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Question: A famous relation in physics relates the moving mass \[{m_0}\]of a particle in terms of its speed v ...

A famous relation in physics relates the moving mass m0{m_0}of a particle in terms of its speed v and the speed of light c. (This relation first arose as a consequence of the special theory due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writesm=m0(1−v2)12m = \dfrac{{{m_0}}}{{{{\left( {1 - {v^2}} \right)}^{\dfrac{1}{2}}}}}. Guess where to put the missing c.

Explanation

Solution

Use the principle of Homogeneity of dimensions, which states that dimensions of each term of a dimensional equation must be the same on both sides of the equation.
The dimensional formula is an expression for the units of a physical quantity in terms of their fundamental quantity. The fundamental quantities are mass (M), length (L), and time (T), and the dimensional formula is expressed in terms of powers of M, L, and T.

Complete step by step answer:
Given in the equationm=m0(1−v2)12m = \dfrac{{{m_0}}}{{{{\left( {1 - {v^2}} \right)}^{\dfrac{1}{2}}}}}that the boy recalls has a missing c in it.
Since by principle of Homogeneity of dimensions, we know the dimension of both sides of the equation must be equal, now let’s check for the dimension of the equation that is given by the boy
m=m0(1−v2)12m = \dfrac{{{m_0}}}{{{{\left( {1 - {v^2}} \right)}^{\dfrac{1}{2}}}}}
The dimension of LHS of the equation
LHS=[m]=MLHS = \left[ m \right] = M
Now for the dimension of RHS of the equation
RHS=m0(1−v2)12=M(1−v2)12RHS = \dfrac{{{m_0}}}{{{{\left( {1 - {v^2}} \right)}^{\dfrac{1}{2}}}}} = \dfrac{M}{{{{\left( {1 - {v^2}} \right)}^{\dfrac{1}{2}}}}}
We can see the dimension of LHS of the equation is equal to the dimension of the numerator of RHS of the equation; hence we need to make the denominator (1−v2)12{\left( {1 - {v^2}} \right)^{\dfrac{1}{2}}}to be dimensional less
As we know the unit of v=msv = \dfrac{m}{s}and also the unit ofc=msc = \dfrac{m}{s}, so if we divide vvwithcc the denominator becomes dimensional less =(1−v2c2)12 = {\left( {1 - \dfrac{{{v^2}}}{{{c^2}}}} \right)^{\dfrac{1}{2}}}
Therefore the correct equation will be
m=m0(1−v2c2)12m = \dfrac{{{m_0}}}{{{{\left( {1 - \dfrac{{{v^2}}}{{{c^2}}}} \right)}^{\dfrac{1}{2}}}}}

Note: It is always to be kept in mind that while writing the dimensional formula only and only SI units of the measuring quantities should be used and should be bifurcated further.