Elastic Collision - Scattering

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In an elastic collision involving a proton and a helium atom, momentum and kinetic energy conservation must be applied in both x and y directions. The problem presents challenges in solving for deflection angles due to the equations being expressed in terms of sine and cosine. Utilizing trigonometric identities can help eliminate variables, allowing for the determination of angles. Alternatively, switching to the center of mass reference frame simplifies the equations, though conversion back to the laboratory frame is necessary for final results. Both methods ultimately yield the same final velocities and angles for the particles involved.
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Elastic Collision -- Scattering

Before I ask my question, here's the problem in full,

"A proton of mass mp, with initial velocity v0 collides with a helium atom, mass 4mp, that is initially at rest. If the proton leaves the point of impact at an angle of 45 degrees with its original line of motion, find the final velocities of each particle. Assume that the collision is perfectly elastic."

I've tried writing the momentum conservation in the x and y directions (in the laboratory frame), but I end up getting two equations with two unknowns (the deflection angles). Normally I could solve this, but the x equations are given in terms of the cosines of the angles, while the y equations are given in terms of the sines of the angles. Can anyone tell me how to write down the conservation equations in a way so that I'll be able to solve for the angles?

Also, would it be a better idea to write the equations in terms of the center of mass reference frame?

Thanks for your help.
 
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Well the conservation of momentum must be conserved in the x and y direction. If you assume the proton had no initial y momentum then the final y momentum must be zero. You can equate the momentum of the two particles that way. Then conservation of momentum in the x-direction still applies. If you still have too many variables, kinetic energy is also conserved.
 


In an elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum before the collision is equal to the total momentum after the collision, and the total kinetic energy before the collision is equal to the total kinetic energy after the collision. In the case of scattering, the particles involved are deflected from their original paths, but their total energy and momentum remain the same.

To solve this problem, you can use the equations for momentum and kinetic energy conservation in the laboratory frame. The momentum conservation equations in the x and y directions will give you two equations with two unknowns (the deflection angles). To solve for these angles, you can use the trigonometric identities that relate cosines and sines. For example, you can use the identity cos^2x + sin^2x = 1 to eliminate one of the unknowns and solve for the other.

Alternatively, you can also use the center of mass reference frame, where the total momentum of the system is zero. In this frame, the equations for momentum and kinetic energy conservation will be simpler and can be solved for the final velocities of the particles. However, you will still need to convert back to the laboratory frame to find the deflection angles.

Ultimately, both approaches will give you the same final velocities and deflection angles. It is up to you to decide which method is easier for you to use. Good luck!
 
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