A 1424-kg railroad car traveling at a speed of 11 m/s strikes an identical car at rest. If the cars lock together as a result of the collision, what is their common speed (in m/s) afterward?
(m1+m2)*V = m1*V1
(1424+1424)*V = 1424*11
2848V = 15,664
V = 5.5 m/s.
To find the common speed of the railroad cars after the collision, we need to 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 given by the product of its mass and velocity:
Momentum = mass × velocity
Let's denote the mass of each railroad car as m (1424 kg in this case) and the initial velocity of the first car as v1 (11 m/s) and the initial velocity of the second car as v2 (0 m/s since it is at rest).
Before the collision, the total momentum is the sum of 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)
= (m × v1) + (m × v2)
= (m × 11 m/s) + (m × 0 m/s)
= (11m^2) kg⋅m/s
After the collision, the two cars will lock together and move as one unit with a common velocity, which we need to find. Let's denote the common velocity as v.
The total momentum after the collision is the product of the combined mass (2m, since each car has the same mass) and the common velocity:
Total final momentum = (combined mass × common velocity)
= (2m × v)
According to the principle of conservation of momentum, the total initial momentum is equal to the total final momentum:
Total initial momentum = Total final momentum
(11m^2) kg⋅m/s = (2m × v)
Now we can solve for v:
v = (11m^2) kg⋅m/s / (2m)
v = 11 m/s
Therefore, the common speed of the railroad cars after the collision is 11 m/s.