To find the total momentum of the system, we can use the formula for momentum: \( p = m \cdot v \).
For Cart 1:
- Mass \( m_1 = 4.0 , \text{kg} \)
- Velocity \( v_1 = 6.0 , \text{m/s} \) (to the right, we consider it positive)
For Cart 2:
- Mass \( m_2 = 2.5 , \text{kg} \)
- Velocity \( v_2 = -3.0 , \text{m/s} \) (to the left, we consider it negative)
Now we calculate the total momentum:
\[ \text{Total Momentum} = m_1 \cdot v_1 + m_2 \cdot v_2 \]
Substituting in the values:
\[ \text{Total Momentum} = (4.0 , \text{kg} \cdot 6.0 , \text{m/s}) + (2.5 , \text{kg} \cdot -3.0 , \text{m/s}) \] \[ = 24.0 , \text{kg m/s} - 7.5 , \text{kg m/s} \] \[ = 16.5 , \text{kg m/s} \]
The total momentum of the system is 16.5.
Now, to find the velocity of Cart 2 after the elastic collision, we will use the principles of conservation of momentum and conservation of kinetic energy.
Let \( v_{2f} \) be the final velocity of Cart 2 after the collision.
Using the conservation of momentum: \[ m_1 v_1 + m_2 v_2 = m_1 v_{1f} + m_2 v_{2f} \] Where:
- \( v_{1f} = 4.5 , \text{m/s} \) (final velocity of Cart 1)
Plugging in the initial values into the momentum equation:
\[ (4.0)(6.0) + (2.5)(-3.0) = (4.0)(4.5) + (2.5)v_{2f} \] \[ 24.0 - 7.5 = 18.0 + 2.5v_{2f} \] \[ 16.5 = 18.0 + 2.5v_{2f} \] Subtracting 18.0 from both sides: \[ 16.5 - 18.0 = 2.5v_{2f} \] \[ -1.5 = 2.5v_{2f} \] Dividing by 2.5: \[ v_{2f} = \frac{-1.5}{2.5} = -0.6 , \text{m/s} \]
The final velocity of Cart 2 after the elastic collision is -0.6.
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