Question

question 10
two bumper cars, each with a mass of 467 kg, collide head - on in an elastic collision. one is initially moving to the right at 6.5 m/s, but rebounds to the left at 3.7 m/s after the collision. the second is initially moving at 6.1 m/s to the left. what is the second bumper cars velocity after the collision?
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Explanation:

Step1: Define variables

Let $m_1 = m_2=467$ kg, $v_{1i}=6.5$ m/s, $v_{2i}=- 6.1$ m/s, $v_{1f}=-3.7$ m/s. The positive - direction is to the right.

Step2: Apply conservation of momentum formula

The conservation of momentum formula is $m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$. Since $m_1 = m_2$, we can cancel out the mass on both sides of the equation, getting $v_{1i}+v_{2i}=v_{1f}+v_{2f}$.

Step3: Solve for $v_{2f}$

Rearrange the equation from Step 2 to solve for $v_{2f}$: $v_{2f}=v_{1i}+v_{2i}-v_{1f}$.
Substitute the values: $v_{2f}=6.5+( - 6.1)-( - 3.7)$.
First, $6.5+( - 6.1)=6.5 - 6.1 = 0.4$ m/s.
Then, $0.4-( - 3.7)=0.4 + 3.7=4.1$ m/s.

Answer:

$4.1$ m/s