Why is the square root of (v^2 + u^2) used in the equation for momentum?

[lordloss]

Homework Statement

a) KE=(1/2)(m)(v-u)2
Conversion: u = (31km/hr)(1hr/3600s)(1000m/km) = 8.61m/s
v = (64km/hr)(1hr/3600s)(1000m/km) = 17.78m/s
KE = (1/2)(1850kg)(17.782-8.612) = The Answer for a)

b)Momentum= m(√(v2+u2)), v=14.17i u=11.39j
= 1850(√(17.782+8.612)) = Answer for b)

c)tan θ = (8.61/17.78) = Degrees South of East

Homework Equations

p=mv

Momentum= m(√(v2+u2))

The Attempt at a Solution

What I can't figure out is why they are taking the square root of the two squared numbers, can anyone help explain this? The question is asking for the magnitude of the momentum, but I thought that was p=mv.

 

[rl.bhat]

When two momentum are perpendicular to each other, the resultant momentum is calculated by the above method.

 

[lordloss]

Wow. One reply can really make a world of difference. Thank You, it makes sense now.