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