Question

A particle of mass m is projected with velocity v making of an angle of 45 degree with the horizontal. when the particle lands on the level ground, then what is the magnitude of the change in its momentum?

Answer 1

Answer:

change in momentum will be only along the y axis. so net change will be

2mv sin45°

Answer 2

Answer:

$\sqrt{2}$ mv

Explanation:

Here the particle follows a projectile motion.

In projectile Motion the velocity along the x-axis does not change.

∴ Change in momentum along x-axis is zero.

we will only consider the velocity along the y-axis to get the change in momentum of the particle.

so,

velocity along the y-axis = vsin45°

|ΔP|= m(Δv)

|ΔP|= m[vsin45°-(-vsin45°)]

|ΔP|=2×mv×sin45°

|ΔP|=2×mv×1/$\sqrt{2}$

|ΔP|= $\sqrt{2}$mv