Best relationship between momentum and kinetic energy possessed by an object in terms of mass is

$m o m e n t u m = {(2 m \times k i n e t i c e n e r g y)}^{\frac{1}{2}}$

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${(m o m e n t u m)}^{\frac{1}{2}} = 2 m \times k i n e t i c e n e r g y$

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${(K i n e t i c e n e r g y)}^{\frac{1}{2}} = 2 \times m o m e n t u m$

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$K i n e t i c e n e r g y = {(2 \times m o m e n t u m)}^{\frac{1}{2}}$

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Solution

The correct option is A
$m o m e n t u m = {(2 m \times k i n e t i c e n e r g y)}^{\frac{1}{2}}$

Step 1: Expression of kinetic energy and momentum
Kinetic energy (K.E) = $\frac{1}{2} m v^{2}$
Momentum (p) = $m v$
where $m$ = mass and $v$ = velocity
Step 2: Relation between momentum and kinetic energy
Considering the expression of kinetic energy,
$K . E = \frac{1}{2} m v^{2}$
Multiplying mass $(m)$ on numerator and denominator,
$K . E = \frac{1}{2} m v^{2} (\frac{m}{m}) K . E = \frac{m^{2} v^{2}}{2 m} K . E = \frac{{(m v)}^{2}}{2 m}$
Substituting $m v = p$ ,
$K . E = \frac{p^{2}}{2 m} 2 m x K . E = p^{2} p = {(2 m x K . E)}^{\frac{1}{2}}$
Therefore $m o m e n t u m = {(2 m \times k i n e t i c e n e r g y)}^{\frac{1}{2}}$
Hence the correct answer is Option A.

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