A 7150 kg truck traveling east at 5 m/s collides with a 1650 kg car traveling 30 degrees south of west at 20 m/s.   After the collision at what angle do the two vehicles move.   Express you answer in degrees from the positive x-axis.  Use a coordinate where north is the +y axis and east is the +x axis.

The answer is 293.4905300644 but I don't know how to get it. Thank you in advance.

initial east,x, momentum = 7150(5) -1650(20)cos 30

= 7,171

initial north,y, momentum = 0 - 1650(20)sin 30
= -16,500

Those will not change during the collision
so
tan of angle south of east = 16500/7171

angle south of east = 66.5 deg

Their answer, is 293.5 COUNTERCLOCKWISE from EAST (x axis direction), which would land on our line (360 - 66.5 = 293.5)

DRAW IT !!!!!
360 -293.5 = 66.5

By the way those of us who do navigation have trouble with x, y rather than east, north

THANK YOU!

To find the angle at which the two vehicles move after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.

First, let's find the initial momentum of each vehicle. The momentum (p) of an object is given by the product of its mass (m) and velocity (v): p = m * v.

For the truck:
Mass (m1) = 7150 kg
Velocity (v1) = 5 m/s
Initial momentum (p1) = m1 * v1 = 7150 kg * 5 m/s = 35750 kg*m/s

For the car:
Mass (m2) = 1650 kg
Velocity (v2) = 20 m/s
Initial momentum (p2) = m2 * v2 = 1650 kg * 20 m/s = 33000 kg*m/s

Next, let's analyze the collision. Since the problem does not provide information about the collision duration or any external forces acting on the system, we assume it is an elastic collision, where both momentum and kinetic energy are conserved.

After the collision, the two vehicles move together with a common final velocity (v_f) at an angle of θ from the positive x-axis. To find θ, we can calculate their final velocity components in the x and y directions.

Let's break down the velocities of the truck and the car into their x and y-components using trigonometry:

For the truck:
vx1 = v1 * cos(0°) = v1 * 1 = 5 m/s
vy1 = v1 * sin(0°) = v1 * 0 = 0 m/s

For the car:
vx2 = v2 * cos(30°)
vy2 = v2 * sin(30°)

Now, let's apply the conservation of momentum in the x and y directions:

In the x-direction:
m1 * vx1 + m2 * vx2 = (m1 + m2) * v_f * cos(θ)

In the y-direction:
m2 * vy2 = (m1 + m2) * v_f * sin(θ)

We can rearrange these equations to solve for v_f * cos(θ) and v_f * sin(θ):

v_f * cos(θ) = (m1 * vx1 + m2 * vx2) / (m1 + m2)
v_f * sin(θ) = m2 * vy2 / (m1 + m2)

Now we can find v_f * cos(θ) and v_f * sin(θ) using the values we have:

v_f * cos(θ) = (7150 kg * 5 m/s + 1650 kg * 20 m/s) / (7150 kg + 1650 kg)
v_f * sin(θ) = 1650 kg * 20 m/s * sin(30°) / (7150 kg + 1650 kg)

Calculating these values:
v_f * cos(θ) ≈ 16.0 m/s
v_f * sin(θ) ≈ 6.6941 m/s

Now, we can use these values to find the final velocity magnitude (v_f). The final velocity magnitude (|v_f|) is given by the Pythagorean theorem:

|v_f|^2 = (v_f * cos(θ))^2 + (v_f * sin(θ))^2

Substituting the values we found:
|v_f|^2 ≈ (16.0 m/s)^2 + (6.6941 m/s)^2
|v_f| ≈ 17.4916 m/s

Finally, we can find the angle θ from the positive x-axis using trigonometry:

θ = arctan(v_f * sin(θ) / v_f * cos(θ)) = arctan(6.6941 m/s / 16.0 m/s)

Evaluating this expression:
θ ≈ 24.6686°

Since the problem specifies that the angle should be expressed from the positive x-axis, we subtract this angle from 360°:

θ = 360° - 24.6686° ≈ 335.3314°

Therefore, the two vehicles move at an angle of approximately 335.3314° from the positive x-axis after the collision.