Elastic Collision Between Two Masses

[derravaragh]

Homework Statement

Consider a 2-D elastic collision between two masses. The first mass is moving at initial speed v0 towards the second mass. The second mass is initially at rest. Mass m1 = 0.1 kg and mass m2 = 0.2 kg. The first mass recoils at 30° above the horizontal at speed v1, and the second mass recoils at θ degrees below the horizontal at speed v2, calculate v2 and θ. The horizontal is the direction parallel to the first mass's initial direction of motion.

Homework Equations

Momentum is constant Pi= Pf
Kinetic Energy is constant Ki = Kf

The Attempt at a Solution

I've worked through this already, but came to a road block. I set up the momentum equations for the x and y directions to solve for the angle θ, which is in terms of v1 and v0, and then squared these two equations and added, then used the equation for the conservation of the kinetic energy, and set both equal to (v2)^2, so I could solve for one of the angles, but I can't actually solve for either v0 or v1 because nothing cancels. I'm at:
(v1)^2 = (v0)^2 + 1.732v0v1.
Is this right, or did I make a mistake somewhere, because from this I can't seem to see how I can solve for v2 because I can't solve for v1 even if I had v0.

My other steps came to:
x-direction => v2cos(θ) = .5v0 - .433v1
y-direction => v2sin(θ) = .25v1
Kinetic Energy => (v2)^2 = .5(v0)^2 - .5(v1)^2

I don't want to have the solution, if you could just let me know whether I'm on the right track or if I'm wrong and where I made a mistake, it would be greatly appreciated.

 

[haruspex]

(v1)^2 = (v0)^2 + 1.732v0v1.
Is this right, or did I make a mistake somewhere, because from this I can't seem to see how I can solve for v2 because I can't solve for v1 even if I had v0.

If v0 is known then the above is a quadratic in one unknown. Just use the usual formula.

 

[derravaragh]

I just realized I gave the wrong set of equations, that one doesn't match the other three I got, but it's the same concept. I see my issue now, pretty simple. Thank you for your help.