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

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The discussion focuses on the use of the square root of the sum of the squares of two velocities in calculating momentum. This method is necessary when dealing with perpendicular components of momentum, as it allows for the determination of the resultant momentum's magnitude. The equation p = mv is standard, but when velocities are in different directions, the Pythagorean theorem is applied to find the overall momentum. The conversion of speeds from kilometers per hour to meters per second is also highlighted as part of the calculations. Understanding this approach clarifies how to accurately compute momentum in multi-dimensional scenarios.
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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.
 
Last edited:
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When two momentum are perpendicular to each other, the resultant momentum is calculated by the above method.
 
Wow. One reply can really make a world of difference. Thank You, it makes sense now.
 

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