Question

?? I got 5 minutes left, please help.

Answer 1

Answer:

Here we know that:

$m(v) = \frac{M_0}{\sqrt{1 - \frac{V}{C} } }$

Where V is the speed, C = 3*10^8 m/s

We want to solve:

$m(v) = \frac{M_0}{\sqrt{1 - \frac{V}{C} } } = 2*M_0$

We can just isolate V from the above equation, so we will get:

$\frac{M_0}{\sqrt{1 - \frac{V}{C} } } = 2*M_0$

$\frac{1}{\sqrt{1 - \frac{V}{C} } } = 2$

$1 = 2\sqrt{1 - \frac{V}{C} }$

$(1/2)^2 = 1 - \frac{V}{C}$

$V = (1 - (1/4))*C = (3/4)*C = (3/4)*3*10^8 m/s = (9/4)*10^8 m/s$

That is the velocity such that the effective mass is twice the rest mass.