Questions: A cue stick has a mass of 0.5 kg. The cue stick hits a ball with a mass of 0.2 kg at a velocity of 2.5 m / s. What is the velocity of the ball after it is hit? (1 point) 2.5 m / s 6.3 m / s 3.6 m / s 8.3 m / s

Solution Steps

We need to find the velocity of the ball after it is hit by the cue stick. The problem involves a collision, and we can assume it is an elastic collision since no other information is provided.

In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. The formula for momentum is:

\[ p = mv \]

where \(m\) is mass and \(v\) is velocity.

Let \(v_{\text{ball}}\) be the velocity of the ball after the collision. The initial momentum of the system is:

\[ p_{\text{initial}} = m_{\text{cue}} \cdot v_{\text{cue}} + m_{\text{ball}} \cdot 0 \]

Given:

• \(m_{\text{cue}} = 0.5 \, \mathrm{kg}\)
• \(v_{\text{cue}} = 2.5 \, \mathrm{m/s}\)
• \(m_{\text{ball}} = 0.2 \, \mathrm{kg}\)

The initial momentum is:

\[ p_{\text{initial}} = 0.5 \cdot 2.5 = 1.25 \, \mathrm{kg \cdot m/s} \]

The final momentum of the system is:

\[ p_{\text{final}} = m_{\text{cue}} \cdot v_{\text{cue, final}} + m_{\text{ball}} \cdot v_{\text{ball}} \]

Assuming the cue stick comes to a stop after hitting the ball (for simplicity), \(v_{\text{cue, final}} = 0\). Therefore:

\[ p_{\text{final}} = 0.2 \cdot v_{\text{ball}} \]

Set the initial and final momentum equal:

\[ 1.25 = 0.2 \cdot v_{\text{ball}} \]

Solve for \(v_{\text{ball}}\):

\[ v_{\text{ball}} = \frac{1.25}{0.2} = 6.25 \, \mathrm{m/s} \]

Final Answer