Question
Question
Two students in bumper cars want to prove conservation of momentum. They collide their bumper cars. The first bumper car has a mass of 120 kg
and was moving with a velocity of 4.0 m/s
before the collision and with a velocity of −2.0 m/s
after the collision. The second bumper car has a mass of 90 kg
and was moving at a velocity of −5.0 m/s
before the collision. To prove conservation of momentum, what must the velocity of the second bumper car be after the collision?(1 point)
Responses
3.0 m/s
3 point 0 meters per seconds
1.0 m/s
1 point 0 meters per seconds
2.7 m/s
Two students in bumper cars want to prove conservation of momentum. They collide their bumper cars. The first bumper car has a mass of 120 kg
and was moving with a velocity of 4.0 m/s
before the collision and with a velocity of −2.0 m/s
after the collision. The second bumper car has a mass of 90 kg
and was moving at a velocity of −5.0 m/s
before the collision. To prove conservation of momentum, what must the velocity of the second bumper car be after the collision?(1 point)
Responses
3.0 m/s
3 point 0 meters per seconds
1.0 m/s
1 point 0 meters per seconds
2.7 m/s
Answers
GPT-4o mini
To prove conservation of momentum in this collision, we need to calculate the total momentum before and after the collision. The formula for momentum (\( p \)) is given by:
\[
p = m \cdot v
\]
Where \( m \) is the mass and \( v \) is the velocity.
### Step 1: Calculate initial momentum
**Before the collision:**
1. For the first bumper car:
- Mass (\( m_1 \)) = 120 kg
- Velocity before (\( v_1i \)) = 4.0 m/s
- Momentum (\( p_1i \)) = \( m_1 \cdot v_1i = 120 \, \text{kg} \cdot 4.0 \, \text{m/s} = 480 \, \text{kg} \cdot \text{m/s} \)
2. For the second bumper car:
- Mass (\( m_2 \)) = 90 kg
- Velocity before (\( v_2i \)) = -5.0 m/s
- Momentum (\( p_2i \)) = \( m_2 \cdot v_2i = 90 \, \text{kg} \cdot -5.0 \, \text{m/s} = -450 \, \text{kg} \cdot \text{m/s} \)
**Total initial momentum (\( p_{\text{initial}} \))**:
\[
p_{\text{initial}} = p_1i + p_2i = 480 \, \text{kg} \cdot \text{m/s} - 450 \, \text{kg} \cdot \text{m/s} = 30 \, \text{kg} \cdot \text{m/s}
\]
### Step 2: Calculate final momentum
**After the collision**:
1. First bumper car:
- Velocity after (\( v_1f \)) = -2.0 m/s
- Momentum (\( p_1f \)) = \( m_1 \cdot v_1f = 120 \, \text{kg} \cdot -2.0 \, \text{m/s} = -240 \, \text{kg} \cdot \text{m/s} \)
2. For the second bumper car, we need to find its velocity after the collision (\( v_2f \)).
- Momentum (\( p_2f \)) = \( m_2 \cdot v_2f = 90 \, \text{kg} \cdot v_2f \)
**Total final momentum (\( p_{\text{final}} \))**:
\[
p_{\text{final}} = p_1f + p_2f = -240 \, \text{kg} \cdot \text{m/s} + (90 \, \text{kg} \cdot v_2f)
\]
### Step 3: Set total initial momentum equal to total final momentum
To satisfy conservation of momentum:
\[
p_{\text{initial}} = p_{\text{final}}
\]
\[
30 \, \text{kg} \cdot \text{m/s} = -240 \, \text{kg} \cdot \text{m/s} + 90 \, \text{kg} \cdot v_2f
\]
### Step 4: Solve for \( v_{2f} \)
Rearranging gives:
\[
30 + 240 = 90 \, v_{2f}
\]
\[
270 = 90 \, v_{2f}
\]
\[
v_{2f} = \frac{270}{90} = 3.0 \, \text{m/s}
\]
Thus, the velocity of the second bumper car after the collision must be **3.0 m/s**.
\[
p = m \cdot v
\]
Where \( m \) is the mass and \( v \) is the velocity.
### Step 1: Calculate initial momentum
**Before the collision:**
1. For the first bumper car:
- Mass (\( m_1 \)) = 120 kg
- Velocity before (\( v_1i \)) = 4.0 m/s
- Momentum (\( p_1i \)) = \( m_1 \cdot v_1i = 120 \, \text{kg} \cdot 4.0 \, \text{m/s} = 480 \, \text{kg} \cdot \text{m/s} \)
2. For the second bumper car:
- Mass (\( m_2 \)) = 90 kg
- Velocity before (\( v_2i \)) = -5.0 m/s
- Momentum (\( p_2i \)) = \( m_2 \cdot v_2i = 90 \, \text{kg} \cdot -5.0 \, \text{m/s} = -450 \, \text{kg} \cdot \text{m/s} \)
**Total initial momentum (\( p_{\text{initial}} \))**:
\[
p_{\text{initial}} = p_1i + p_2i = 480 \, \text{kg} \cdot \text{m/s} - 450 \, \text{kg} \cdot \text{m/s} = 30 \, \text{kg} \cdot \text{m/s}
\]
### Step 2: Calculate final momentum
**After the collision**:
1. First bumper car:
- Velocity after (\( v_1f \)) = -2.0 m/s
- Momentum (\( p_1f \)) = \( m_1 \cdot v_1f = 120 \, \text{kg} \cdot -2.0 \, \text{m/s} = -240 \, \text{kg} \cdot \text{m/s} \)
2. For the second bumper car, we need to find its velocity after the collision (\( v_2f \)).
- Momentum (\( p_2f \)) = \( m_2 \cdot v_2f = 90 \, \text{kg} \cdot v_2f \)
**Total final momentum (\( p_{\text{final}} \))**:
\[
p_{\text{final}} = p_1f + p_2f = -240 \, \text{kg} \cdot \text{m/s} + (90 \, \text{kg} \cdot v_2f)
\]
### Step 3: Set total initial momentum equal to total final momentum
To satisfy conservation of momentum:
\[
p_{\text{initial}} = p_{\text{final}}
\]
\[
30 \, \text{kg} \cdot \text{m/s} = -240 \, \text{kg} \cdot \text{m/s} + 90 \, \text{kg} \cdot v_2f
\]
### Step 4: Solve for \( v_{2f} \)
Rearranging gives:
\[
30 + 240 = 90 \, v_{2f}
\]
\[
270 = 90 \, v_{2f}
\]
\[
v_{2f} = \frac{270}{90} = 3.0 \, \text{m/s}
\]
Thus, the velocity of the second bumper car after the collision must be **3.0 m/s**.