question_answer

A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. 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 \[mv\sin \theta -(-mv\sin \theta )=2mv\frac{1}{\sqrt{2}}=\sqrt{2}mv\]