To determine the velocity of the ball after being hit by the cue stick, we can apply the principle of conservation of momentum. The total momentum before the collision should equal the total momentum after the collision.
Let's denote:
- Mass of the cue stick, \( m_1 = 0.5 , \text{kg} \)
- Velocity of the cue stick before the collision, \( v_1 = 2.5 , \text{m/s} \)
- Mass of the ball, \( m_2 = 0.2 , \text{kg} \)
- Velocity of the ball after the collision, \( v_2 \)
Initially, we assume the ball is at rest before being hit, so its initial velocity \( v_{2,\text{initial}} = 0 , \text{m/s} \).
The total initial momentum of the system is: \[ p_{\text{initial}} = m_1 \cdot v_1 + m_2 \cdot v_{2,\text{initial}} = (0.5 , \text{kg}) \cdot (2.5 , \text{m/s}) + (0.2 , \text{kg}) \cdot (0 , \text{m/s}) = 1.25 , \text{kg} \cdot \text{m/s} \]
After the collision, the total momentum will be: \[ p_{\text{final}} = m_1 \cdot v_{1,\text{final}} + m_2 \cdot v_2 \]
For simplicity, we can assume that after hitting the ball, the cue stick's speed is reduced but for calculation, let's just look at the momentum transfer to the ball and disregard the cue stick's final speed.
Setting initial momentum equal to final momentum: \[ 1.25 , \text{kg} \cdot \text{m/s} = 0.5 , \text{kg} \cdot v_{1,\text{final}} + 0.2 , \text{kg} \cdot v_2 \]
If we assume the cue stick comes to a stop (ideal case) right after hitting the ball: \[ 1.25 , \text{kg} \cdot \text{m/s} = 0 + 0.2 , \text{kg} \cdot v_2 \]
Thus, solving for \( v_2 \): \[ v_2 = \frac{1.25 , \text{kg} \cdot \text{m/s}}{0.2 , \text{kg}} = 6.25 , \text{m/s} \]
Rounding gives us approximately \( 6.3 , \text{m/s} \).
So, the correct answer is \( 6.3 , \text{m/s} \).