To find the final speed of the two cars after the collision, we can use the principle of conservation of momentum. The total momentum before the collision should equal the total momentum after the collision.
The momentum before the collision is given by:
\[ \text{Momentum}_{\text{initial}} = m_1 \cdot v_1 + m_2 \cdot v_2 \]
where:
- \( m_1 = 1250 , \text{kg} \) (mass of the first car),
- \( v_1 = 2.24 , \text{m/s} \) (velocity of the first car),
- \( m_2 = 1300 , \text{kg} \) (mass of the second car),
- \( v_2 = 0 , \text{m/s} \) (velocity of the second car, which is stationary).
Calculating the initial momentum:
\[ \text{Momentum}_{\text{initial}} = 1250 , \text{kg} \cdot 2.24 , \text{m/s} + 1300 , \text{kg} \cdot 0 , \text{m/s} \] \[ = 2800 , \text{kg m/s} \]
After the collision, the two cars stick together, so their combined mass is:
\[ m_{\text{total}} = m_1 + m_2 = 1250 , \text{kg} + 1300 , \text{kg} = 2550 , \text{kg} \]
Let \( v_f \) be the final velocity after the collision. The momentum after the collision is:
\[ \text{Momentum}{\text{final}} = m{\text{total}} \cdot v_f \]
Equating the initial and final momentum:
\[ 2800 , \text{kg m/s} = 2550 , \text{kg} \cdot v_f \]
Now, solve for \( v_f \):
\[ v_f = \frac{2800 , \text{kg m/s}}{2550 , \text{kg}} \approx 1.10 , \text{m/s} \]
Therefore, the final speed of both cars moving forward is approximately 1.10 m/s.