Two arrows are fired horizontally with the same speed of 31.8 m/s. Each arrow has a mass of 0.149 kg. One is fired due east and the other due south. Find the magnitude of the total momentum of the two arrow system in kg-m/s. (Hint: How do you combine vector arrows at right angles?)
each vector has two components, most commonly in the x and y directions. So, you have
v = <31.8,0> + <0,-31.8> = <31.8,-31.8>
|v| = 31.8√2 = 44.97 m/s in a southeast direction, since tanθ = -31.8/31.8
so, the momentum p = mv
To find the total momentum of the two arrow system, we need to combine the momentum of each arrow separately.
Let's first calculate the momentum of the arrow fired due east. Since the arrow is fired horizontally, its vertical velocity component is zero. Thus, its momentum only has a horizontal component.
The momentum of an object is given by the formula: momentum = mass x velocity
For the arrow fired due east:
mass = 0.149 kg (given)
velocity = 31.8 m/s (given)
momentum of arrow fired east = mass x velocity
= 0.149 kg x 31.8 m/s
= 4.7322 kg·m/s (rounded to 4 decimal places)
Now, let's calculate the momentum of the arrow fired due south. Since the arrow is fired horizontally, its horizontal velocity component is zero. Thus, its momentum only has a vertical component.
momentum of arrow fired south = mass x velocity
= 0.149 kg x 31.8 m/s
= 4.7322 kg·m/s (rounded to 4 decimal places)
Now, since the arrows are fired at right angles to each other (east and south), their momentum vectors can be combined using the Pythagorean theorem.
To find the magnitude of the total momentum, we use the formula: magnitude = √(momentum east)^2 + (momentum south)^2
magnitude = √(4.7322 kg·m/s)^2 + (4.7322 kg·m/s)^2
= √(22.40929 kg^2·m^2/s^2) + (22.40929 kg^2·m^2/s^2)
= √(44.81858 kg^2·m^2/s^2)
= 6.694 kg·m/s (rounded to 3 decimal places)
So, the magnitude of the total momentum of the two arrow system is 6.694 kg·m/s.
To find the magnitude of the total momentum of the two arrow system, we need to calculate the momentum of each arrow and then combine them using vector addition.
The momentum of an object is given by the product of its mass and velocity: p = m * v.
For the arrow fired due east, its velocity vector will have a magnitude of 31.8 m/s in the east direction (positive x-axis) and a y-component of 0 m/s.
The momentum of this arrow can be calculated as follows:
p1 = m * v1 = (0.149 kg) * (31.8 m/s) = 4.7382 kg·m/s (in the positive x-direction).
For the arrow fired due south, its velocity vector will have a magnitude of 31.8 m/s in the south direction (negative y-axis) and an x-component of 0 m/s.
The momentum of this arrow can be calculated as follows:
p2 = m * v2 = (0.149 kg) * (-31.8 m/s) = -4.7382 kg·m/s (in the negative y-direction).
To combine the momenta of the two arrows at right angles, we use vector addition. Since the arrows are fired at right angles to each other, the magnitude of the total momentum will be the square root of the sum of the squares of the individual momenta.
Magnitude of the total momentum (ptotal) = sqrt((p1^2) + (p2^2))
= sqrt((4.7382^2) + (-4.7382^2))
= sqrt(22.4748 + 22.4748)
= sqrt(44.9496)
= 6.7066 kg·m/s.
Therefore, the magnitude of the total momentum of the two arrow system is 6.7066 kg·m/s.