Recoil velocity after collision

[diffusion]

Homework Statement

A particle of mass 1.00u traveling 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?

Homework Equations

F=ma
P=mv

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 transferred 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?

 

[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

 

[diffusion]

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

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).

 

[sArGe99]

Homework Statement

A particle of mass 1.00u traveling 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?

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

 

[sArGe99]

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).

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..

 

[diffusion]

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

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

 

[sArGe99]

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

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

 

[diffusion]

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

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.

 

[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.