To prove conservation of momentum, we can use the principle that the total momentum before the collision must equal the total momentum after the collision.
Let's denote:
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Mass of the first bumper car \( m_1 = 120 , \text{kg} \)
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Initial velocity of the first bumper car \( u_1 = 4.0 , \text{m/s} \)
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Final velocity of the first bumper car \( v_1 = -2.0 , \text{m/s} \)
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Mass of the second bumper car \( m_2 = 90 , \text{kg} \)
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Initial velocity of the second bumper car \( u_2 = -5.0 , \text{m/s} \)
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Final velocity of the second bumper car \( v_2 \) (this is what we need to find)
First, we calculate the total momentum before the collision:
\[ \text{Total momentum before} = m_1 u_1 + m_2 u_2 \]
\[ = (120 , \text{kg} \times 4.0 , \text{m/s}) + (90 , \text{kg} \times -5.0 , \text{m/s}) \]
\[ = 480 , \text{kg m/s} - 450 , \text{kg m/s} \]
\[ = 30 , \text{kg m/s} \]
Next, we calculate the total momentum after the collision:
\[ \text{Total momentum after} = m_1 v_1 + m_2 v_2 \]
\[ = (120 , \text{kg} \times -2.0 , \text{m/s}) + (90 , \text{kg} \times v_2) \]
\[ = -240 , \text{kg m/s} + 90 v_2 \]
Now, we set the total momentum before equal to the total momentum after:
\[ 30 , \text{kg m/s} = -240 , \text{kg m/s} + 90 v_2 \]
Solving for \( v_2 \):
\[ 30 + 240 = 90 v_2 \]
\[ 270 = 90 v_2 \]
\[ v_2 = \frac{270}{90} = 3.0 , \text{m/s} \]
Therefore, the velocity of the second bumper car after the collision must be 3.0 m/s.