Two cars collided at an intersection. Car A (2000. [kg]), is going from west to east, while car B (1500. [kg]),

is going from north to south at 15.0 [m/s]. As a result, the two cars become enmeshed and moved as one.

You inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65.0°

south of east from the point of impact. How fast was car A going just before the collision?

To determine the speed of Car A just before the collision, we can apply the principle of conservation of momentum. Since the cars become enmeshed after the collision, the final momentum of the system is the combined momentum of both cars.

### Step 1: Set up the momentum equations
Momentum is conserved in both the east-west (x) and north-south (y) directions.

Let:
- \( m_A = 2000 \, \text{kg} \) (mass of car A),
- \( v_A \) (speed of car A, which we need to find),
- \( m_B = 1500 \, \text{kg} \) (mass of car B),
- \( v_B = 15.0 \, \text{m/s} \) (speed of car B),
- \( v_f \) be the final velocity of the enmeshed cars after the collision,
- \( \theta = 65^\circ \) (angle at which the cars move after the collision, south of east).

The x-direction momentum equation (east-west) before the collision:
\[
m_A v_A = (m_A + m_B) v_f \cos(\theta)
\]

The y-direction momentum equation (north-south) before the collision:
\[
m_B v_B = (m_A + m_B) v_f \sin(\theta)
\]

### Step 2: Solve for the final velocity \(v_f\)
From the second equation, we can solve for \( v_f \) as follows:
\[
v_f = \frac{m_B v_B}{(m_A + m_B) \sin(\theta)}
\]

Substituting the given values:
\[
v_f = \frac{1500 \times 15.0}{(2000 + 1500) \sin(65^\circ)} 
= \frac{22500}{3500 \times 0.9063} 
= \frac{22500}{3172.05} \approx 7.09 \, \text{m/s}
\]

### Step 3: Solve for the speed of Car A
Now, use this \( v_f \) value in the first equation to solve for \( v_A \):
\[
m_A v_A = (m_A + m_B) v_f \cos(\theta)
\]
\[
v_A = \frac{(m_A + m_B) v_f \cos(\theta)}{m_A}
\]

Substitute the known values:
\[
v_A = \frac{(2000 + 1500) \times 7.09 \times \cos(65^\circ)}{2000}
\]
\[
v_A = \frac{3500 \times 7.09 \times 0.4226}{2000}
\]
\[
v_A \approx \frac{10507.9}{2000} \approx 5.25 \, \text{m/s}
\]

### Conclusion:
The speed of Car A just before the collision was approximately **5.25 m/s**.

Would you like more details or have any questions?

Here are 5 related questions for you to explore:

1. How would the outcome change if Car A had a different mass?
2. What if the cars had bounced off each other instead of becoming enmeshed?
3. How is momentum conserved in real-world collisions involving deformation?
4. Could the angle of collision change if one car were turning?
5. How would the calculation differ if one car were accelerating?

**Tip:** Always check the direction of velocities in momentum problems; vectors play a crucial role in solving them.

Two cars collided at an intersection. Car A (2000 [kg]), is going from west to east, while car B (1500 [kg]), is going from north to south at 15.0 [m/s]. As a result, the two cars become enmeshed and moved as one. You inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65.0° south of east from the point of impact. How fast was car A going just before the collision?

Calculate the speed of Car A before a collision using conservation of momentum. Car A (2000 kg) collided with Car B (1500 kg) at an intersection, and both moved together at an angle of 65° south of east. Learn how to apply vector components and momentum equations to solve this real-world physics problem.

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