Question

engineers are using computer models to study train collisions to design safer train cars. they start by modeling an elastic collision between two train cars traveling toward each other. car 1 is traveling north at 20 m/s and has a mass of 12,745 kg. car 2 is traveling south at 15 m/s and has a mass of 4,125 kg. after the collision, car 1 has a final velocity of 3 m/s north. what is the final velocity of car 2?

a. 38 m/s north

b. 38 m/s south

c. 56 m/s north

d. 56 m/s south

Explanation:

Step1: Define Variables and Direction

Let north be positive. So, \( m_1 = 12745 \, \text{kg} \), \( v_{1i} = 20 \, \text{m/s} \), \( m_2 = 4125 \, \text{kg} \), \( v_{2i} = -15 \, \text{m/s} \) (south is negative), \( v_{1f} = 3 \, \text{m/s} \). For elastic collision, \( m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \).

Step2: Plug in Values

Substitute into the momentum conservation equation: \( 12745(20) + 4125(-15) = 12745(3) + 4125v_{2f} \).

Step3: Calculate Left Side

\( 12745\times20 = 254900 \), \( 4125\times(-15) = -61875 \). Sum: \( 254900 - 61875 = 193025 \).

Step4: Calculate Right Side

\( 12745\times3 = 38235 \). So, \( 193025 = 38235 + 4125v_{2f} \).

Step5: Solve for \( v_{2f} \)

Subtract 38235: \( 193025 - 38235 = 4125v_{2f} \Rightarrow 154790 = 4125v_{2f} \). Divide: \( v_{2f} = \frac{154790}{4125} \approx 37.5 \approx 38 \, \text{m/s} \). Positive means north.

Answer:

A. 38 m/s north