Question
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
Answers
GPT-4o mini
To determine the velocity of the ball after being hit by the cue stick, we can apply the principle of conservation of momentum. The momentum before the collision must equal the momentum after the collision.
1. **Calculate the momentum of the cue stick and the ball before the collision:**
- The cue stick is moving and the ball is stationary before the impact.
- Momentum of the cue stick before the collision \( p_{cue\_before} = m_{cue} \times v_{cue} \)
- Since the ball is stationary, its momentum before the collision is 0.
- Total initial momentum \( p_{initial} = p_{cue\_before} + p_{ball\_before} = (0.5 \, \text{kg}) \times (2.5 \, \text{m/s}) + 0 = 1.25 \, \text{kg m/s} \)
2. **Apply the conservation of momentum:**
The total momentum after the collision \(p_{initial} = p_{final}\).
- Let \( v_{ball\_after} \) be the velocity of the ball after the collision.
- Let \( v_{cue\_after} \) be the velocity of the cue stick after the collision (this is not given, but we could assume some final velocity or inelastic condition).
For elastic collisions, we would also need to account for the velocity of the cue stick after the collision, but not all collisions are elastic. Without a specific final velocity for the cue stick after impact, we'll assume perfect transfer of momentum with only the ball being solved.
Assuming an elastic collision, we can use the equations reflecting conservation of momentum and conservation of kinetic energy; however, this could complicate the calculations without more information.
If we assume a perfectly inelastic condition (which is generally less common but simpler), we can then state:
\[
m_{cue} \cdot v_{cue} + m_{ball} \cdot v_{ball} = m_{cue} \cdot v_{cue\_after} + m_{ball} \cdot v_{ball\_after}
\]
To simplify assume that the cue stick transfers its momentum perfectly to the ball for this calculation, making \( v_{cue\_after} = 0 \):
\[
1.25 \, \text{kg m/s} = 0.2 \, \text{kg} \cdot v_{ball\_after}
\]
Solving for \( v_{ball\_after} \):
\[
v_{ball\_after} = \frac{1.25 \, \text{kg m/s}}{0.2 \, \text{kg}} = 6.25 \, \text{m/s}
\]
Thus, under these assumptions, the velocity of the ball after it is hit would be **6.25 m/s**.
1. **Calculate the momentum of the cue stick and the ball before the collision:**
- The cue stick is moving and the ball is stationary before the impact.
- Momentum of the cue stick before the collision \( p_{cue\_before} = m_{cue} \times v_{cue} \)
- Since the ball is stationary, its momentum before the collision is 0.
- Total initial momentum \( p_{initial} = p_{cue\_before} + p_{ball\_before} = (0.5 \, \text{kg}) \times (2.5 \, \text{m/s}) + 0 = 1.25 \, \text{kg m/s} \)
2. **Apply the conservation of momentum:**
The total momentum after the collision \(p_{initial} = p_{final}\).
- Let \( v_{ball\_after} \) be the velocity of the ball after the collision.
- Let \( v_{cue\_after} \) be the velocity of the cue stick after the collision (this is not given, but we could assume some final velocity or inelastic condition).
For elastic collisions, we would also need to account for the velocity of the cue stick after the collision, but not all collisions are elastic. Without a specific final velocity for the cue stick after impact, we'll assume perfect transfer of momentum with only the ball being solved.
Assuming an elastic collision, we can use the equations reflecting conservation of momentum and conservation of kinetic energy; however, this could complicate the calculations without more information.
If we assume a perfectly inelastic condition (which is generally less common but simpler), we can then state:
\[
m_{cue} \cdot v_{cue} + m_{ball} \cdot v_{ball} = m_{cue} \cdot v_{cue\_after} + m_{ball} \cdot v_{ball\_after}
\]
To simplify assume that the cue stick transfers its momentum perfectly to the ball for this calculation, making \( v_{cue\_after} = 0 \):
\[
1.25 \, \text{kg m/s} = 0.2 \, \text{kg} \cdot v_{ball\_after}
\]
Solving for \( v_{ball\_after} \):
\[
v_{ball\_after} = \frac{1.25 \, \text{kg m/s}}{0.2 \, \text{kg}} = 6.25 \, \text{m/s}
\]
Thus, under these assumptions, the velocity of the ball after it is hit would be **6.25 m/s**.