Question

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

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

Answer 1

The correct option is D $mv\sqrt{2}$

We know,

The initial angle and final Angle with the horizontal in a projectile

motion, they are same in magnitude and opposite in direction.

Hence, the initial and final velocity are at angle of $(45-(-45\left)\right)={90}^0$.

The difference between the momentums is the change in momentum,

$=m(\overrightarrow{v_1}-\overrightarrow{v_2})$

Where, $|v_1|=|v_2|$

Change in momentum$=mv\sqrt{2}$

Option $\text{D}$ is the correct answer