question_answer

A particle of mass m is projected with velocity\[v\]making an angle of \[{{45}^{o}}\] with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be

A) \[2\,\,mv\]                        

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

C) \[mv\sqrt{2}\]                  

D)  zero

Correct Answer: C

Solution :

The horizontal momentum does not change. The change in vertical momentum is                 \[mv\sin \theta -(-mv\sin \theta )=2mv\frac{1}{\sqrt{2}}\]                 \[=\sqrt{2}mv\]