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5.Work, Energy, Power and Collision
medium
If momentum is increased by $20\%$ , then K.E. increases by ........... $\%$
A
$44$
B
$55$
C
$66$
D
$77$
Solution
(a)$E = \frac{{{P^2}}}{{2m}}.$ If m is constant then $E \propto {P^2}$
$⇒$$\frac{{{E_2}}}{{{E_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^2} = {\left( {\frac{{1.2P}}{P}} \right)^2} = 1.44$
$⇒$ ${E_2} = 1.44{E_1} = {E_1} + 0.44{E_1}$
${E_2} = {E_1} + 44\% $ of ${E_1}$
i.e. the kinetic energy will increase by $44\%$
Standard 11
Physics
