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

  • Thread starter Thread starter lordloss
  • Start date Start date
  • Tags Tags
    Momentum
AI Thread Summary
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.
lordloss
Messages
7
Reaction score
1

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:
Physics news on Phys.org
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.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Conservation of momentum & energy
Replies
11
Views
2K
Homework in physics (average force and momentum)
Replies
3
Views
3K
Two dimensional momentum problem
Replies
20
Views
3K
Calculate change in KE and momentum (impulse)
Replies
7
Views
4K
Why Use Kinetic Energy Instead of Momentum to Stop a Car?
Replies
6
Views
2K
Unit Conversion: 5m/s^2 to Km/hr^2 & ft/hr^2
Replies
1
Views
2K
2-D Linear Momentum, Collision Problem
Replies
1
Views
2K
Speeds regarding 2D inelastic collision
Replies
8
Views
3K
Momentum - two cars crashing at an angle
Replies
2
Views
3K
Momentum collision in two dimensions
Replies
6
Views
2K

Hot Threads

Recent Insights

Back
Top