A particle of mass M decays into two identical particles each of mass m, where m

Question

A particle of mass M decays into two identical particles each of mass m, where m = 0.2M. Prior to the decay, the particle of mass M has a total energy of 4Mc2 in the laboratory reference frame. The velocities of the decay product are along the direction of motion M. Find the velocities of the decay products in the laboratory reference frame. (Round your answer to three decimal places.)

______ c (higher speed product)
______ c (lower speed product)

I got the higher speed product to be .999 c and it is correct but I cannot get the lower speed product to be correct

Explanation / Answer

total energy of a particle=Moc^2+.5mV^2 M=Mo(1-v^2/c^2)^.5 4Moc^2=Moc^2+.5Mo/(1-v^2/c^2)^.5*v^2=? v^2/(1-v^2/c^2)^.5=6c^2 mv1+mv2=Mv Solving it we get m=.2M/(1-v^2/c^2)^.5 =>v=.987c

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