In a physics lab, a cube slides down a frictionless incline as shown in the figure , and elastically strikes another cube at the bottom that is only 1/6 its mass. If the incline is 30 cm high and the table is 90 cm off the floor, where does each cube land?
mgh = 1/2mu^2
mu1 + mu2 = mv1 + mv2, or...
u1 = v1 + 1/6v2
The Attempt at a Solution
I'm having some difficult with this problem.
I first figured out the velocity, u1, of the first block using mgh=.5mu^2. I ended up with 2.42, which should be the speed at which the second block is hit.
Conservation of momentum:
m1u1 + (1/6)m2u2 = m1v1 + (1/6)m2v2
Masses cancel out, and the (1/6)m2u2 is zero, since the second block is at rest at the moment of impact from the first block. So...
u1 = v1 + (1/6)v2. Since u1 = 2.42...
2.42 = v1 + (1/6v2), or...
14.52 = 6v1 + v2
Which SHOULD mean...
v1 = 2.42 - (1/6)v2 and
v2 = 14.52-6v1.
That's where I stop. Since the definition of each velocity references itself, then I end up getting v1 = v1, and v2 = v2. That's obviously not helpful. What's the next step?