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 find the velocity of the ball after it is hit by the cue stick, we can apply the law of conservation of momentum.
The formula for momentum is:
\[
\text{momentum} = \text{mass} \times \text{velocity}
\]
Before the collision, only the cue stick is moving (assuming the ball is initially at rest). The initial momentum (\(p_{\text{initial}}\)) can be calculated as follows:
- Mass of the cue stick (\(m_{c}\)) = 0.5 kg
- Velocity of the cue stick before hitting the ball (\(v_{c}\)) = 2.5 m/s
- Mass of the ball (\(m_{b}\)) = 0.2 kg
- Velocity of the ball before being hit (\(v_{b}\)) = 0 m/s (since it's at rest)
Calculating the initial momentum:
\[
p_{\text{initial}} = m_{c} \cdot v_{c} + m_{b} \cdot v_{b} = (0.5 \, \text{kg} \times 2.5 \, \text{m/s}) + (0.2 \, \text{kg} \times 0 \, \text{m/s}) = 1.25 \, \text{kg m/s}
\]
After the cue stick hits the ball, we need to denote the final velocities:
- Final velocity of the cue stick = \(v'_{c}\)
- Final velocity of the ball = \(v'_{b}\)
Using the conservation of momentum:
\[
p_{\text{initial}} = p_{\text{final}}
\]
So,
\[
1.25 \, \text{kg m/s} = m_{c} \cdot v'_{c} + m_{b} \cdot v'_{b}
\]
Without additional information about how the collision occurs (elastic, inelastic, etc.), it's impossible to determine the final velocities of both objects individually without making assumptions.
### Simple Assumption (Perfectly Elastic Collision)
If we assume that the collision is perfectly elastic (meaning both momentum and kinetic energy are conserved), we can use the following formulas:
1. Conservation of momentum:
\[
m_{c} \cdot v_{c} + m_{b} \cdot v_{b} = m_{c} \cdot v'_{c} + m_{b} \cdot v'_{b}
\]
2. Conservation of kinetic energy:
\[
\frac{1}{2} m_{c} v_{c}^2 + \frac{1}{2} m_{b} v_{b}^2 = \frac{1}{2} m_{c} {v'_{c}}^2 + \frac{1}{2} m_{b} {v'_{b}}^2
\]
However, to solve for the velocity of the ball \(v'_{b}\) after the collision without more information, we often assume the cue stick comes to a stop (for simplicity).
If we assume the cue stick stops:
- Then momentum conservation gives us:
\[
1.25 \, \text{kg m/s} = 0.5 \, \text{kg} \cdot 0 + 0.2 \, \text{kg} \cdot v'_{b}
\]
\[
1.25 = 0.2 \cdot v'_{b}
\]
\[
v'_{b} = \frac{1.25}{0.2} = 6.25 \, \text{m/s}
\]
Therefore, if we assume the cue stick stops after the collision, the velocity of the ball after it is hit would be **6.25 m/s**.
### Note:
This is a simplified scenario. Actual results may vary based on the nature of the collision. If you have more details on whether the collision is elastic or inelastic or other specific conditions, we can obtain a more accurate answer.
The formula for momentum is:
\[
\text{momentum} = \text{mass} \times \text{velocity}
\]
Before the collision, only the cue stick is moving (assuming the ball is initially at rest). The initial momentum (\(p_{\text{initial}}\)) can be calculated as follows:
- Mass of the cue stick (\(m_{c}\)) = 0.5 kg
- Velocity of the cue stick before hitting the ball (\(v_{c}\)) = 2.5 m/s
- Mass of the ball (\(m_{b}\)) = 0.2 kg
- Velocity of the ball before being hit (\(v_{b}\)) = 0 m/s (since it's at rest)
Calculating the initial momentum:
\[
p_{\text{initial}} = m_{c} \cdot v_{c} + m_{b} \cdot v_{b} = (0.5 \, \text{kg} \times 2.5 \, \text{m/s}) + (0.2 \, \text{kg} \times 0 \, \text{m/s}) = 1.25 \, \text{kg m/s}
\]
After the cue stick hits the ball, we need to denote the final velocities:
- Final velocity of the cue stick = \(v'_{c}\)
- Final velocity of the ball = \(v'_{b}\)
Using the conservation of momentum:
\[
p_{\text{initial}} = p_{\text{final}}
\]
So,
\[
1.25 \, \text{kg m/s} = m_{c} \cdot v'_{c} + m_{b} \cdot v'_{b}
\]
Without additional information about how the collision occurs (elastic, inelastic, etc.), it's impossible to determine the final velocities of both objects individually without making assumptions.
### Simple Assumption (Perfectly Elastic Collision)
If we assume that the collision is perfectly elastic (meaning both momentum and kinetic energy are conserved), we can use the following formulas:
1. Conservation of momentum:
\[
m_{c} \cdot v_{c} + m_{b} \cdot v_{b} = m_{c} \cdot v'_{c} + m_{b} \cdot v'_{b}
\]
2. Conservation of kinetic energy:
\[
\frac{1}{2} m_{c} v_{c}^2 + \frac{1}{2} m_{b} v_{b}^2 = \frac{1}{2} m_{c} {v'_{c}}^2 + \frac{1}{2} m_{b} {v'_{b}}^2
\]
However, to solve for the velocity of the ball \(v'_{b}\) after the collision without more information, we often assume the cue stick comes to a stop (for simplicity).
If we assume the cue stick stops:
- Then momentum conservation gives us:
\[
1.25 \, \text{kg m/s} = 0.5 \, \text{kg} \cdot 0 + 0.2 \, \text{kg} \cdot v'_{b}
\]
\[
1.25 = 0.2 \cdot v'_{b}
\]
\[
v'_{b} = \frac{1.25}{0.2} = 6.25 \, \text{m/s}
\]
Therefore, if we assume the cue stick stops after the collision, the velocity of the ball after it is hit would be **6.25 m/s**.
### Note:
This is a simplified scenario. Actual results may vary based on the nature of the collision. If you have more details on whether the collision is elastic or inelastic or other specific conditions, we can obtain a more accurate answer.