Question
Help please. I know how to compute for the rest mass but how can you solve this problem. Solve for v in m = m0/(sqrt of 1 - (v^2/c^2)
Answers
Reiny
m = m0/(sqrt (1 - (v^2/c^2)) ---> notice I fixed your mismatched brackets
m(sqrt (1 - (v^2/c^2)) = m0
square both sides
m^2(1 - v^2/c^2) = m0^2
expand
m^2 - m^2 v^2/c^2 = mo^2
multiply each term by c^2
c^2 m^2 - m^2 v^2 = m0^2 c^2
c^2 m^2 - m0^2 c^2 = m^2 v^2
divide by m^2
v^2 = (c^2 m^2 - m0^2)/m^2
v^2 = 1 - m0^2/m^2
v = √(1 - m0^2/m^2)
check my algebra steps, I should have written it out on paper first.
It is easy to make errors when just typing it on here as you go along.
m(sqrt (1 - (v^2/c^2)) = m0
square both sides
m^2(1 - v^2/c^2) = m0^2
expand
m^2 - m^2 v^2/c^2 = mo^2
multiply each term by c^2
c^2 m^2 - m^2 v^2 = m0^2 c^2
c^2 m^2 - m0^2 c^2 = m^2 v^2
divide by m^2
v^2 = (c^2 m^2 - m0^2)/m^2
v^2 = 1 - m0^2/m^2
v = √(1 - m0^2/m^2)
check my algebra steps, I should have written it out on paper first.
It is easy to make errors when just typing it on here as you go along.
Steve
Aside from the typo where the c^2 gets lost, I'd just do it like this:
m^2(1 - v^2/c^2) = m0^2
1 - (v/c)^2 = (m0/m)^2
(v/c)^2 = 1 - (m0/m)^2
v/c = √(1 - (m0/m)^2)
To me, that makes it easier to see how the ratio of v to c is related to the ratio of the masses. Usually in this kind of relativistic stuff, v is expressed as 0.8c or some such value, anyway.
m^2(1 - v^2/c^2) = m0^2
1 - (v/c)^2 = (m0/m)^2
(v/c)^2 = 1 - (m0/m)^2
v/c = √(1 - (m0/m)^2)
To me, that makes it easier to see how the ratio of v to c is related to the ratio of the masses. Usually in this kind of relativistic stuff, v is expressed as 0.8c or some such value, anyway.