How Does Special Relativity Explain Momentum in Particle Disintegration?

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thenewbosco
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A particle disintegrates into two pieces: the first has mass 1.00 MeV/c^2 and momentum 1.75MeV/c
the second has mass 1.50 MeV/c^2 and momentum 2.00 MeV/c.

find the mass and speed of the original particle.

What i have done is used the fact that [tex]p=\gamma m v[/tex] as well as [tex]E^2 = p^2c^2 + (mc^2)^2[/tex] to derive that for the original particle: [tex]\gamma = m*v[/tex] where [tex]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
however this is two unknowns and one equation.

any help on this...
 
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You have enough information about the pieces after the decay, to find their energies. Then you can apply conservation of energy and conservation of momentum to find the energy and momentum of the original particle, and from those, you can find the quantities that you're asked for.
 
i solved for the energies and got for the particle traveling in x: 1.79MeV and for the one in y:2.5 MeV.

now what i have done is set up the following:
for energies
[tex]E_{init}=\frac{mc^2}{\sqrt{1-\frac{u^2}{c^2}}}=(1.79+2.5)[/tex]
for x momentum:
[tex]\frac{mu_{x}{\sqrt{1-\frac{(u_{x})^2}{c^2}}}}=1.75=p_{xf}[/tex]
and y momentum:
[tex]\frac{mu_{y}{\sqrt{1-\frac{(u_{y})^2}{c^2}}}}=2.00=p_{yf}[/tex]
(the tex code is wrong but the square roots should be in the denominator)
now by solving these three using [tex](u_{x})^2+(u_{y})^2=u^2[/tex]
i should be able to find the mass and speed...

is this correct?
thanks
 
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Yes, your method should work, although it might not be the simplest one in terms of the math involved.

You might consider calculating the magnitude of the initial momentum first, from the x and y components (which you already know). Note that the problem asks only for the speed of the original particle, and not its direction of motion.