Questions: Two cars are about to collide. Car A has a mass of 1240 kg, and is traveling north at a speed of 18.3 m / s. Car B has a mass of 1990 kg, and is traveling south at a speed of 22.5 m / s. After the collision, the two cars become entangled, and the cars move as a single unit. Immediately after the collision, what will be the velocity of the combined unit? -5.04 m / s 4.56 m / s 3.16 m / s 9.49 m / s

Solution Steps

• The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.

• Calculate the momentum of Car A: \( p_A = m_A \cdot v_A \) \[ p_A = 1240 \, \text{kg} \times 18.3 \, \text{m/s} = 22692 \, \text{kg} \cdot \text{m/s} \]
• Car B is initially at rest, so its momentum is zero: \( p_B = 0 \)

• Total initial momentum: \[ p_{\text{total initial}} = p_A + p_B = 22692 \, \text{kg} \cdot \text{m/s} + 0 = 22692 \, \text{kg} \cdot \text{m/s} \]

• Combined mass of Car A and Car B: \[ m_{\text{combined}} = m_A + m_B = 1240 \, \text{kg} + 1990 \, \text{kg} = 3230 \, \text{kg} \]

• Using the conservation of momentum: \[ p_{\text{total initial}} = m_{\text{combined}} \cdot v_{\text{final}} \] \[ 22692 \, \text{kg} \cdot \text{m/s} = 3230 \, \text{kg} \cdot v_{\text{final}} \]
• Solve for \( v_{\text{final}} \): \[ v_{\text{final}} = \frac{22692 \, \text{kg} \cdot \text{m/s}}{3230 \, \text{kg}} = 7.02 \, \text{m/s} \]

• The calculated final velocity \( 7.02 \, \text{m/s} \) does not match any of the provided options exactly. Therefore, there might be an error in the provided options or additional context needed.

Final Answer