Question

linear momentum of body is increased by 50%, find % change in E.​

Answer 1

Topic :- Work , Energy and Power

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$\bullet \: \sf\footnotesize P_{(initial)} = P$

$\bullet \: \sf\footnotesize P_{(final)} = P + 50\%$

$\bullet \: \sf\footnotesize P_{(final)} = P + \dfrac{50}{100} P$

$\bullet \: \sf\footnotesize P_{(final)} = P + \dfrac{5}{10} P$

$\bullet \: \sf\footnotesize P_{(final)} = 10 \times P + \dfrac{5P}{10}$

$\bullet \: \sf\footnotesize P_{(final)} = \dfrac{10P + 5P}{10}$

$\bullet \: \sf\footnotesize P_{(final)} = \dfrac{15P}{10}$

$\bullet \: \sf\footnotesize P_{(final)} = \dfrac{15P}{10}$

$\bullet \: \sf\footnotesize P_{(final)} = \dfrac{15P \div 5}{10 \div 5}$

$\bullet \: \sf\footnotesize P_{(final)} = \dfrac{3P }{2}$

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Relation between Linear momentum (P) and Kinetic energy (K.E) is given as :

$\longrightarrow\:\sf P = \sqrt{2mk}$

$\longrightarrow\:\sf P^{2} = 2mk$

$\longrightarrow\:\sf k = \dfrac{P^{2}}{ 2m}...(i)$

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$\longrightarrow\:\sf \dfrac{k_{(final)} - k_{(initial)}}{k_{(initial)} } \times 100$

From equation (i) we can write Final and initial Kinetic energy as :

$\longrightarrow\:\sf \dfrac{ \dfrac{P_{(final)} ^{2} }{2m} - \dfrac{P_{(initial)} ^{2} }{2m} }{ \dfrac{P_{(initial)} ^{2}}{2m}}\times 100$

$\longrightarrow\:\sf \dfrac{P_{(final)} ^{2}- P_{(initial)} ^{2}}{P_{(initial) }^{2} }\times 100$

$\longrightarrow\:\sf \dfrac{ \bigg( \dfrac{3P}{2} \bigg) ^{2} - P^{2}}{P^{2} }\times 100$

$\longrightarrow\:\sf \dfrac{ \dfrac{9P}{4}- P^{2}}{P^{2} }\times 100$

$\longrightarrow\:\sf \dfrac{ 5}{4}\times 100$

$\longrightarrow\:\sf \dfrac{ 500}{4}$

$\longrightarrow\: \underline{ \underline{\sf 125\%}}$

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