help is much needed?

Question

In Figure 9-22, projectile particle 1 is an alpha particle and target particle 2 is an oxygen nucleus. The alpha particle is scattered at angle 1 = 64.0° and the oxygen nucleus recoils with speed 1.10 105 m/s at angle 2 = 47.0°. In atomic mass units, the mass of an alpha particle is 4.00 u and the mass of an oxygen nucleus is 16.0 u.

(a) Find the final speed of the alpha particle.
(b) Find the initial speed of the alpha particle.

http://i175.photobucket.com/albums/w138/kacortez/hrw7_9-22.gif

Answer 1

use momentum conservation
m1*v1i = m1*v1f + m2*v2f (vector form)
X component: m1*v1i = m1*v1f*cosθ1 + m2*v2f*cosθ2 (1)
Y component: 0 = -m1*v1f*sinθ1 + m2*v2f*sinθ2 (2)
from (2) we have v1f = m2*v2f*sinθ2/(m1*sinθ1)
sub m1 = 4.00 u, m2 = 16.0 u, v2f = 1.10x10^5 m/s, θ1 = 64.0°, θ2 = 47.0° into it we can get the value of v1f.
After getting v1f, we may get v1i from (1)
v1i = (m1*v1f*cosθ1 + m2*v2f*cosθ2)/m1

Answer 2

Conservation of linear momentum:
http://en.wikipedia.org/wiki/Momentum

momentum = mass x velocity
like velocity, momentum is a vector

Conservation of momentum means the momentum before the collision must equal the momentum after:

(m1 u1) + (m2 u2) = (m1 v1) + (m2 v2)

In this case, u1 is toward the right and u2 is zero so:

(m1)(|v1|)(cos @1) + (m2)(|v2|)(cos @2) = m1 |u1|

(m1)(|v1|)(sin @1) - (m2)(|v2|)(sin @2) = 0

two equations in two unknowns.