Prove that the maximum velocity imparted by an α-particle to a proton during their collision is 1.6 of the initial velocity of the α-particle.
Answers
Solve two simultaneous equations for energy and momentum conservation. By intuition you know that the max proton energy is going to come from being hit head on where the alpha follows the proton after the collision.
mv^2/2 for energy and mv for momentum. Assuming that the mass of the alpha is four times the mass of the proton.
4*v^2 + 1*0^2 = 4*v2^2 + 1*vp^2 where all the 2’s in the denominators cancel
4*v + 1*0 = 4*v2 + 1*vp where v is the initial alpha velocity, 0 is the initial proton velocity, v2 is the reduced alpha velocity after the collision, and vp is the proton velocity after the collision.
So I assumed that v = 1 to make the math easy. vp of 1.6 does work. I suppose that to get this you’d have to solve a quadratic. If you do, tell me what the second root is? I wonder what it represents???
If you look at the equations, I think you can convince yourself that the ratio does not depend on v. E.g., if I have a solution and I double everything in the first and second equations, the equalities will still hold.