Determine the change in direction of a particle

[MaestroBach]

Homework Statement

A particle of mass m is moving with velocity*~v*1. It leaves the half-space z <0 in which its potential energy is U1 and enters the half-space z >0 where its potential energy is U2. Determine the change in the direction of the motion of the particle. (Hint: The particle is free in the x and y directions because the potential energy is constant in the (x, y) plane.)

Relevant Equations

Relevant equations: Conservation of momentum, Conservation of energy

Note: I don't know if this actually qualifies as advanced physics, it probably doesn't. It's a review problem in a non-introductory class but I can't solve it so...

Beginning with the hint, I know that the x and y components of velocity don't change when the particle moves from z < 0 to z > 0 because the potential energy is constant in the x,y plane.

Therefore, using conservation of energy I wrote:

U1 + (1/2)mV1^2 = U2 + (1/2)mV2^2, and I substituted v1^2 with (Vx^2 + Vy^2 + Vzo^2) and v2^2 with (Vx^2 + Vy^2 + Vzf^2), where Vzo and Vzf are the initial and final components of velocity in the z direction. After doing some cancelling out, I get

U1 + (1/2)mVzo^2 = U2 + (1/2)mVzf^2

This is where I'm stuck, I don't know where to get another equation from.

(If my notation doesn't make sense ask me, thanks)

 

[mfb]

Looks good. You can solve it for Vzf, the last unknown.

If you want you can also express everything in terms of angles and speed instead of velocity components, and find the outgoing angle as function of the incoming angle (relative to the x-y-plane).

 

[MaestroBach]

Looks good. You can solve it for Vzf, the last unknown.

If you want you can also express everything in terms of angles and speed instead of velocity components, and find the outgoing angle as function of the incoming angle (relative to the x-y-plane).

Haha, thanks, I guess I was overcomplicating things in my head