Question

An empty 2,500 kg train car is headed northbound at a velocity of 5 m/s. Ahead of the first car, an empty 1,500 kg car is headed northbound on the same track at 1 m/s. The faster train bumps into the slower one, and they combine into one train. What is the velocity of the two-car train? 1 m/s 3 m/s 3. 5 m/s 5 m/s.

Answer 1

The velocity of the two-car train is 3.5 m/s.

To solve the problem above, we use the law of conservation of momentum

Law of conservation of momentum: States that, for a closed system, the total momentum before collision is equal to the total momentum after collision.

⇒ Formula:

mu+m'u' = V(m+m')..................... Equation 1

⇒ Where:

• m = mass of the faster train
• m' = mass of the slower train
• u = initial velocity of the faster train
• u' = initial velocity of the slower train
• V = common velocity of the two trains

⇒ make V the subject of the equation

• V = (mu+m'u')/(m+m')....................... Equation 2

From the question,

⇒ Given:

• m = 2500 kg
• m' = 1500 kg
• u = 5 m/s
• u' = 1 m/s

⇒ Substitute these values into equation 2

• V = [(2500×5)+(1500×1)]/(2500+1500)
• V = [12500+1500]/(4000)
• V = 14000/4000
• V = 3.5 m/s.

Hence, The velocity of the two-car train is 3.5 m/s

Learn more about the law of conservation of momentum: brainly.com/question/9408577