Elastic collision between an unknown mass and an alpha particle

[pentazoid]

Homework Statement

In an elastic collision between an alpha particle and an unknown nucleus at rest, the alpha particle was deflected through a right angle and lost 40 percent of its energy . Identify the mysterious nucleus.

Homework Equations

(E₂)/(E₀)=[4*gamma/(gamma+1)^2]*(sin(phi/2))^2

$$\theta$$₂=.5*(pi-phi)

gamma is the ratio between the alpha mass particle and the unknown mass

The Attempt at a Solution

Is the deflected angle the recoil angle? If so then I can use the recoil angle to find phi. Then I can use phi to find gamma and with gamma since I know the mass of an alpha particle, I can find the unknown mass. When the problem says the recoil angle loses 40 % of its energy , doesn't that mean E₂=.4E₀

 

[tiny-tim]

Is the deflected angle the recoil angle? If so then I can use the recoil angle to find phi. Then I can use phi to find gamma and with gamma since I know the mass of an alpha particle, I can find the unknown mass. When the problem says the recoil angle loses 40 % of its energy , doesn't that mean E₂=.4E₀

Hi pentazoid!

Yes, the deflected angle is the recoil angle …

but no, E₂=.6E₀

 

[pentazoid]

Hi pentazoid!

Yes, the deflected angle is the recoil angle …

but no, E₂=.6E₀

if my deflected angle is 90 degrees that means my phi is zero. which means sin(phi/2) is zero, which means then that there is no way to determine what the unknown mass is

 

[tiny-tim]

if my deflected angle is 90 degrees that means my phi is zero. which means sin(phi/2) is zero, which means then that there is no way to determine what the unknown mass is

I'm confused …

isn't φ = π/2?

 

[turin]

Is the deflected angle the recoil angle?

Not if you mean the angle that the nucleas recoils. Conservation of energy and momentum determine this recoil angle.