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:
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For the first bumper car:
- Mass (\( m_1 \)) = 120 kg
- Velocity before (\( v_1i \)) = 4.0 m/s
- Momentum (\( p_1i \)) = \( m_1 \cdot v_1i = 120 , \text{kg} \cdot 4.0 , \text{m/s} = 480 , \text{kg} \cdot \text{m/s} \)
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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:
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First bumper car:
- Velocity after (\( v_1f \)) = -2.0 m/s
- Momentum (\( p_1f \)) = \( m_1 \cdot v_1f = 120 , \text{kg} \cdot -2.0 , \text{m/s} = -240 , \text{kg} \cdot \text{m/s} \)
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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.