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Let us first write the dimension of each of the quantities mentioned in the question.

Dimension of kinetic energy are $\left[ \text{M}{{\text{L}}^{2}}{{\text{T}}^{-2}} \right]$

Dimension of velocity are $\left[ {{\text{M}}^{0}}\text{L}{{\text{T}}^{-1}} \right]$

Dimensions of mass of a body is $\left[ \text{M}{{\text{L}}^{0}}{{\text{T}}^{0}} \right]$

It is given to us that the kinetic energy is dependent on mass and the velocity of a body. Hence we can say that $K.E=k{{v}^{a}}{{m}^{b}}....(1)$where k,a and b are all constants

Now let us replace the quantities in the above expression with their dimensions

$\begin{align}

& K.E=k{{v}^{a}}{{m}^{b}} \\

&\left[\text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-2}}}\right]\text{=k}{{\left[ {{\text{M}}^{\text{0}}}\text{L}{{\text{T}}^{\text{-1}}} \right]}^{\text{a}}}{{\left[ \text{M}{{\text{L}}^{\text{0}}}{{\text{T}}^{\text{0}}} \right]}^{\text{b}}}

\end{align}$

By using law of exponents i.e. ${{\left( {{\text{A}}^{\text{m}}} \right)}^{\text{n}}}\text{=}{{\text{A}}^{\text{mn}}}$ we get,

$\left[ \text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-2}}} \right]\text{=k}\left[ {{\text{M}}^{\text{0}}}{{\text{L}}^{\text{a}}}{{\text{T}}^{\text{-a}}} \right]\left[ {{\text{M}}^{\text{b}}}{{\text{L}}^{\text{0}}}{{\text{T}}^{\text{0}}} \right]$

Again we will use one more law of exponents i.e. ${{\text{A}}^{\text{m}}}{{\text{A}}^{\text{n}}}\text{=}{{\text{A}}^{\text{m+n}}}$

$\left[ \text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-2}}} \right]\text{=k}\left[ {{\text{M}}^{b}}{{\text{L}}^{\text{a}}}{{\text{T}}^{\text{-a}}} \right]$

After comparing the powers of the same fundamental quantities we get,

b= 1 and a=2.

Now let us substitute the values in equation 1 and we get,

$ \text{K}\text{.E=k}{{\text{v}}^{\text{2}}}\text{m}$

Hence,

$\text{K}\text{.E=k m}{{\text{v}}^{\text{2}}}$

From the above expression we can conclude that kinetic energy of a body in motion is directly proportional to its mass and directly proportional to the square of its velocity.

It is to be noted that the value of k is experimentally found to be half. The above expression is for a particle in motion. T any instant its kinetic energy is given by the above obtained expression. This method is actually useful to relate quantities when we cannot come to a final expression between the physical quantities.