Question

a body recall the relative stick mass wrongly
$m= m0 \div \sqrt{1 - {v}^{2} }$
using the dimensional method put the missing C in the proper place​

Answer 1

We're given the mistaken equation,

$m=\dfrac{m_0}{\sqrt{1-v^2}}$

In this equation, the velocity of light, 'c', is forgot to be given.

Well, a power of 'c' is forgot to be given here.

So we have to find the actual position of the power of 'c'.

We take our equation dimensionally.

$[m]=\left[\dfrac{m_0}{\sqrt{1-v^2}}\right]$

Both the dimensions of  $m$  and  $m_0$  are same, M. So,

$M=\dfrac{M}{\left[\sqrt{1-v^2}\right]}\ \implies\ \left[\sqrt{1-v^2}\right]=\dfrac{M}{M}=M^0L^0T^0$

So we get that  $\sqrt{1-v^2}$  is dimensionless. From this, we get that the power of 'c' is somewhere in this term!

Let's go on!

$\left[\sqrt{1-v^2}\right]=M^0L^0T^0\ \implies\ \left[1-v^2\right]=(M^0L^0T^0)^2\\ \\ \implies\ \left[v^2\right]=[v]^2=M^0L^0T^0$

This is because constants are dimensionless, we know it.

But  $[v]^2=M^0L^0T^0$  is a contradictional equation.

$[v]^2=(LT^{-1})^2=L^2T^{-2}=M^0L^0T^0=L^0T^0$

Now consider the dimension of the velocity of light, 'c'. Well, it's a velocity!

$[c]=LT^{-1}\quad\quad\text{so}\quad\quad[c]^2=L^2T^{-2}$

So, in our equation  $L^2T^{-2}=L^0T^0,$  to get both sides equal, or to get the LHS dimensionless, we can divide the LHS by LHS itself, and we can guess that either LHS is divided by dimension of the power of 'c', or vice versa.

Means,

$\dfrac{L^2T^{-2}}{L^2T^{-2}}=L^0T^0\ \implies\ \dfrac{[v]^2}{[c]^2}=M^0L^0T^0\quad\text{OR}\quad\dfrac{[c]^2}{[v]^2}=M^0L^0T^0$

Hence, dimensionally, our real equation may be,

$m=\dfrac{m_0}{\sqrt{1-\left(\dfrac{v^2}{c^2}\right)}}\quad\quad\text{OR}\quad\quad m=\dfrac{m_0}{\sqrt{1-\left(\dfrac{c^2}{v^2}\right)}}$