Question

what is the change in momentum between the initial and final points of the projectile path,if the range is maximum​

Answer 1

To find:

Change in momentum between the initial and final points of the projectile path,if the range is maximum.

Calculation:

For max range: $\theta$ = 45°

Initial momentum in y axis:

$= m \{v \sin( \theta) \} \: \hat{j}$

$= m \{v \sin( {45}^{ \circ} ) \} \: \hat{j}$

$= \dfrac{mv}{ \sqrt{2} } \: \hat{j}$

Initial momentum in x axis:

$= m \{v \cos( \theta) \} \: \hat{i}$

$= m \{v \cos( {45}^{ \circ} ) \} \: \hat{i}$

$= \dfrac{mv}{ \sqrt{2} } \: \hat{i}$

Final momentum in Y axis:

$= m \{v \sin( \theta) \} \: (- \hat{j})$

$= m \{v \sin( {45}^{ \circ} ) \} \: (- \hat{j})$

$= \dfrac{mv}{ \sqrt{2} } \: (- \hat{j})$

Final momentum in x axis:

$= m \{v \cos( \theta) \} \: \hat{i}$

$= m \{v \cos( {45}^{ \circ} ) \} \: \hat{i}$

$= \dfrac{mv}{ \sqrt{2} } \: \hat{i}$

So, change in momentum

$\Delta P = ( \dfrac{mv}{ \sqrt{2} } - \dfrac{mv}{ \sqrt{2} } ) \: \hat{i} + ( - \dfrac{mv}{ \sqrt{2} } - \dfrac{mv}{ \sqrt{2} } ) \: \hat{j}$

$= > \Delta P = ( 0 ) \: \hat{i} + ( - \dfrac{2mv}{ \sqrt{2} } ) \: \hat{j}$

$= > \Delta P = \dfrac{2mv}{ \sqrt{2} } \: ( - \hat{j})$

$=> \Delta P = \sqrt{2}mv (-\hat{j})$

So, final answer is:

$\boxed{ \bf{\Delta P = \sqrt{2}mv \: ( - \hat{j})}}$