To solve for the velocity of ball B after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision must equal the total momentum after the collision.
First, we calculate the momentum before the collision:
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Calculate the momentum of ball A before the collision: \[ p_A = m_A \cdot v_A = 12.0 , \text{kg} \cdot 2.0 , \text{m/s} = 24.0 , \text{kg m/s} \]
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Calculate the momentum of ball B before the collision: \[ p_B = m_B \cdot v_B = 3.5 , \text{kg} \cdot 1.0 , \text{m/s} = 3.5 , \text{kg m/s} \]
Now, we find the total momentum before the collision: \[ p_{\text{initial}} = p_A + p_B = 24.0 , \text{kg m/s} + 3.5 , \text{kg m/s} = 27.5 , \text{kg m/s} \]
After the collision, we know that:
- The mass of ball A (\(m_A\)) remains 12.0 kg and its velocity (\(v_A'\)) after the collision is 1.0 m/s.
- We need to find the velocity of ball B after the collision (\(v_B'\)).
The momentum after the collision can be calculated as: \[ p_{\text{final}} = m_A \cdot v_A' + m_B \cdot v_B' \] Substituting in the known values: \[ p_{\text{final}} = 12.0 , \text{kg} \cdot 1.0 , \text{m/s} + 3.5 , \text{kg} \cdot v_B' = 12.0 , \text{kg m/s} + 3.5 , \text{kg} \cdot v_B' \]
Setting the initial momentum equal to the final momentum gives us: \[ 27.5 , \text{kg m/s} = 12.0 , \text{kg m/s} + 3.5 , \text{kg} \cdot v_B' \]
Now we can solve for \(v_B'\): \[ 27.5 , \text{kg m/s} - 12.0 , \text{kg m/s} = 3.5 , \text{kg} \cdot v_B' \] \[ 15.5 , \text{kg m/s} = 3.5 , \text{kg} \cdot v_B' \] \[ v_B' = \frac{15.5 , \text{kg m/s}}{3.5 , \text{kg}} \approx 4.43 , \text{m/s} \]
Thus, the velocity of ball B after the collision is approximately 4.43 m/s westward.
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