To solve the problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the collision must equal the total momentum after the collision.
Let's denote:
- \( m_1 = 8.0 , \text{kg} \) (mass of the first cart),
- \( v_{1i} = 4.0 , \text{m/s} \) (initial velocity of the first cart, positive to the right),
- \( m_2 = 4.0 , \text{kg} \) (mass of the second cart),
- \( v_{2i} = -6.0 , \text{m/s} \) (initial velocity of the second cart, negative because it moves to the left),
- \( v_{1f} \) (final velocity of the first cart, after the collision),
- \( v_{2f} = 3.0 , \text{m/s} \) (final velocity of the second cart, after the collision).
The total momentum before the collision is: \[ p_\text{before} = m_1 v_{1i} + m_2 v_{2i} = (8.0 , \text{kg})(4.0 , \text{m/s}) + (4.0 , \text{kg})(-6.0 , \text{m/s}) \] Calculating this: \[ p_\text{before} = 32.0 , \text{kg m/s} - 24.0 , \text{kg m/s} = 8.0 , \text{kg m/s} \]
The total momentum after the collision is: \[ p_\text{after} = m_1 v_{1f} + m_2 v_{2f} \] Substituting the known values gives: \[ p_\text{after} = (8.0 , \text{kg}) v_{1f} + (4.0 , \text{kg})(3.0 , \text{m/s}) \] Calculating this: \[ p_\text{after} = 8.0 v_{1f} + 12.0 , \text{kg m/s} \]
Setting the total momentum before the collision equal to the total momentum after the collision: \[ 8.0 , \text{kg m/s} = 8.0 v_{1f} + 12.0 , \text{kg m/s} \]
Now, we can solve for \( v_{1f} \): \[ 8.0 v_{1f} = 8.0 - 12.0 \] \[ 8.0 v_{1f} = -4.0 \] \[ v_{1f} = -\frac{4.0}{8.0} = -0.5 , \text{m/s} \]
Thus, the velocity of the 8.0-kilogram cart after the collision is \( \boxed{-0.5 , \text{m/s}} \), indicating it moves to the left after the collision.