# Elastic collisions between proton and helium nucleus

## Homework Statement

In an elastic collision between a proton and a helium nucleus at rest, the proton was scattered through an angle of 45 degrees. What proportion of its initial energy did it lose? what was the recoil angle of the helium nucleus?

## Homework Equations

tan$$\theta$$1=(sin$$\Psi$$))/((cos($$\Psi$$)+$$\gamma$$)

$$\theta$$2=1/2(pi-$$\Psi$$)

tan$$\theta$$=(($$\gamma$$)+1)/($$\gamma$$-1)*cot(.5*$$\gamma$$)

E2/E0=4$$\gamma$$(sin(1/2*$$\gamma$$)^2/($$\gamma$$+1)^2

phi is the scaterttering angle in the ZM frame and gamma=m2/m1, the mass ratio of the two particles. $$\theta$$2 is the recoil angle and $$\theta$$1 is the scatter angle.

## The Attempt at a Solution

mhelium=4*mproton , therefore gamma=1/4
$$\theta$$1=45 degrees

tan(45 degrees)=sin(phi)/(cos(phi)+.25)
1=sin(phi)/(cos(phi)+.25)==> cos(phi)+.25=sin (phi)

not sure how I can determine phi with sin(phi)-cos*(phi)=.25 ; I know I need phi to determine the recoil angle.

not sure how to determined how much initial energy was lost but I know it probably will have to apply this elastic formulae:

E2/E0=4$$\gamma$$(sin(1/2*$$\gamma$$)^2/($$\gamma$$+1)^2

Does E0 represent the initial energy?

tiny-tim
Hi pentazoid! (have a phi: φ )
Hint: sin(φ - 45º) = sinφcos45º - cosφsin45º 