What Is the Velocity of Particles After Decay?

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The discussion centers on the decay of a particle with mass M into two identical particles, each with mass 0.48M, and the resulting velocities. Participants highlight that the sum of the masses after decay does not equal the initial mass, raising questions about the conversion of "missing" rest mass energy. The relationship between rest mass energy and total energy is also explored, particularly in the context of a particle with a rest mass energy of 100 MeV and a total energy of 200 MeV. Clarifications are sought on how to calculate the velocities of the particles post-decay and the implications of energy conservation. The conversation emphasizes the need to understand energy transformations during particle decay.
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1.) A particle with mass M initially at rest decays into two identical particles each of mass 0.48M, the speed of the identical particles after the decay is :



after decaying , the sum of two mass doesn't equal to initial mass...


can somebody help me how to solve this qusiton


2.)if a particle whose rest mass energy equals 100MeV has a total energy of 200Mev, its velocity is :


i think there is some kind of relation between 1 and 2.

please help me ?

thx
 
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Hi wowolala,

1. Now that the particle has decayed and the two identical particles are moving away, in what form of energy has the "missing" rest mass energy been converted to?

2. How is the total energy related to the rest mass energy?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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