Two arrows are fired simultaneously with the same speed of 30.0 m/s. Each arrow has a mass of 0.100 kg. One is fired due east and the other due south. What is the magnitude of the total momentum of this two-arrow system?
V = Sqrt(30^2+30^2) =
Momentum = (M1+M2)*V.
Well, 30.0 m/s might sound fast for arrows, but I guess superheroes have to start somewhere. Anyway, to find the total momentum of the two-arrow system, we need to calculate the momentum of each arrow and then add them up.
For the arrow fired due east, we have:
Mass of arrow = 0.100 kg
Speed of arrow = 30.0 m/s
The momentum of this arrow can be calculated using the formula:
Momentum = mass × velocity
So, the momentum of the arrow fired due east is:
Momentum_east = 0.100 kg × 30.0 m/s
Now, for the arrow fired due south, since it's perpendicular to the east direction, the magnitudes of the momenta of the two arrows will simply add up, as momentum is a vector quantity.
So, the magnitude of the total momentum of the two-arrow system is:
Total_momentum = |Momentum_east| + |Momentum_south|
Momentum_east = 0.100 kg × 30.0 m/s
And as the magnitude of southward momentum would be equal to the eastward momentum:
Momentum_south = Momentum_east
Therefore,
Total_momentum = 2 × Momentum_east
Total_momentum = 2 × (0.100 kg × 30.0 m/s)
Total_momentum = 2 × 3.00 kg·m/s
Total_momentum = 6.00 kg·m/s
So, the magnitude of the total momentum of this two-arrow system will be 6.00 kg·m/s. I hope these arrows hit their targets with that kind of momentum!
To find the total momentum of the two-arrow system, we need to calculate the momentum of each arrow separately and then add them together.
The momentum of an object is given by the formula:
Momentum (p) = mass (m) × velocity (v)
For the arrow fired due east:
Mass (m1) = 0.100 kg
Velocity (v1) = 30.0 m/s
Momentum of arrow 1 (p1) = m1 × v1
Momentum of arrow 1 (p1) = 0.100 kg × 30.0 m/s
Momentum of arrow 1 (p1) = 3.00 kg·m/s
For the arrow fired due south:
Mass (m2) = 0.100 kg
Velocity (v2) = 30.0 m/s
Momentum of arrow 2 (p2) = m2 × v2
Momentum of arrow 2 (p2) = 0.100 kg × 30.0 m/s
Momentum of arrow 2 (p2) = 3.00 kg·m/s
Now, to find the total momentum, we add the momentum of both arrows:
Total momentum = p1 + p2
Total momentum = 3.00 kg·m/s + 3.00 kg·m/s
Total momentum = 6.00 kg·m/s
Therefore, the magnitude of the total momentum of the two-arrow system is 6.00 kg·m/s.
To find the magnitude of the total momentum of the two-arrow system, we need to calculate the momentum for each arrow and then add them up.
The momentum of an object is given by the product of its mass and velocity. The formula for momentum is:
Momentum = mass x velocity
For the arrow fired due east, we have:
Mass (m1) = 0.100 kg
Velocity (v1) = 30.0 m/s
Using the formula, the momentum of this arrow is:
Momentum1 = m1 x v1 = (0.100 kg) x (30.0 m/s) = 3.00 kg·m/s
For the arrow fired due south, we have:
Mass (m2) = 0.100 kg
Velocity (v2) = 30.0 m/s
Using the formula, the momentum of this arrow is:
Momentum2 = m2 x v2 = (0.100 kg) x (30.0 m/s) = 3.00 kg·m/s
Now, to find the total momentum of the two-arrow system, we add the individual momenta together:
Total momentum = Momentum1 + Momentum2 = 3.00 kg·m/s + 3.00 kg·m/s = 6.00 kg·m/s
Therefore, the magnitude of the total momentum of this two-arrow system is 6.00 kg·m/s.