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**1. Homework Statement**

In [itex]\alpha[/itex] decay a nucleus X at rest decays to a daughter nucleus Y and an [itex]\alpha[/itex] particle. Conservation of momentum and kinetic energy gives:

[itex]M_{\alpha}v_{\alpha}+M_{Y}v_{Y}=0[/itex]

[itex]\frac{1}{2}M_{\alpha}v_{\alpha}^{2}+\frac{1}{2}M_{Y}v_{Y}^{2}=Q[/itex]

Where the Q value is the available energy found through [itex]Q=(M(X)-M(Y)-M(\alpha))c^{2}[/itex]

Show the kinetic energy of the two decay products are given by

[itex]E^{\alpha}_{k}=\frac{M_{Y}}{(M_{Y}+M_{\alpha})}Q[/itex]

[itex]E^{Y}_{k}=\frac{M_{\alpha}}{(M_{Y}+M_{\alpha})}Q[/itex]

**2. Homework Equations**

[itex]\frac{1}{2}M_{\alpha}v_{\alpha}^{2}+\frac{1}{2}M_{Y}v_{Y}^{2}=Q[/itex]

[itex]M_{\alpha}v_{\alpha}+M_{Y}v_{Y}=0[/itex]

[itex]Q=(M(X)-M(Y)-M(\alpha))c^{2}[/itex]

[itex]E^{\alpha}_{k}=\frac{M_{Y}}{(M_{Y}+M_{\alpha})}Q[/itex]

[itex]E^{Y}_{k}=\frac{M_{\alpha}}{(M_{Y}+M_{\alpha})}Q[/itex]

**3. The Attempt at a Solution**

I have tried rearranging the energy equation to get [itex]E^{\alpha}_{k}=Q-\frac{1}{2}M_{Y}v_{Y}^{2}[/itex]

since

[itex]E^{\alpha}_{k}=\frac{1}{2}M_{\alpha}v_{\alpha}^{2}[/itex]

Then rearranging the conservation of momentum equation and substituting in for various variables but I can't get anything that looks like the required expressions. I know this is a fairly simple algebra exercise but I just can't figure out what to do so any advice or suggestions would be appreciated.