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    Stuck on this question:

    A slowly moving alpha particle collides elastically with a stationary proton. Condider the collision in the zero momentum frame to show that the alpha particle cannot be deflected by more than arcsin(1/4)=14.5 degrees from its initial line of flight.

    Ive drawn diagrams involving the lab and ZM frames, but I don't understand what to do from there.
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    (Original post by jmz34)
    Stuck on this question:

    A slowly moving alpha particle collides elastically with a stationary proton. Condider the collision in the zero momentum frame to show that the alpha particle cannot be deflected by more than arcsin(1/4)=14.5 degrees from its initial line of flight.

    Ive drawn diagrams involving the lab and ZM frames, but I don't understand what to do from there.
    Well you need a nice picture really, which I don't want to draw for you and that will explain it all nicely.

    You could probably think of it using the sine rule as well, from your diagram you should see that:
    (u/5)/(4u/5) = (sin x)/(sin y) where x is the angle you want and y is just the angle between u/5 and the final velocity.

    And from your diagram you could say that max value of x is when y is a right angle - this might take a bit of thinking about, but it is true, think about what happens as the angle varies in the zmf (it can be anything)

    and from this we get max value sin x = 1/4

    It is hard to explain without a diagram though, so sorry
 
 
 
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