Recoil velocity after collision

1. Mar 19, 2009

diffusion

1. The problem statement, all variables and given/known data
A particle of mass 1.00u travelling at speed v, collides with a stationary Cu nucleus of mass 62.93u, and rebounds in the exact opposite direction with a speed of .9687v. What is the recoil velocity of the Cu atom, in terms of v?

2. Relevant equations
F=ma
P=mv

3. The attempt at a solution
If the recoil of the proton is .9687v and the initial velocity was v, then in order to satisfy conservation of momentum, .0313u/v must be transfered to the Cu atom. To solve for velocity of the Cu atom, use p=mv; 0.0313 = 62.93 x v. Therefore v = 0.0005v.

Am I correct or just going in the completely wrong direction here?

2. Mar 19, 2009

sArGe99

Use conservation of linear momentum, you find that the total linear momentum before collision = v
Total linear momentum after collision remains the same and the sum equals v.
Hint : Account for the direction of the particle after collision

3. Mar 19, 2009

diffusion

I'm getting the same answer, but still feel like I'm on the wrong track.

Total linear momentum = v = 1.0(.9687) + 62.93(0.00050v).

4. Mar 19, 2009

sArGe99

What would be the direction of velocity of the particle after the collision?

5. Mar 19, 2009

sArGe99

Wouldn't it be 1.0(.9687 v)? Velocity is a vector, so we have to take into account the direction the body is moving in..

6. Mar 19, 2009

diffusion

Negative along x-axis for the proton. Positive for the Cu atom.

7. Mar 19, 2009

sArGe99

Yes.Correct.
So wouldn't one take the velocity of the proton to be negative?
Now, you can write the momentum conservation equation.

8. Mar 19, 2009

diffusion

So if the velocity and therefore momentum of the proton is -.9687, the momentum of the Cu particle would have to be... .9687 + .9687 + 0.0313 = 2.2504. Divide by 62.93 = velocity of 0.0357v.

I'm so far off.

9. Mar 19, 2009

sArGe99

The momentum conservation equation would be
v = -0.9687v + 62.93 u
Express u, the recoil velocity of Cu atom in terms of v. That is the way its found out, I believe.