Question

A projectile of mass $m$ is fired with velocity a $v$ from a point $P$ making an angle of $45^\circ $ with the horizontal. Neglecting air resistance, the magnitude of the change in momentum between the points $P$ and arriving at $Q$ is

• A. zero
• B. $\frac{1}{2}mv$
• C. $mv\sqrt{2}$
• D. $2 \, mv$

Answer 1

Answer: (C)

Since there is no resistance, the trajectory of motion will be parabolic, and as a result, the angle of projection will be equal to the angle of striking the ground.
Also, we know that the horizontal component of velocity remains unaffected throughout the motion. So it will not contribute to the change in momentum. The components contributing to the change in momentum will be the vertical components.
Initial vertical momentum component is $mvsin45^\circ $ and the final vertical momentum component is $-\left(mvsin 45 ^\circ \right)$ .
Change in momentum, $\Delta p=mvsin 45^\circ -\left(- mvsin ⁡ 45 ^\circ \right)=2mvsin⁡45^\circ = \, 2 \, mv\frac{1}{\sqrt{2}}=\sqrt{2}mv$ .