question_answer

A particle of mass m is projected with velocity v making an angle of \[{{45}^{o}}\] with the horizontal. [AIPMPT (S) 2008]

When the particle lands on the level ground the magnitude of the change in its momentum will be

A)                                  2mv     

B) \[mv/\sqrt{2}\]

C) \[mv\sqrt{2}\] 

D) zero

Correct Answer: C

Solution :

Key Idea: Required momentum is the difference of final and initial momentum.

The situation is shown in figure.

Change in momentum

\[\Delta \vec{P}={{\vec{P}}_{f}}-{{\vec{P}}_{i}}\]

\[=m({{\vec{v}}_{f}}-{{\vec{v}}_{i}})\]

\[=m[v\cos {{45}^{o}}\hat{i}-v\sin {{45}^{o}}\hat{j})\]

\[-(v\cos {{45}^{o}}\hat{i}+v\sin {{45}^{o}}\hat{j})]\]

\[=m\left[ \left( \frac{v}{\sqrt{2}}\hat{i}-\frac{v}{\sqrt{2}}\hat{j} \right)-\left( \frac{v}{\sqrt{2}}\hat{i}+\frac{v}{\sqrt{2}}\hat{j} \right) \right]\]

\[=-\sqrt{2}mv\,\,\hat{j}\]

\[\therefore \]      \[[\Delta \vec{P}]=\sqrt{2}mv\]

Alternative:

The horizontal momentum does not change.

The change in vertical momentum is

\[mv\sin \theta -(-mv\sin \theta )=2mv\frac{1}{\sqrt{2}}=\sqrt{2}mv\]