Question

Using dimensional analysis, show that the kinetic energy of a body of mass m moving with a velocity v varies as $m{{v}^{2}}$

Answer 1

Hint: For the given quantities of mass and velocity assign some variable powers to it to form a general equation for kinetic energy. Solve the equations and equate the powers for each of the dimensions from both the sides of the equation to get the value of the assumed variables and hence the relation.

Formula used: Dimensional formula of energy $\left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]$, mass $\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]$ and velocity $\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]$.

Complete step-by-step answer:
Every quantity can be expressed in the terms of the following seven dimensions
Dimension Symbol
Length L
Mass M
Time T
Electric charge Q
Luminous intensity C
Temperature $\theta $
Angle None

We are asked to derive the relationship between kinetic energy mass and velocity.
Kinetic energy is one of the types of energy and it is defined as the energy in an object due to its virtue of motion. The dimension of kinetic energy will be the same as that of energy. Therefore, kinetic energy is measured in Joules (SI unit of energy) and its dimensional formula will be $\left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]$.
Since, we are asked to draw a relationship between mass of the body and its velocity. We will take the dimensions of these two quantities and assume a variable measure which varies it.
The dimensional formula of mass is $\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]$.
The dimensional formula of velocity is $\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]$
We assume that kinetic energy depends on mass and velocity according to the following general equation:
$\text{kinetic energy }={{m}^{x}}{{v}^{y}}$(1)
Let us equate the dimensions on both the sides of equality
\[\begin{align}
  & \left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]={{\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]}^{x}}{{\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]}^{y}} \\
 & \left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]=\left[ {{M}^{x}}{{L}^{0}}{{T}^{0}} \right]\left[ {{M}^{0}}{{L}^{y}}{{T}^{-y}} \right] \\
 & \left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]=\left[ {{M}^{x}}{{L}^{y}}{{T}^{-y}} \right] \\
\end{align}\]
Powers for every dimension, M, L and T are equated from both the sides to get the following equations:
$\begin{align}
  & x=1 \\
 & y=2 \\
\end{align}$
Without much calculations, we have found the values of x and y. We plug these results in equation (1)
To get
$\text{kinetic energy}=m{{v}^{2}}$
Hence proved.

Note: For this particular question, we could find the dimensions as a product. The dimensional formulae for which are similar to powers of a base formulae. For example, ${{a}^{{{x}_{1}}}}{{b}^{{{y}_{1}}}}\times {{a}^{{{x}_{2}}}}{{b}^{{{y}_{2}}}}={{a}^{{{x}_{1}}+{{x}_{2}}}}{{b}^{{{y}_{1}}+{{y}_{2}}}}$ is similar to $[{{M}^{{{x}_{1}}}}{{L}^{{{y}_{1}}}}][{{M}^{{{x}_{2}}}}{{L}^{{{y}_{2}}}}]=[{{M}^{{{x}_{1}}+{{x}_{2}}}}{{L}^{{{y}_{1}}+{{y}_{2}}}}]$.