Question
A baseball bat has a mass of 0.94 kg, and a baseball has a mass of 0.14 kg. If the magnitude of the velocity of the bat is 35.0 m/s when it hits the ball, what will the magnitude of the velocity of the ball be immediately following the collision with the bat?(1 point)
Responses
35 m/s
35 m/s
32.9 m/s
32.9 m/s
235 m/s
235 m/s
250 m/s
250 m/s
Remaining Attempts : 3
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Responses
35 m/s
35 m/s
32.9 m/s
32.9 m/s
235 m/s
235 m/s
250 m/s
250 m/s
Remaining Attempts : 3
Skip to navigation
page 2 of 2
Answers
GPT-4o mini
To solve this problem, we can use the principle of conservation of linear momentum, which states that the total momentum of a closed system before a collision is equal to the total momentum after the collision, provided no external forces act on it.
In this case, the system consists of the baseball bat and the baseball. We can denote the following:
- Mass of the bat \( m_b = 0.94 \, \text{kg} \)
- Velocity of the bat before the collision \( v_b = 35.0 \, \text{m/s} \)
- Mass of the baseball \( m_{ball} = 0.14 \, \text{kg} \)
- Velocity of the baseball before the collision \( v_{ball} = 0 \, \text{m/s} \) (assuming the baseball is at rest)
Let \( v'_{ball} \) be the velocity of the baseball after the collision.
Initially, the momentum of the system before the collision is:
\[
\text{Initial Momentum} = m_b \cdot v_b + m_{ball} \cdot v_{ball} = (0.94 \, \text{kg}) \cdot (35.0 \, \text{m/s}) + (0.14 \, \text{kg}) \cdot (0 \, \text{m/s})
\]
Calculating the initial momentum:
\[
\text{Initial Momentum} = 0.94 \cdot 35.0 = 32.9 \, \text{kg m/s}
\]
After the collision, the momentum of the system is:
\[
\text{Final Momentum} = m_b \cdot v_b' + m_{ball} \cdot v'_{ball}
\]
where \( v_b' \) is the velocity of the bat after the collision (which we don’t know). However, for a simple analysis, if we assume that the bat comes to a stop after hitting the ball (which may not be true in real life, but simplifies our calculations), then \( v_b' = 0 \).
Thus, the final momentum simplifies to:
\[
\text{Final Momentum} = 0 + m_{ball} \cdot v'_{ball} = 0.14 \cdot v'_{ball}
\]
By conservation of momentum:
\[
\text{Initial Momentum} = \text{Final Momentum}
\]
\[
32.9 = 0.14 \cdot v'_{ball}
\]
To find \( v'_{ball} \):
\[
v'_{ball} = \frac{32.9}{0.14} \approx 235 \, \text{m/s}
\]
Therefore, the magnitude of the velocity of the baseball immediately following the collision with the bat will be approximately **235 m/s**.
The correct response is **235 m/s**.
In this case, the system consists of the baseball bat and the baseball. We can denote the following:
- Mass of the bat \( m_b = 0.94 \, \text{kg} \)
- Velocity of the bat before the collision \( v_b = 35.0 \, \text{m/s} \)
- Mass of the baseball \( m_{ball} = 0.14 \, \text{kg} \)
- Velocity of the baseball before the collision \( v_{ball} = 0 \, \text{m/s} \) (assuming the baseball is at rest)
Let \( v'_{ball} \) be the velocity of the baseball after the collision.
Initially, the momentum of the system before the collision is:
\[
\text{Initial Momentum} = m_b \cdot v_b + m_{ball} \cdot v_{ball} = (0.94 \, \text{kg}) \cdot (35.0 \, \text{m/s}) + (0.14 \, \text{kg}) \cdot (0 \, \text{m/s})
\]
Calculating the initial momentum:
\[
\text{Initial Momentum} = 0.94 \cdot 35.0 = 32.9 \, \text{kg m/s}
\]
After the collision, the momentum of the system is:
\[
\text{Final Momentum} = m_b \cdot v_b' + m_{ball} \cdot v'_{ball}
\]
where \( v_b' \) is the velocity of the bat after the collision (which we don’t know). However, for a simple analysis, if we assume that the bat comes to a stop after hitting the ball (which may not be true in real life, but simplifies our calculations), then \( v_b' = 0 \).
Thus, the final momentum simplifies to:
\[
\text{Final Momentum} = 0 + m_{ball} \cdot v'_{ball} = 0.14 \cdot v'_{ball}
\]
By conservation of momentum:
\[
\text{Initial Momentum} = \text{Final Momentum}
\]
\[
32.9 = 0.14 \cdot v'_{ball}
\]
To find \( v'_{ball} \):
\[
v'_{ball} = \frac{32.9}{0.14} \approx 235 \, \text{m/s}
\]
Therefore, the magnitude of the velocity of the baseball immediately following the collision with the bat will be approximately **235 m/s**.
The correct response is **235 m/s**.