To prove conservation of momentum, we need to ensure that the total momentum before the collision is equal to the total momentum after the collision.
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Calculate the initial momentum of each bumper car:
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For the first bumper car: \[ m_1 = 120 , \text{kg}, \quad v_{1i} = 4.0 , \text{m/s} \ p_{1i} = m_1 \cdot v_{1i} = 120 , \text{kg} \cdot 4.0 , \text{m/s} = 480 , \text{kg m/s} \]
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For the second bumper car: \[ m_2 = 90 , \text{kg}, \quad v_{2i} = -5.0 , \text{m/s} \ p_{2i} = m_2 \cdot v_{2i} = 90 , \text{kg} \cdot (-5.0 , \text{m/s}) = -450 , \text{kg m/s} \]
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Total initial momentum (before collision): \[ p_{initial} = p_{1i} + p_{2i} = 480 , \text{kg m/s} - 450 , \text{kg m/s} = 30 , \text{kg m/s} \]
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Calculate the final momentum of each bumper car:
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The first bumper car after the collision: \[ v_{1f} = -2.0 , \text{m/s} \ p_{1f} = m_1 \cdot v_{1f} = 120 , \text{kg} \cdot (-2.0 , \text{m/s}) = -240 , \text{kg m/s} \]
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Let the final velocity of the second bumper car be \( v_{2f} \). Then the final momentum of the second bumper car is: \[ p_{2f} = m_2 \cdot v_{2f} = 90 , \text{kg} \cdot v_{2f} \]
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Total final momentum (after collision): \[ p_{final} = p_{1f} + p_{2f} = -240 , \text{kg m/s} + 90 , \text{kg} \cdot v_{2f} \]
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Set the initial momentum equal to the final momentum: \[ p_{initial} = p_{final} \ 30 , \text{kg m/s} = -240 , \text{kg m/s} + 90 , \text{kg} \cdot v_{2f} \]
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Solve for \( v_{2f} \): \[ 30 + 240 = 90 , \text{kg} \cdot v_{2f} \ 270 = 90 , \text{kg} \cdot v_{2f} \ v_{2f} = \frac{270}{90} = 3.0 , \text{m/s} \]
Thus, the velocity of the second bumper car after the collision must be \( \boxed{3.0 , \text{m/s}} \) to prove conservation of momentum.
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