If momentum is increased by 20% , then K.E. i | vedclass

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Question

(a)$E = \frac{{{P^2}}}{{2m}}.$ If m is constant then $E \propto {P^2}$
$&rArr;$$\frac{{{E_2}}}{{{E_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^2

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(a)$E = \frac{{{P^2}}}{{2m}}.$ If m is constant then $E \propto {P^2}$
$&rArr;$$\frac{{{E_2}}}{{{E_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^2} = {\left( {\frac{{1.2P}}{P}} \right)^2} = 1.44$
$&rArr;$ ${E_2} = 1.44{E_1} = {E_1} + 0.44{E_1}$
${E_2} = {E_1} + 44\% $ of ${E_1}$
i.e. the kinetic energy will increase by $44\%$

If momentum is increased by $20\%$ , then K.E. increases by ........... $\%$

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