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.
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)
2 answers
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.