How Far Can a Particle Travel at 0.99c Before Decaying?

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To determine how far a particle travels at 0.99c before decaying, the time dilation effect must be considered using the Lorentz transformation. The particle's rest lifetime is 1*10^(-7) seconds, but this lifetime appears longer to a stationary observer due to time dilation. The formula T' = T/sqrt(1 - (v/c)^2) calculates the dilated lifetime. By multiplying the dilated time by the particle's speed, the distance traveled before decay can be found. This approach effectively accounts for relativistic effects on the particle's decay.
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can someone help with this question?


A certain particle has a lifetime of 1*10^(-7) sec when measured at rest. How far does it go before decaying if its speed is 0.99c when it is created?
 
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Well... I'd start this problem by putting it into the Lorentz formula and seeing the amount of time it would take for it to decay as observed by a stationary observer.

I'd then multiply this time with the speed at which it is travelling.
 
@@a
why'd you use the lorentz formula?
 
He means the time dilation application of the Lorentz transformation. The particle still decays at the same rate in its own reference frame, but in an observer's frame in which it is traveling at 0.99c, this time interval is dilated: T' = T/root(1 - (v/c)^2).
 
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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