A particle of mass m is projected with velocity v . . .

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

a) mv√2

b) Zero

c) 2mv

d) mv/√2

Doubt by Jaskirat

Solution : 

Momentum is a vector quantity. It is the product of mass and velocity. In angular projectile, there are two components of velocity one is horizontal component while other is vertical component. 

Change in Momentum is Horizontal direction : 

In horizontal the direction, the component of velocity (ucosθ) remains the same. So final velocity and initial velocity in the horizontal direction will remain same. So

Δp_x = mΔv
Δp_x = m(v_x-u_x)

Δp_x = m(ucosθ-ucosθ)
Δp_x = 0

Now, in vertical direction the initial component of velocity will be usinθ, but when the projectile hits back to the ground then it's will strike with the same vertical velocity usinθ but in the opposite direction. So
v_y=-usinθ
v_x=usinθ

Now change in momentum in vertical direction will be 

Δp_y = mΔv
Δp_y = m(v_y-u_y)

Δp_y = m(-usinθ-usinθ)
Δp_y = m[-2usinθ]
Δp_y = m[-2usin45°] [θ=45°]
Δp_y = m[-2u(1/√2)]
Δp_y = m[-u√2]
Δp_y = -mu√2
|Δp_y| = mu√2

Hence, a) would be the correct option.