To prove conservation of momentum, we will use the principle that the total momentum before the collision equals the total momentum after the collision.
First, we will calculate the momentum before the collision for both bumper cars.
Momentum Before Collision:
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For the first bumper car: \[ p_1\text{(before)} = m_1 \cdot v_1 = 120 , \text{kg} \cdot 4.0 , \text{m/s} = 480 , \text{kg m/s} \]
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For the second bumper car: \[ p_2\text{(before)} = m_2 \cdot v_2 = 90 , \text{kg} \cdot (-5.0) , \text{m/s} = -450 , \text{kg m/s} \]
Total Momentum Before Collision: \[ p_{\text{total (before)}} = p_1\text{(before)} + p_2\text{(before)} = 480 , \text{kg m/s} + (-450 , \text{kg m/s}) = 30 , \text{kg m/s} \]
Momentum After Collision:
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For the first bumper car after the collision: \[ p_1\text{(after)} = m_1 \cdot v_1\text{(after)} = 120 , \text{kg} \cdot (-2.0) , \text{m/s} = -240 , \text{kg m/s} \]
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Let \( v_2\text{(after)} \) be the velocity of the second bumper car after the collision. Then its momentum will be: \[ p_2\text{(after)} = m_2 \cdot v_2\text{(after)} = 90 , \text{kg} \cdot v_2\text{(after)} \]
Total Momentum After Collision: \[ p_{\text{total (after)}} = p_1\text{(after)} + p_2\text{(after)} = -240 , \text{kg m/s} + 90 , \text{kg} \cdot v_2\text{(after)} \]
Setting Up the Equation: According to conservation of momentum: \[ p_{\text{total (before)}} = p_{\text{total (after)}} \] So we set up the equation: \[ 30 = -240 + 90 \cdot v_2\text{(after)} \]
Now, we solve for \( v_2\text{(after)} \): \[ 30 + 240 = 90 \cdot v_2\text{(after)} \] \[ 270 = 90 \cdot v_2\text{(after)} \] \[ v_2\text{(after)} = \frac{270}{90} = 3.0 , \text{m/s} \]
Therefore, the velocity of the second bumper car after the collision must be 3.0 m/s.