A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be:

A Particle of Mass m is Projected with Velocity v Making an Angle of 45°

The magnitude of the resultant velocity at the point of projection and the landing point is same.

Clearly, change in momentum along horizontal (i.e along x-axis)

$= m v \cos θ - m v \cos θ = 0$

Change in momentum along vertical (i.e. along y–axis) =

$m v \sin θ - (- m v \sin θ)$

$= 2 m v \sin θ = 2 m v \times \sin 45 °$

$= 2 mv \times \frac{1}{\sqrt{2}} = \sqrt{2} mv$

Hence, resultant change in momentum $= \sqrt{2} mv$