A car with a mass of 980 kg is initially traveling east toward an intersection with a speed of vc = 18.9 m/s and a 1500 kg pickup is traveling north toward the same intersection. The car and truck collide at the intersection and stick together. After the collision, the wreckage (car and truck) moves off in a direction of 40.0° above the x-axis. Determine the initial speed of the truck and the final speed of the wreckage.

initial speed of the truck m/s
final speed of the wreckage m/s

sex

17 m/s

To solve this problem, we can use the principle of conservation of momentum. Before the collision, the momentum in the x and y directions are:

px_car = m_car * V_carx
py_car = m_car * V_cary

px_truck = m_truck * V_truckx
py_truck = m_truck * V_trucky

After the collision, the momentum in the x and y directions will be:

px_wreckage = (m_car + m_truck) * V_wreckagex
py_wreckage = (m_car + m_truck) * V_wreckagey

According to the principle of conservation of momentum, the total momentum in the x and y directions is conserved. Therefore, we can write:

px_car + px_truck = px_wreckage
py_car + py_truck = py_wreckage

Using the given values:

m_car = 980 kg
V_carx = 18.9 m/s
m_truck = 1500 kg
V_trucky = ?

First, let's calculate V_truckx:

px_car = m_car * V_carx
px_truck = m_truck * V_truckx

px_car + px_truck = px_wreckage
m_car * V_carx + m_truck * V_truckx = (m_car + m_truck) * V_wreckagex

Plugging in the values:

980 kg * 18.9 m/s + 1500 kg * V_truckx = (980 kg + 1500 kg) * V_wreckagex

From this equation, we can solve for V_truckx:

V_truckx = (980 kg * 18.9 m/s + 1500 kg * V_truckx) / (980 kg + 1500 kg)

Simplifying this equation gives:

V_truckx = (18642 kg·m/s + 1500 kg * V_truckx) / 2480 kg

Now, let's calculate V_wreckagey:

py_car = m_car * V_cary
py_truck = m_truck * V_trucky

py_car + py_truck = py_wreckage
m_car * V_cary + m_truck * V_trucky = (m_car + m_truck) * V_wreckagey

From the given problem statement, we know that V_wreckagey = 0 (the wreckage moves in the x direction only). Therefore, we can write:

980 kg * V_cary + 1500 kg * V_trucky = 0

Plugging in the values:

980 kg * 0 + 1500 kg * V_trucky = 0

Simplifying this equation gives:

1500 kg * V_trucky = 0

From this equation, we can see that V_trucky = 0.

Therefore, the initial speed of the truck is 0 m/s (since it is stationary) and the final speed of the wreckage is equal to the magnitude of V_wreckagex.

To solve this problem, we need to use principles of conservation of momentum and analyze the collision using vector components.

Let's break down the information given:

Car mass (m1) = 980 kg
Car speed (v1) = 18.9 m/s (Eastward)
Pickup truck mass (m2) = 1500 kg

We need to determine the initial speed of the truck (v2) and the final speed of the wreckage (vf).

Now, let's approach this problem using the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.

Before the collision:
Total momentum = m1 * v1 (momentum of the car) + m2 * v2 (momentum of the truck)

After the collision:
Total momentum = (m1 + m2) * vf (momentum of the wreckage)

Now, let's calculate the momentum before and after the collision using vector components.

Before the collision:
The momentum of the car (m1 * v1) is directed only in the East direction, while the momentum of the truck (m2 * v2) is directed only in the North direction.

m1 * v1 = (980 kg) * (18.9 m/s) (East)
m2 * v2 = (1500 kg) * v2 (North)

After the collision, the wreckage moves off in a direction 40.0° above the x-axis. We can represent vf as vx and vy (horizontal and vertical components).

vf = vx + vy

Now we can set up equations to solve for v2 and vf.

In the x-direction:
m1 * v1 = (m1 + m2) * vx

In the y-direction:
0 = (m1 + m2) * vy

Now we can solve for v2 and vf using the given information.

Step 1: Solve for v2
From the x-direction equation:
m1 * v1 = (m1 + m2) * vx
(980 kg) * (18.9 m/s) = (980 kg + 1500 kg) * vx
18612 kg*m/s = 2480 kg * vx
vx = 18612 kg*m/s / 2480 kg
vx ≈ 7.50 m/s

Step 2: Solve for vy
From the y-direction equation:
0 = (m1 + m2) * vy
0 = (980 kg + 1500 kg) * vy
0 = 2480 kg * vy
vy = 0 m/s

Step 3: Solve for vf
Using vf = vx + vy:
vf = 7.50 m/s + 0 m/s
vf = 7.50 m/s

Therefore, the initial speed of the truck (v2) is approximately 7.50 m/s, and the final speed of the wreckage (vf) after the collision is approximately 7.50 m/s.