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