Question

How will the momentum of a body changes if its K.E. is doubled

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the Concept of Kinetic Energy and Momentum Ratio has been used. Firstly we need to find the relation between Momentum and Kinetic Energy of the body with normal Kinetic Energy. After that we shall find relation between Momentum when Kinetic Energy us doubled. Finally we can find the increase in Momentum.

______________________________________

★ Formula Used :-

$\\\;\boxed{\sf{Kinetic\;Energy,\;K.E.\;=\;\bf{\dfrac{1}{2}\;\times\;mv^{2}}}}$

$\\\;\boxed{\sf{Momentum,\;p\;=\;\bf{mv}}}$

______________________________________

★ Solution :-

• Let the mass of the body be 'm'

• Let the velocity of the body be 'v'

______________________________________

~ Original Kinetic Energy and Momentum ::

$\\\;\;\sf{:\rightarrow\;\;Kinetic\;Energy,\;K.E.\;=\;\bf{\dfrac{1}{2}\;\times\;mv^{2}}}$

$\\\;\;\bf{:\rightarrow\;\;K.E._{(Original)}\;=\;\bf{\dfrac{1}{2}\;\times\;mv^{2}}}$

$\\\;\;\sf{:\Longrightarrow\;\;Momentum,\;p\;=\;\bf{mv}}$

$\\\;\;\bf{:\Longrightarrow\;\;p_{(Original)}\;=\;\bf{mv}}$

→ Relation :-

$\\\;\;\sf{:\rightarrow\;\;K.E._{(Original)}\;=\;\bf{\dfrac{1}{2}\;\times\;mv\;\times\;v}}$

By applying Momentum, it can be written as,

$\\\;\;\sf{:\rightarrow\;\;K.E._{(Original)}\;=\;\bf{\dfrac{1}{2}\;\times\;p_{(Original)}\;\times\;v}}$

$\\\;\;\bf{:\rightarrow\;\;p_{(Original)}\;=\;\bf{\dfrac{K.E._{(Original)}\;\times\;2}{v}}}$

$\\\;\;\bf{:\rightarrow\;\;p_{(Original)}\;\times\;v\;=\;\bf{K.E._{(Original)}\;\times\;2}}$

______________________________________

~ Momentum when Kinetic Energy is Doubled ::

$\\\;\;\sf{:\rightarrow\;\;Kinetic\;Energy,\;K.E.\;=\;\bf{\dfrac{1}{2}\;\times\;mv^{2}}}$

Now multiplying and dividing all the terms by m, we get

$\\\;\;\sf{:\rightarrow\;\;Kinetic\;Energy,\;K.E.\;=\;\bf{\dfrac{1}{2}\;\times\;mv^{2}\;\times\;\dfrac{m}{m}}}$

$\\\;\;\sf{:\rightarrow\;\;Kinetic\;Energy,\;K.E.\;=\;\bf{\dfrac{1}{2m}\;\times\;m^{2}\:v^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;Kinetic\;Energy,\;K.E.\;=\;\bf{\dfrac{1}{2m}\;\times\;(mv)^{2}}}$

We know that, p = mv

$\\\;\;\sf{:\rightarrow\;\;Kinetic\;Energy,\;K.E.\;=\;\bf{\dfrac{1}{2m}\;\times\;(p)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;p^{2}\;=\;\bf{2\:m\:K.E.}}$

$\\\;\;\sf{:\rightarrow\;\;p\;=\;\bf{\sqrt{2\:m\:K.E.}}}$

Now when the Kinetic Energy is doubled that is 2KE.

$\\\;\;\sf{:\rightarrow\;\;p_{(changed)}\;=\;\bf{\sqrt{2\:m\:2\:K.E.}}}$

$\\\;\;\sf{:\rightarrow\;\;p_{(changed)}\;=\;\bf{\sqrt{4\:m\:\;\times\;\dfrac{1}{2}\;\times\;mv^{2}}}}$

$\\\;\;\sf{:\rightarrow\;\;p_{(changed)}\;=\;\bf{\sqrt{2\:m^{2}v^{2}}}}$

$\\\;\;\sf{:\rightarrow\;\;p_{(changed)}\;=\;\bf{\sqrt{2\:(mv)^{2}}}}$

$\\\;\;\sf{:\rightarrow\;\;p_{(changed)}\;=\;\bf{\sqrt{2\:(p_{(original)})^{2}}}}$

$\\\;\;\sf{:\rightarrow\;\;p_{(changed)}\;=\;\bf{p_{(original)}\;\sqrt{2}}}$

This means, Momentum increases by the rate of √2 when Kinetic Energy is doubled.

$\\\;\underline{\boxed{\tt{Thus,\;\;increase\;\;in\;\;Momentum\;\;=\;\bf{\blue{\sqrt{2}\;\;times}}}}}$

______________________________________

★ More Formulas to know :

$\\\;\sf{\leadsto\;\;\green{Force\;=\;ma}}$

$\\\;\sf{\leadsto\;\;\green{Potential\;Energy\;=\;mgh}}$

$\\\;\sf{\leadsto\;\;\green{Workdone\;=\;Force\;\times\;Distance}}$

$\\\;\sf{\leadsto\;\;\green{Power\;=\;\dfrac{Workdone}{Time}}}$

$\\\;\sf{\leadsto\;\;\green{Pressure\;=\;\dfrac{Force}{Area}}}$

Answer 2

yo yo honey singh ah uh bzshajw.

$\frac{4}{8} = 0.5$

i am right answer na brooo