Two motorcycles of equal mass collide at a 90� intersection. If the momentum of motorcycle
A is 450 kg km/h west and the momentum of motorcycle B is 725 kg km/h
south, what is the magnitude of the resulting momentum of the final mass?
To find the magnitude of the resulting momentum of the final mass after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Let's break down the given momenta of motorcycle A and B into their horizontal (x) and vertical (y) components:
Momentum of motorcycle A:
- Magnitude: 450 kg km/h
- Direction: West
Momentum of motorcycle B:
- Magnitude: 725 kg km/h
- Direction: South
To find the x-component of momentum A, we need to multiply its magnitude by the cosine of the angle between its direction and the x-axis:
Momentum A (x-component) = 450 kg km/h * cos(90°) = 0 kg km/h
Similarly, to find the y-component of momentum B, we multiply its magnitude by the cosine of the angle between its direction and the y-axis:
Momentum B (y-component) = 725 kg km/h * cos(90°) = -725 kg km/h
Now, let's calculate the total x-component and y-component of momentum before the collision:
Total momentum (x-component) = Momentum A (x-component) + Momentum B (x-component) = 0 kg km/h + 0 kg km/h = 0 kg km/h
Total momentum (y-component) = Momentum A (y-component) + Momentum B (y-component) = 0 kg km/h + (-725 kg km/h) = -725 kg km/h
Since momentum is a vector, we need to calculate the magnitude of the resulting momentum vector after the collision:
Magnitude of the resulting momentum = √((Total momentum (x-component))^2 + (Total momentum (y-component))^2)
Magnitude of the resulting momentum = √((0 kg km/h)^2 + (-725 kg km/h)^2)
Magnitude of the resulting momentum = √(0 + 526,562.5) = √526,562.5 ≈ 725 kg km/h
Therefore, the magnitude of the resulting momentum of the final mass after the collision is approximately 725 kg km/h.