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tmlrlz
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Homework Statement
In the special theory of relativity the mass of a particle moving at speed v is given by the expression
m/(1-(v2/c2))
where m is the mass at rest and c is the speed of light. At what rate is the mass of the particle changing when the speed of the particle is (1/2)c and is increasing at the rate of 0.01c per second?
Homework Equations
The Attempt at a Solution
For this problem i tried using related rates. The question is asking for dm/dt when v = (1/2)c and they give dv/dt = 0.01c
dm/dt = (dm/dv)*(dv/dt)
Suppose that y = m/√(1-(v2/c2)) i took the derivative of y
the one assumption i made which is what I'm confused about is that the derivative of y = 0
[(dm/dv)*(√(1-(v2/c2))) - 1/2*((1-(v2/c2))-1/2*(-2v/c2)*(m)]/(1-(v2/c2) = 0
dm/dv(√(1-(v2/c2))) = -m(1-(v2/c2))-1/2(v/c2)
dm/dv = [-m(v/c2)]/(1-(v2/c2))
dm/dt = [-m(v/c2)]/(1-(v2/c2)) * (0.01c)
= -0.01mv/(c-v2/c)
At v = 1/2c
dm/dt = -mc/200 * 4/3c
= -m/150
I would just like to know if my approach is correct and specifically, if i was correct in assuming that the derivative of y = 0. Thank You.