A 10,000 kg railroad car traveling at a speed of 24.0 m/s strikes an identical car at rest. If the cars lock together as a result of the collision, what is their common speed afterward?

To determine the common speed of the two railroad cars after the collision, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

The momentum of an object is defined as the product of its mass and velocity.

Let's denote the mass of each railroad car as m and the initial velocity of the first car as v1 and the initial velocity of the second car as v2. Since the second car is at rest, v2 = 0.

The momentum before the collision is given by the sum of the individual momenta of each car:

Initial momentum = m1 * v1 + m2 * v2

Since the second car is at rest, the equation simplifies to:

Initial momentum = m1 * v1 + 0

The total mass of the two cars is equal to the sum of their individual masses:

Total mass = m1 + m2

After the collision, the cars lock together and move as a single object, so they will have a common velocity, which we'll call v.

Therefore, the final momentum is given by:

Final momentum = (m1 + m2) * v

According to the principle of conservation of momentum, the initial momentum and final momentum are equal:

m1 * v1 + 0 = (m1 + m2) * v

Now we can rearrange the equation to solve for v, the common velocity of the cars:

v = (m1 * v1) / (m1 + m2)

Let's substitute the given values:

m1 = m2 = 10,000 kg
v1 = 24.0 m/s

Plug these values into the equation and calculate to find their common velocity:

v = (10,000 kg * 24.0 m/s) / (10,000 kg + 10,000 kg)

v = (240,000 kg·m/s) / 20,000 kg

v = 12.0 m/s

Therefore, the common speed of the two railroad cars after the collision is 12.0 m/s.

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