Posted by Please Help! on .
An alpha particle collides with an oxygen nucleus, initially at rest. The alpha particle is scattered at an angle of 60.9° above its initial direction of motion, and the oxygen nucleus recoils at an angle of 51.9° on the opposite side of that initial direction. The final speed of the nucleus is 1.32 105 m/s. In atomic mass units, the mass of an alpha particle is 4.0 u. The mass of an oxygen nucleus is 16 u.
(a) Find the final speed of the alpha particle.
(b) Find the initial speed of the alpha particle.
The final momentum components of the two particles perpendicular to the initial direction must cancel. Therefore
1.32*10^5*sin51.9 = Va'*sin60.9
Solve for Va', the final speed of the alpha particle.
Get the initial speed of the alpha particle, Va, by applying conservation of momentum to the intial direction of motion.
Va*4 = 16*1.32*10^5*cos51.0 + 4*Va'*cos60.9
You know Va' from part a. Solve for Va
I tried this and got
118882 for Va'
and 390097 for Va
but it was wrong?