Elastic Collisons (Conservation of Ke and P)

[notagenius08]

Homework Statement

A proton is traveling to the right at 2.0x10^7 m/s. It has a head on perfectly elastic collision with a carbon atom. The Mass of the Carbon atom is 12 times the mass of the proton. What are the speed and direction of each after collision?

Homework Equations

Ke=1/2M*V^2
P(momentum)=M*V

The Attempt at a Solution

p=proton
c=carbon

After the collision:

1/2MpVpf^2 + 1/2McVcf^2=1/2MpVp^2

MpVpf+McVcf=MpVp


Solved for:

Vcf=Vp-Vpf/12
Vpf=Vp-12Vcf
Vcf^2=Vp^2-Vpf^2/12
Vpf^2=Vp^2-12Vcf

I've tried a few times but can figure out how to solve the rest of the equation. Every time I make a substitution I get MpVp=MpVp.

Which I already knew, so I guess my algebra is a bit crap.(spelling too!)

What is the next step?

Thanks to anyone that helps.

 

[physics girl phd]

I'll walk you through one way to solve:

You have four equations.
1) Take the first (for Vcf), square it to find Vcf^2.
2) Now plug it into the third equation instead of Vcf^2 (on the left side).
3) This result now needs to be reorganized so everything is on the right side or left (so one side is zero.
4) Note: the above result is likely a quadratic equation with the unknown "x" being Vpf.
5) Solve the quadratic equation for Vpf.
6) now plug that back into any of your original four equations to find Vcf.

why did I chose 1 and 3?

they come from different original equations (momentum conservation and energy conservation). These equations are therefore "independent" of each other. You can't chose 1 and 2, or 3 and 4 for the process... but you could have also chosen 2 and 4, or 1 and 4 or 2 and 3.

 

[notagenius08]

Thanks for your help.

It actually does simplify down, and so you don't have to use the quadratic equations.

Thanks again for your help!