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

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SUMMARY

The equation for momentum, expressed as Momentum = m(√(v² + u²)), is utilized to calculate the magnitude of momentum when two velocities are perpendicular. In this case, the velocities are converted from kilometers per hour to meters per second, with u = 8.61 m/s and v = 17.78 m/s. The calculation involves using the mass of 1850 kg to derive the momentum. Understanding the use of the square root in this context is crucial for grasping how to combine perpendicular vectors in physics.

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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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