<PRE><B>Question 1057844</B>
let u be the mass of the alpha particle, then 4u is the mass of the oxygen nucleus 
:
vi is the initial velocity of the alpha particle and 0 is the initial velocity of the oxygen nucleus
:
vf1 is the final velocity of the alpha particle
vf2 is the final velocity of the oxygen nucleus
:
conservation of linear momentum for the x axis is
:
(u * vi) + (4u * 0) = u*vf1*cos(66) + 4u*vf2*cos(57)
:
conservation of linear momentum along the y axis is
:
0 + 0 = u*vf1*sin(66) - 4u*vf2*sin(57)
:
0.9135 * vf1 = 3.3547 * vf2
:
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we use the result from the conservation of momentum along the y axis
:
vf1 / vf2 = 3.3547 / 0.9135 = 3.6724
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:
Note that we can not use the x-axis since we do not know the initial
velocity of the alpha particle
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