Question

question 7. two identical cars, each with a mass of 1,000 kg, are moving towards each other. car a has a velocity of 10 m/s to the right, and car b has a velocity of 5 m/s to the left. after the collision they move off separately. what are the final velocities of each car? a. car a - 5 m/s, car b - 5 m/s b. car a - 5 m/s, car b 10 m/s c. car a 10 m/s, car b 10 m/s d. car a 10 m/s, car b - 10 m/s

Explanation:

Step1: Set up the momentum conservation equation

Let the mass of each car \(m = 1000\space kg\), \(v_{A1}=10\space m/s\), \(v_{B1}=- 5\space m/s\) (taking right - direction as positive). According to the law of conservation of momentum \(m_{A}v_{A1}+m_{B}v_{B1}=m_{A}v_{A2}+m_{B}v_{B2}\). Since \(m_{A} = m_{B}=m\), the equation simplifies to \(v_{A1}+v_{B1}=v_{A2}+v_{B2}\). Substituting the values: \(10+( - 5)=v_{A2}+v_{B2}\), so \(v_{A2}+v_{B2}=5\).

Step2: Consider the nature of elastic collision (since cars are identical and move off separately, assume elastic collision where relative speed of approach equals relative speed of separation)

The relative speed of approach is \(v_{A1}-v_{B1}=10-( - 5)=15\space m/s\). The relative speed of separation is \(v_{B2}-v_{A2}\). For elastic collision \(v_{B2}-v_{A2}=v_{A1}-v_{B1} = 15\space m/s\).

Step3: Solve the system of equations

We have the system of equations \(

$$\begin{cases}v_{A2}+v_{B2}=5\\v_{B2}-v_{A2}=15\end{cases}$$

\). Add the two equations: \((v_{A2}+v_{B2})+(v_{B2}-v_{A2})=5 + 15\). \(2v_{B2}=20\), so \(v_{B2}=10\space m/s\). Substitute \(v_{B2}=10\) into \(v_{A2}+v_{B2}=5\), we get \(v_{A2}=- 5\space m/s\).

Answer:

B. Car A - 5 m/s, Car B 10 m/s