Question

idmf the kinetic energy of a body is halved what will be the change in momentum ​

Answer 1

Given:

Kinetic Energy of a body is halved.

To find:

Change in momentum of the object.

Calculation:

The general expression relating kinetic energy with Momentum is as follows:

$\boxed{ \sf{KE = \dfrac{ {p}^{2} }{2m} }}$

Here, p is momentum and m is mass of body.

So, for constant mass:

$\boxed{ \sf{KE \propto {p}^{2} }}$

Therefore, we can say that:

$\therefore \: \dfrac{KE_{2}}{KE_{1}} = \dfrac{ {(p_{2})}^{2} }{ {(p_{1})}^{2} }$

$= > \: \dfrac{ \frac{1}{2} \times KE_{1}}{KE_{1}} = \dfrac{ {(p_{2})}^{2} }{ {(p_{1})}^{2} }$

$= > \: \dfrac{1}{2} = \dfrac{ {(p_{2})}^{2} }{ {(p_{1})}^{2} }$

$= > \: \dfrac{1}{ \sqrt{2} } = \dfrac{ p_{2} }{ p_{1} }$

$= > \: p_{2} = \dfrac{p_{1}}{ \sqrt{2} }$

So, the change in momentum

$= > \:p_{1} - p_{2} =p_{1} ( 1 - \dfrac{1}{ \sqrt{2} } )$

$= > \:p_{1} - p_{2} =p_{1} ( \dfrac{ \sqrt{2} - 1}{ \sqrt{2} } )$

$= > \:\Delta P =p_{1} ( \dfrac{ \sqrt{2} - 1}{ \sqrt{2} } )$

So, final answer is:

$\boxed{ \bf{ \:\Delta P =p_{1} ( \dfrac{ \sqrt{2} - 1}{ \sqrt{2} } )}}$