3. Two vehicles collide at a 90 degree intersection. If the momentum of vehicle A is
6.10 x10^5 kg km/h south and the momentum of vehicle B is 7.20 x 10^5 kg km/h
east, what is the magnitude of the resulting momentum of the final mass?
M1*V1 + M2*V2 = -6.10*10^5i + 7.20*10^5.
Tan A = Y/X = (-6.10*10^5)/7.20*10^5 = -0.84722, A = -40.3o = 40.3o S. of E.
Resulting momentum =
X/Cos A = 7.20^10^5/Cos(-40.3) =
To find the magnitude of the resulting momentum of the final mass, we need to combine the momenta of the two vehicles. Since the momenta of the two vehicles are given in different directions, we need to use vector addition.
Let's break down the given momenta into their corresponding x and y components:
For vehicle A:
Momentum of vehicle A = 6.10 x 10^5 kg km/h south
To convert this momentum to x and y components, we need to consider that south is in the negative y direction. So, the y component of the momentum of vehicle A is -6.10 x 10^5 kg km/h.
For vehicle B:
Momentum of vehicle B = 7.20 x 10^5 kg km/h east
To convert this momentum to x and y components, we need to consider that east is in the positive x direction. So, the x component of the momentum of vehicle B is 7.20 x 10^5 kg km/h.
Now, we can add the x and y components of the momenta to find the resulting momentum.
Sum of x components = 7.20 x 10^5 kg km/h
Sum of y components = -6.10 x 10^5 kg km/h
To find the magnitude of the resulting momentum, we can use the Pythagorean theorem:
Magnitude of the resulting momentum = √((Sum of x components)^2 + (Sum of y components)^2)
Magnitude of the resulting momentum = √((7.20 x 10^5)^2 + (-6.10 x 10^5)^2)
Magnitude of the resulting momentum ≈ 9.49 x 10^5 kg km/h
So, the magnitude of the resulting momentum of the final mass is approximately 9.49 x 10^5 kg km/h.