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    Can someone help me on where I can start to derive this expression I'm really curious and I haven't been able to find a way. (The expression that I should have been lead to was A).
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    Conservation of momentum.

    The initially momentum of the system is zero, as the parent nucleus is stationary. Let v_1 be the velocity of the daughter nucleus, and v_2 be the velocity of the \alpha particle. The mass of the daughter nucleus is M-m.

    Thus,

    0 = (M-m)v_1 - mv_2

    We can derive an expression for v_2 in terms of the kinetic energy E:

    E = \dfrac{1}{2}mv_2^2 \, \, \, \Rightarrow \, \, v_2 = \sqrt{\dfrac{2E}{m}}

    Subbing this into the first equation:

    0 = (M-m)v_1 - m\sqrt{\dfrac{2E}{m}} \, \, \, \Rightarrow \, \, (M-m)v_1 = \sqrt{2mE}

    v_1 = \dfrac{\sqrt{2mE}}{(M-m)}

    The negative on the RHS of the first equation is due to their opposite directions of motion.
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    (Original post by Phichi)
    Conservation of momentum.

    The initially momentum of the system is zero, as the parent nucleus is stationary. Let v_1 be the velocity of the daughter nucleus, and v_2 be the velocity of the \alpha particle. The mass of the daughter nucleus is M-m.

    Thus,

    0 = (M-m)v_1 - mv_2

    We can derive an expression for v_2 in terms of the kinetic energy E:

    E = \dfrac{1}{2}mv_2^2 \, \, \, \Rightarrow \, \, v_2 = \sqrt{\dfrac{2E}{m}}

    Subbing this into the first equation:

    0 = (M-m)v_1 - m\sqrt{\dfrac{2E}{m}} \, \, \, \Rightarrow \, \, (M-m)v_1 = \sqrt{2mE}

    v_1 = \dfrac{\sqrt{2mE}}{(M-m)}

    The negative on the RHS of the first equation is due to their opposite directions of motion.
    what happened to the two m's when you subbed in v2 into the original equations im confused on how it became Root*(2mE)?

    Oh wait never mind I can see how it happens using laws of Indices. Thanks for your help cleared a lot of confusion for me
 
 
 
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