Keeping the mass of object same if the velocity of object is reduced to 1-4 th of its original velocity-Find the ratio of changed kinetic energy to original kinetic energy-

If V^' = V/4 K.E_1/K.E_2=V^2/V_1^2=V^2/ ( V/4 )^2=V^2/V^2/16 =16 V̸^̸2̸/V̸^̸2̸ =16/1=16:1 ...

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Byju's Answer

Standard XII

Physics

Momentum

Keeping the m...

Question

Keeping the mass of object same if the velocity of object is reduced to ${\frac{1}{4}}^{t h}$ of its original velocity.Find the ratio of changed kinetic energy to original kinetic energy?

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Solution

If $V^{^{'}} = \frac{V}{4}$
$\frac{K . E_{1}}{K . E_{2}} = \frac{V^{2}}{V_{1}^{2}} = \frac{V^{2}}{{(\frac{V}{4})}^{2}} = \frac{V^{2}}{\frac{V^{2}}{16}} = 16 \frac{/ V^{2}}{/ V^{2}} = \frac{16}{1} = 16 : 1$

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Keeping the mass of object same if the velocity of object is reduced to 14th of its original velocity.Find the ratio of changed kinetic energy to original kinetic energy?

If V′=V4 K.E1K.E2=V2V21=V2(V4)2=V2V216=16/V2/V2=161=16:1

Keeping the mass of object same if the velocity of object is reduced to ${\frac{1}{4}}^{t h}$ of its original velocity.Find the ratio of changed kinetic energy to original kinetic energy?

If $V^{^{'}} = \frac{V}{4}$
$\frac{K . E_{1}}{K . E_{2}} = \frac{V^{2}}{V_{1}^{2}} = \frac{V^{2}}{{(\frac{V}{4})}^{2}} = \frac{V^{2}}{\frac{V^{2}}{16}} = 16 \frac{/ V^{2}}{/ V^{2}} = \frac{16}{1} = 16 : 1$