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 \[mv\sin \theta -(-mv\sin \theta )=2mv\frac{1}{\sqrt{2}}=\sqrt{2}mv\]You need to login to perform this action.
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