Question
Rule out or accept the following formulae for kinetic energy on the basis of dimensional arguments:
- $\frac{3}{16}\text{mv}^2$
- $\frac{1}{2}\text{mv}^2+\text{ma}$
✓
Answer
$\text{K.E}=\frac{1}{2}\text{mv}^2$ Dimension of $\text{K}.\text{E}=[\text{M}\text{L}^2\text{T}^{-2}]$
Since Dimension of $K.E. =$ Dimension of $\frac{3}{16}\text{mv}^2$
It is dimensionally corrct. $\frac{1}{2}\text{mv}^2+\text{ma}$
Dimensions $=[\text{ML}^2\text{T}^{-2}]+[\text{MLT}^{-2}]$
Which is dimensionally incorrect.
Since Dimension of $K.E. =$ Dimension of $\frac{3}{16}\text{mv}^2$
It is dimensionally corrct. $\frac{1}{2}\text{mv}^2+\text{ma}$
Dimensions $=[\text{ML}^2\text{T}^{-2}]+[\text{MLT}^{-2}]$
Which is dimensionally incorrect.
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(i) Pilot of an aircraft in flight.
(ii) A driver driving at a high steady speed of 200 km/h in a sports car on a straight road.
(iii) A cyclist taking a scissor turn.
(iv) The guard of a train slowing down and stopping at the station.
(v) A child moving in a big circle