Question
A ball with a mass of 0.9 kg and a velocity of 10 m/s hits another ball with the same mass. The first ball stops moving, and the second ball then moves forward and hits a third ball with a mass of 0.3 Kg. If the system is closed, what is the velocity of the third ball?(1 point)
Responses
36 m/s
36 m/s
1.44 m/s
1.44 m/s
4 m/s
4 m/s
30 m/s
Responses
36 m/s
36 m/s
1.44 m/s
1.44 m/s
4 m/s
4 m/s
30 m/s
Answers
GPT-4o mini
To solve for the velocity of the third ball after the first two balls collide, we can use the principle of conservation of momentum.
1. **Initial Momentum Calculation**:
- The first ball has a mass of \( m_1 = 0.9 \) kg and an initial velocity \( v_1 = 10 \) m/s.
- The second ball, having the same mass, is at rest, so \( v_2 = 0 \) m/s.
The total initial momentum of the system is:
\[
p_{\text{initial}} = m_1 \cdot v_1 + m_2 \cdot v_2 = 0.9 \, \text{kg} \cdot 10 \, \text{m/s} + 0.9 \, \text{kg} \cdot 0 \, \text{m/s} = 9 \, \text{kg m/s}
\]
2. **Outcome of the First Collision**:
After the collision, the first ball stops moving:
- Final velocity of the first ball, \( v_1' = 0 \) m/s.
- Let \( v_2' \) be the final velocity of the second ball.
By conservation of momentum:
\[
p_{\text{initial}} = p_{\text{final}}
\]
\[
9 \, \text{kg m/s} = 0.9\, \text{kg} \cdot 0 + 0.9\, \text{kg} \cdot v_2'
\]
This gives:
\[
v_2' = \frac{9 \, \text{kg m/s}}{0.9\, \text{kg}} = 10 \, \text{m/s}
\]
Thus, the second ball starts moving with a velocity of 10 m/s.
3. **Collision between the second ball and the third ball**:
Now the second ball (mass \( m_2 = 0.9 \) kg) moving at 10 m/s hits the third ball (mass \( m_3 = 0.3 \) kg).
Let \( v_3' \) be the final velocity of the third ball after the collision.
Assuming the collision is perfectly elastic, we apply conservation of momentum:
\[
p_{\text{initial}} = p_{\text{final}}
\]
\[
m_2 \cdot v_2' + m_3 \cdot 0 = m_2 \cdot v_2'' + m_3 \cdot v_3'
\]
If we assume the second ball comes to a stop after hitting the third ball (which is the case for a perfectly elastic collision with a much lighter ball), we have:
\[
0.9\, \text{kg} \cdot 10\, \text{m/s} = 0 + 0.3\, \text{kg} \cdot v_3'
\]
\[
9\, \text{kg m/s} = 0.3\, \text{kg} \cdot v_3'
\]
\[
v_3' = \frac{9 \, \text{kg m/s}}{0.3\, \text{kg}} = 30 \, \text{m/s}
\]
Thus, the velocity of the third ball after being struck is **30 m/s**.
**Final Answer: 30 m/s**
1. **Initial Momentum Calculation**:
- The first ball has a mass of \( m_1 = 0.9 \) kg and an initial velocity \( v_1 = 10 \) m/s.
- The second ball, having the same mass, is at rest, so \( v_2 = 0 \) m/s.
The total initial momentum of the system is:
\[
p_{\text{initial}} = m_1 \cdot v_1 + m_2 \cdot v_2 = 0.9 \, \text{kg} \cdot 10 \, \text{m/s} + 0.9 \, \text{kg} \cdot 0 \, \text{m/s} = 9 \, \text{kg m/s}
\]
2. **Outcome of the First Collision**:
After the collision, the first ball stops moving:
- Final velocity of the first ball, \( v_1' = 0 \) m/s.
- Let \( v_2' \) be the final velocity of the second ball.
By conservation of momentum:
\[
p_{\text{initial}} = p_{\text{final}}
\]
\[
9 \, \text{kg m/s} = 0.9\, \text{kg} \cdot 0 + 0.9\, \text{kg} \cdot v_2'
\]
This gives:
\[
v_2' = \frac{9 \, \text{kg m/s}}{0.9\, \text{kg}} = 10 \, \text{m/s}
\]
Thus, the second ball starts moving with a velocity of 10 m/s.
3. **Collision between the second ball and the third ball**:
Now the second ball (mass \( m_2 = 0.9 \) kg) moving at 10 m/s hits the third ball (mass \( m_3 = 0.3 \) kg).
Let \( v_3' \) be the final velocity of the third ball after the collision.
Assuming the collision is perfectly elastic, we apply conservation of momentum:
\[
p_{\text{initial}} = p_{\text{final}}
\]
\[
m_2 \cdot v_2' + m_3 \cdot 0 = m_2 \cdot v_2'' + m_3 \cdot v_3'
\]
If we assume the second ball comes to a stop after hitting the third ball (which is the case for a perfectly elastic collision with a much lighter ball), we have:
\[
0.9\, \text{kg} \cdot 10\, \text{m/s} = 0 + 0.3\, \text{kg} \cdot v_3'
\]
\[
9\, \text{kg m/s} = 0.3\, \text{kg} \cdot v_3'
\]
\[
v_3' = \frac{9 \, \text{kg m/s}}{0.3\, \text{kg}} = 30 \, \text{m/s}
\]
Thus, the velocity of the third ball after being struck is **30 m/s**.
**Final Answer: 30 m/s**