To find the velocity of the ball after it is hit by the cue stick, we can use the conservation of momentum principle, assuming that there are no external forces acting on the system.
The momentum before the collision must equal the momentum after the collision. We can express this as:
\[ m_{\text{cue}} \cdot v_{\text{cue}\text{initial}} + m{\text{ball}} \cdot v_{\text{ball}\text{initial}} = m{\text{cue}} \cdot v_{\text{cue}\text{final}} + m{\text{ball}} \cdot v_{\text{ball}_\text{final}} \]
In this scenario:
- The mass of the cue stick, \( m_{\text{cue}} = 0.5 , \text{kg} \)
- The initial velocity of the cue stick, \( v_{\text{cue}_\text{initial}} = 2.5 , \text{m/s} \)
- The mass of the ball, \( m_{\text{ball}} = 0.2 , \text{kg} \)
- The initial velocity of the ball, \( v_{\text{ball}_\text{initial}} = 0 , \text{m/s} \) (assuming it is at rest)
Assuming the cue stick comes to a stop after hitting the ball (which is a simplifying assumption):
- \( v_{\text{cue}_\text{final}} = 0 \)
Using these values, we can set up the equation according to the conservation of momentum:
\[ (0.5 , \text{kg} \cdot 2.5 , \text{m/s}) + (0.2 , \text{kg} \cdot 0) = (0.5 , \text{kg} \cdot 0) + (0.2 , \text{kg} \cdot v_{\text{ball}_\text{final}}) \]
This simplifies to:
\[ 1.25 , \text{kg m/s} = 0.2 , \text{kg} \cdot v_{\text{ball}_\text{final}} \]
To find \( v_{\text{ball}_\text{final}} \), we divide both sides by \( 0.2 , \text{kg} \):
\[ v_{\text{ball}_\text{final}} = \frac{1.25 , \text{kg m/s}}{0.2 , \text{kg}} = 6.25 , \text{m/s} \]
Rounding this value results in \( v_{\text{ball}_\text{final}} \approx 6.3 , \text{m/s} \).
Therefore, the velocity of the ball after it is hit is approximately 6.3 m/s.