A subatomic particle X spontaneously decaysinto two particles, A and B , eachof

Question

A subatomic particle X spontaneously decaysinto two particles, A and B, eachof rest energy 1.40 × 102 MeV. The particles flyoff in opposite directions, each with speed 0.827crelative to an inertial reference frame S. Reference: Ref 28-1
Which expression gives the momentum of particleA (relative to frame S)?

a. 314 MeV/c
b. 109 MeV/c
c. 206 MeV/c
d. 140 MeV/c
e. 249 MeV/c A subatomic particle X spontaneously decaysinto two particles, A and B, eachof rest energy 1.40 × 102 MeV. The particles flyoff in opposite directions, each with speed 0.827crelative to an inertial reference frame S. Reference: Ref 28-1 Which expression gives the momentum of particleA (relative to frame S)? 314 MeV/c 109 MeV/c 206 MeV/c 140 MeV/c 249 MeV/c

Explanation / Answer

the momentum of particle A (relativeto frame S)         p = mu/[1 -(u/c)2] Here m = 140 MeV/c2         u = 0.827 c        On substitution             p = 206 MeV/c         u = 0.827 c        On substitution             p = 206 MeV/c

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