Question

A particle of mass m is projected with velocity v making an angle of 450 with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be :

Answer 1

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