question_answer

A body of mass m is thrown with a velocity v making an angle \[45{}^\circ \] with horizontal. The magnitude of change in momentum between starting point and at maximum height is (neglect air resistance)

A)  \[\sqrt{2}\,m\,v\]                      

B)  2m v

C)  \[\frac{\sqrt{2}\,m\,v}{2}\]                   

D)  None of these

Correct Answer: C

Solution :

At point O,   \[{{\operatorname{v}}_{1}} = v cos\,\hat{i}\,\,+\,\,v\,sin\,\theta \,\hat{j}\] At point O, \[{{\operatorname{v}}_{f}} = v cos\,\hat{i}\,\,\theta \,\hat{i}\] Now,    \[\Delta p=m\left( {{v}_{f}}-{{v}_{i}} \right)\,\,=\,\,-mv\,\,sin\,\theta \,\hat{j}\] \[\left| \Delta p \right|\,\,=\,\,m v\,sin \theta \,\hat{j} =\,\,m\,v\,\,sin 45{}^\circ \] \[=\,\,\,\frac{mv}{\sqrt{2}}\,\,=\,\,\frac{\sqrt{2}mv}{2}\]