I assume you mean:

3v1^2 - 4vv1 + v^2 = 0

[-b +/- sqrt(b^2 - 4ac)]/2a = v1

[4v +/- sqrt([-4v]^2 - 4(3v)(v^2))]/2(3v)

[4v +/- sqrt(16v^2 - 12v^3)]/6v

But now I'm left with

(2/3)v +/- [sqrt(16v^2 - 12v^3)/6v]

I can't combine terms in the square root because they have different exponents, so this doesn't seem like it makes much sense.

 

all the info is given. i derived v1 and v2 from conservation of momentum. i have to do each individually, so really there are only two unknowns with two equations.

i see what you mean, hage, about the extra v's i put in. ill try that.

 

that works. i got (1/3)v = v1. Thanks everyone for all of your help