Question

A moving particle fragments or decays into a particle moving at .53c, mass 135 MeV/c2, and a particle moving at .98c, mass 938 MeV/c2 - both going in the same direction. Calculate the mass and speed of the initial particle.

Answer 1

Answer:

M = 1073 Mev/c2

u = 0.95 C        

Explanation:

given data:

m1 =135 Mev/c2

v1 = 0.53 c

m2 = 938 Mev/c2

v2 = 0.98 c

from conservation of momentum principle we have

$\frac{mu}{\sqrt{1-\frac{u^2}{C^2}}} = \frac{m1v1}{\sqrt{1-\frac{V1^2}{C^2}}} +\frac{m1v1}{\sqrt{1-\frac{V2^2}{C^2}}}$

$\frac{mu}{\sqrt{1-\frac{u^2}{C^2}}} = \frac{135*0.53c}{0.848} +\frac{938*0.98c}{0.2}$

$\frac{mu}{\sqrt{1-\frac{u^2}{C^2}}} = 4680.6 C$   ...............1

Total mass of INITIAL particle M   =m1+m2 = 1073 Mev/c2

using equation 1

$\frac{1073 u}{\sqrt{1-\frac{u^2}{C^2}}}= 4680.6C$

solving for u we get

u = 0.95 C