A 20,000 kg railroad car is traveling at 3 m/s when it collides and couples with a second, identical car at rest. What is the resulting speed of the combined cars?
starting momentum=final momentum
20,000*3+0=40,000V solve for v.
1.5
To find the resulting speed of the combined cars, we can use the law of conservation of momentum, which states that the total momentum of an isolated system remains constant before and after a collision.
The equation for momentum is given by:
Momentum = mass * velocity
Let's denote the mass of each car as m (which is 20,000 kg in this case), and the initial velocity of the first car as v1 (which is 3 m/s).
Since the second car is at rest initially, its initial velocity v2 is 0 m/s.
We can calculate the total momentum before the collision by adding the individual momenta of the two cars:
Total initial momentum = (mass of first car * velocity of first car) + (mass of second car * velocity of second car)
Total initial momentum = (m * v1) + (m * v2)
= (20,000 kg * 3 m/s) + (20,000 kg * 0 m/s)
= 60,000 kg∙m/s
According to the law of conservation of momentum, the total momentum after the collision will also be 60,000 kg∙m/s.
Now, let's assume the resulting speed of the combined cars after the collision is v_r.
The mass of the combined cars would be the sum of the individual masses:
Mass of combined cars = mass of first car + mass of second car
= m + m
= 2m
Using the equation for momentum, we can find the momentum of the combined cars after the collision:
Total final momentum = (mass of combined cars * resulting speed of combined cars)
= (2m * v_r)
Since the initial momentum and final momentum are equal (due to conservation of momentum), we have:
Total initial momentum = Total final momentum
(m * v1) + (m * v2) = 2m * v_r
Substituting the given values:
60,000 kg∙m/s = (2 * 20,000 kg) * v_r
Simplifying the equation, we can solve for v_r:
v_r = 60,000 kg∙m/s / (40,000 kg)
= 1.5 m/s
Therefore, the resulting speed of the combined cars after the collision is 1.5 m/s.