Question 1057844: An alpha-particle collides with an oxygen nucleus which is initially at rest. The alpha-particle is scattered at an angle of 66.0 degrees from its initial direction of motion, and the oxygen nucleus recoils at an angle of 57.0 degrees on the other side of this initial direction. What is the ratio, for alpha-particle to oxygen nucleus, of the final speeds of these particles? The mass of the oxygen nucleus is four times that of the alpha particle.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! let u be the mass of the alpha particle, then 4u is the mass of the oxygen nucleus
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vi is the initial velocity of the alpha particle and 0 is the initial velocity of the oxygen nucleus
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vf1 is the final velocity of the alpha particle
vf2 is the final velocity of the oxygen nucleus
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conservation of linear momentum for the x axis is
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(u * vi) + (4u * 0) = u*vf1*cos(66) + 4u*vf2*cos(57)
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conservation of linear momentum along the y axis is
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0 + 0 = u*vf1*sin(66) - 4u*vf2*sin(57)
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0.9135 * vf1 = 3.3547 * vf2
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we use the result from the conservation of momentum along the y axis
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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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