Question

What happenes to the kinetic energy of a body if it’s momentum is doubled (with working)

Answer 1

$\Huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

As, we know formula for Kinetic Energy

$\LARGE{\underline{\boxed{\sf{K.E \: = \: \frac{1}{2} \: mv^2}}}}$

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And formula for Momentum is

$\Large{\boxed{\sf{p \: = \: mv}}}$

Where,

p is momentum

m is mass

v is velocity ,

$\Large \longrightarrow {\sf{v \: = \: \frac{p}{m}}}$

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Put This value in formula of Kinetic Energy,

$\large \rightarrow {\sf{K.E \: = \: \frac{1}{2} \: m \: (\frac{p}{m})^2}}$

$\large \rightarrow {\sf{K.E \: = \: \frac{1}{2} \: \cancel{m} \: (\frac{p^2}{m^{\cancel{2}}})}}$

$\large \rightarrow {\sf{K.E \: = \: \frac{1}{2} \: \times \: \frac{p^2}{m}}}$

$\Large \implies {\sf{K.E \: = \: \frac{p^2}{2m}}}$

Where,

1/2m is constant

$\Large \implies {\boxed{\red{\sf{K.E \: \propto \: p^2}}}}$

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If Kinetic Energy is Directly proportional to p² , then if momentum is doubled then Kinetic Energy will also double .

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$\rule{200}{2}$

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Answer 2

As, we know formula for Kinetic Energy And formula for Momentum is :–

${\underline{\boxed{\tt{K.E \: = \: \frac{1}{2} \: mv^2}}}}$

$\green{\boxed{\tt{p \: = \: mv}}}$

Where,

p is momentum

m is mass

m is massv is velocity ,

$\bold \longrightarrow {\tt{v \: = \: \frac{p}{m}}}$

Put This value in formula of Kinetic Energy,

$\large \rightarrow {\tt{K.E \: = \: \frac{1}{2} \: m \: (\frac{p}{m})^2}}$

$\large \rightarrow {\tt{K.E \: = \: \frac{1}{2} \: \cancel{m} \: (\frac{p^2}{m^{\cancel{2}}})}}$

$\large {\tt{K.E \: = \: \frac{1}{2} \: \times \: \frac{p^2}{m}}}$

$\Large \implies {\tt{K.E \: = \: \frac{p^2}{2m}}}$

Where, 1/2m is constant

$\red \implies {\boxed{\red{\tt{K.E \: \propto \: p^2}}}}$

If Kinetic Energy is Directly proportional to p² , then if momentum is doubled then Kinetic Energy will also double .