Question

Derive an expression of kinetic energy of a body of mass ‘m’ and moving with velocity ‘v’, using dimensional analysis.

Answer 1

In order to perform the dimensional analysis, we have to understand the two types of units – Base and Derived. The basic units are a set of 7 quantities which are regarded as fundamental quantities by the International System of Units. Every other quantity is called derived quantity, since it can be derived from the base units.

Complete step by step answer:
Dimensional analysis is the process of expressing any derived quantity in the form of one of the 7 base units established by the International System of Units.
The 7 base units are:
A. Length – metre (m)
B. Mass – kilogram (kg)
C. Time – second (sec)
D. Current – ampere(A)
E. Temperature – kelvin (K)
F. Amount of substance – mole(mol)
G. Luminous Intensity – candela(cd)
Consider the units, energy E, mass m and velocity v.
Energy is measured in the derived unit joule (J).
Mass is measured in the base unit kilogram (kg).
Velocity is measured in the derived unit metre-per-second $\left( {m{s^{ - 1}}} \right)$
The dimensions of these units expressed in the basic units are:
Energy, $E = 1J = 1N - m = 1\dfrac{{kg - m}}{{{s^2}}}m = kg{m^2}{s^{ - 2}} = \left[ M \right]{\left[ L \right]^2}{\left[ T \right]^{ - 2}}$
Mass, $M = 1kg = \left[ M \right]$
Velocity, $V = 1m{s^{ - 1}} = {\left[ M \right]^0}{\left[ L \right]^1}{\left[ T \right]^{ - 1}}$
Expressing the energy in terms of mass and velocity, we have –
$E \propto {m^a}{v^b}$
$\left[ M \right]{\left[ L \right]^2}{\left[ T \right]^{ - 2}} \propto {\left[ M \right]^a} \times {\left[ {{{\left[ M \right]}^0}{{\left[ L \right]}^1}{{\left[ T \right]}^{ - 1}}} \right]^b}$
$\Rightarrow \left[ M \right]{\left[ L \right]^2}{\left[ T \right]^{ - 2}} \propto {\left[ M \right]^a} \times {\left[ M \right]^0}{\left[ L \right]^b}{\left[ T \right]^{ - b}}$
$\Rightarrow \left[ M \right]{\left[ L \right]^2}{\left[ T \right]^{ - 2}} \propto {\left[ M \right]^a}{\left[ L \right]^b}{\left[ T \right]^{ - b}}$
Equating the terms on the power, we have –
$\Rightarrow a = 1,b = 2$
Substituting in the equation for energy, we have –
$\Rightarrow E \propto {m^1}{v^2}$
By experimental values, it has been found out that the value of constant to be substituted to remove the proportionality is equal to –
$\Rightarrow E = \dfrac{1}{2}m{v^2}$
Thus, we have established the relationship between the kinetic energy, mass and velocity through the process of the dimensional analysis.

Note: Dimensional analysis is very helpful in experimental physics where an unknown quantity in study must be correlated to the known quantities to establish a relationship between the unknown and known.