Question

A particle of mass m is projected with a velocity v making an angle 45 with the horizontal. the particle lands on ground magnitude of change in momentum

Answer 1

Horizontal velocity does not undergo change  during projectile motion.

So, required change in momentum will be of that of vertical momentum.

Let

ΔPₓ be  change in vertical motion

ΔPₓ = Pₓ (final) - Pₓ (initial)

Initial momentum = mv sinθ

Final momentum = -mvsinθ (as direction of vertical velocity changes)

ΔPₓ = -mvsinθ - mvsinθ

      = -2mvsinθ

      = -2mvsin45⁰

      = -√2 mv

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