Question

The SI unit of energy is $J=kg{m}^2{s}^{-2}$; that of speed $v$ is $m{s}^{-1}$ and of acceleration $a$ is $m{s}^{-2}$. Which of the formula for kinetic energy ($K$) given below can you rule out on the basis of dimensional arguments ($m$ stands for the mass of the body):

• A. $K=m^2{v}^3$
• B. $K=\left(1/2\right)m{v}^2$
• C. $K=ma$
• D. $K=\left(3/16\right)m{v}^2$
• E. $K=\left(1/2\right)m{v}^2+ma$

Answer 1

Answer: (B), (D)

$K=\left(1/2\right)m{v}^2$
$K=\left(3/16\right)m{v}^2$

The dimensional formula for K.E is $\left[M^1{L}^2{T}^{-2}\right]$

According to the principle of homogeneity of dimensional analysis, the dimensions of each term on both the sides of correct formula or equation will be the same.

The dimensions of quantity on R.H.S of different quantities are

$\left(a\right)\left[M^2{L}^3{T}^{-3}\right]$

$\left(b\right)\left[M^1{L}^2{T}^{-2}\right]$

$\left(c\right)\left[M^1{L}^1{T}^{-2}\right]$

$\left(d\right)\left[M^1{L}^2{T}^{-2}\right]$

$\left(e\right)\left[M^1{L}^2{T}^{-2}\right]+\left[M^1{L}^1{T}^{-2}\right]$

Since dimensional formula of the R.H.S of $\left(a\right),\left(c\right)$ and both the terms of R.H.S of $\left(e\right)$ are not the same as that of L.H.S i.e.,$\left[M^1{L}^2{T}^2\right]$

$\therefore$ the formulae $\left(a\right),\left(c\right)$ and $\left(e\right)$ are ruled out dimensionally.