A 1,500-kg car traveling east with a speed of 25.0 m/s collides at an intersection with a 2,500-kg van traveling north at a speed of 20.0 m/s. Find the direction and magnitude of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together). and assuming that friction between the vehicles and the road can be neglected. (v=15.6 m/s, = θ=53.1°)

Question

A 1,500-kg car traveling east with a speed of 25.0 m/s collides at an intersection with a 2,500-kg van traveling north at a speed of 20.0 m/s. Find the direction and magnitude of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together). and assuming that friction between the vehicles and the road can be neglected. (v=15.6 m/s, = θ=53.1°)

Answer 1

To find the direction and magnitude of the velocity of the wreckage after the collision, we can use the principles of conservation of momentum and the equations of vector addition.

1. Start by drawing a diagram to visualize the situation. Label the car's initial velocity as "v1" (east) and the van's initial velocity as "v2" (north).

2. Calculate the momentum (p1) of the car before the collision by multiplying its mass (m1 = 1500 kg) by its initial velocity (v1 = 25.0 m/s):
p1 = m1 * v1

3. Similarly, calculate the momentum (p2) of the van before the collision by multiplying its mass (m2 = 2500 kg) by its initial velocity (v2 = 20.0 m/s):
p2 = m2 * v2

4. Since the collision is perfectly inelastic, the two vehicles stick together after the collision. This means that their combined mass (m_combined) is the sum of the individual masses:
m_combined = m1 + m2

5. Next, calculate the total momentum (p_combined) after the collision using the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision:
p_combined = p1 + p2

6. Divide the total momentum after the collision (p_combined) by the combined mass (m_combined) to find the final velocity (v_combined) of the wreckage:
v_combined = p_combined / m_combined

7. Finally, use the Pythagorean theorem and trigonometry to calculate the magnitude (|v_combined|) and direction (θ) of the final velocity. The magnitude is given by:
|v_combined| = √(v_combined_x^2 + v_combined_y^2)
where v_combined_x is the horizontal component of the velocity and v_combined_y is the vertical component.

8. The direction (θ) can be found using the inverse tangent (arctan) function:
θ = arctan(v_combined_y / v_combined_x)

Substituting the given values:
- v1 = 25.0 m/s (east)
- v2 = 20.0 m/s (north)
- m1 = 1500 kg
- m2 = 2500 kg

Let's now calculate the final velocity and its direction:

First, calculate the momenta:
p1 = (1500 kg) * (25.0 m/s) = 37500 kg·m/s (east)
p2 = (2500 kg) * (20.0 m/s) = 50000 kg·m/s (north)

Next, find the total momentum after the collision:
p_combined = p1 + p2 = 37500 kg·m/s (east) + 50000 kg·m/s (north)

Then, calculate the combined mass:
m_combined = m1 + m2 = 1500 kg + 2500 kg = 4000 kg

Now, find the final velocity:
v_combined = p_combined / m_combined = 37500 kg·m/s (east) + 50000 kg·m/s (north) / 4000 kg
= (37500 kg·m/s + 50000 kg·m/s) / 4000 kg

Simplifying, we get:
v_combined = 87500 kg·m/s / 4000 kg = 21.875 m/s

Using the Pythagorean theorem and trigonometry, we can find the magnitude and direction of the velocity:

|v_combined| = √(v_combined_x^2 + v_combined_y^2)
= √((21.875 m/s)^2 + (21.875 m/s)^2)
= √(478.52 m^2/s^2 + 478.52 m^2/s^2)
= √957.04 m^2/s^2
≈ 30.94 m/s

θ = arctan(v_combined_y / v_combined_x) = arctan(21.875 m/s / 21.875 m/s) = arctan(1) = 45°

Therefore, the magnitude of the velocity of the wreckage after the collision is approximately 30.94 m/s, and the direction is 45° with respect to the east.