Two objects are approaching an intersection. One is a 50 kg traveling at 8

m/s from east to west and the other is a 40 kg going from south to north at 12 m/s.
(a) Find the x and y components of the net momentum of this system.
(b) What are the magnitude and direction of the net momentum

a. X = m1V1 = 50 * -8 = -400.

Y = m2V2 =40 * 12 = 480.

b. tanA = Y/X = 480/-400 = -1.20.
Ar = -50.19 = Reference angle.
A = -50.19 + 180 = 129.8o = Direction.

Magnitude = X/cosA = -400/cos129.8 = 625.

To solve this problem, we need to understand the concepts of momentum and vector components.

(a) Find the x and y components of the net momentum of this system:
The momentum of an object is given by the product of its mass and velocity. In this case, we have two objects, so we need to calculate the momentum for each object separately.

Momentum (p1) of the first object:
Mass (m1) = 50 kg
Velocity (v1) = 8 m/s

To find the x-component, we multiply the velocity by the cosine of the angle between the velocity vector and the x-axis. Since the object is traveling from east to west, the angle is 180 degrees or π radians.

x-component of p1 = m1 * v1 * cos(π) = 50 kg * 8 m/s * cos(π) = -400 kg·m/s

To find the y-component, we multiply the velocity by the sine of the angle between the velocity vector and the y-axis. Since the object is not moving in the y-direction, the angle is 90 degrees or π/2 radians.

y-component of p1 = m1 * v1 * sin(π/2) = 50 kg * 8 m/s * sin(π/2) = 400 kg·m/s

Momentum (p2) of the second object:
Mass (m2) = 40 kg
Velocity (v2) = 12 m/s

For the second object, the x-component is 0 because it is not moving in the x-direction.

x-component of p2 = m2 * v2 * cos(0) = 0

To find the y-component, we multiply the velocity by the sine of the angle between the velocity vector and the y-axis. Since the object is traveling from south to north, the angle is 90 degrees or π/2 radians.

y-component of p2 = m2 * v2 * sin(π/2) = 40 kg * 12 m/s * sin(π/2) = 480 kg·m/s

Now, we can calculate the net momentum by summing up the x and y components of the momentum of both objects.

Net x-component of momentum = x-component of p1 + x-component of p2 = -400 kg·m/s + 0 = -400 kg·m/s
Net y-component of momentum = y-component of p1 + y-component of p2 = 400 kg·m/s + 480 kg·m/s = 880 kg·m/s

Therefore, the x and y components of the net momentum of the system are -400 kg·m/s and 880 kg·m/s, respectively.

(b) What are the magnitude and direction of the net momentum:
To find the magnitude of the net momentum, we can use the Pythagorean theorem:

Magnitude of net momentum = √(Net x-component of momentum)^2 + (Net y-component of momentum)^2
= √((-400 kg·m/s)^2 + (880 kg·m/s)^2)
= √(160000 kg^2·m^2/s^2 + 774400 kg^2·m^2/s^2)
= √(934400 kg^2·m^2/s^2)
≈ 966.28 kg·m/s

To find the direction of the net momentum, we can use trigonometry:

Direction of net momentum = atan(Net y-component of momentum / Net x-component of momentum)
= atan(880 kg·m/s / -400 kg·m/s)
≈ atan(-2.2)

The magnitude of the net momentum is approximately 966.28 kg·m/s, and the direction is approximately atan(-2.2).