Q. A projectile of mass $m$ is thrown with a velocity $v$ making an angle of $45{^\circ}$ with the horizontal. The change in momentum from departure to arrival along vertical direction, is
A) $2mv$
B) $\sqrt{2}\ mv$
C) $mv$
D) $\frac{mv}{2}$
Explanation
(b) Change in momentum is the product of force and time
$\mathrm{\Delta}p = mg \times \frac{2\sin\theta}{g}$
$\text{=}2mv\sin\theta = 2mv\sin{45{^\circ}} = \frac{2mv}{\sqrt{2}} = \sqrt{2}\ mv$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.