A 40,000 kg railroad car initially traveling at 10 m/s collides in elastically with a 20,000 kg railroad car intially at rest. The cars stick together. What is their final speed?
If they stick together, the collision is NOT elastic. Recheck the wording of your question. The question can be answered using conservation of momentum.
To find the final speed of the two cars after the collision, we can use the law of conservation of momentum. According to this law, the total momentum before the collision should be equal to the total momentum after the collision.
The momentum (p) of an object is calculated by multiplying its mass (m) by its velocity (v):
p = m * v
Let's label the first railroad car as Car A with a mass of 40,000 kg and an initial velocity of 10 m/s (vA = 10 m/s), while the second railroad car is Car B with a mass of 20,000 kg and an initial velocity of 0 m/s (vB = 0 m/s).
The total momentum before the collision is the sum of the individual momenta of the two cars:
p_initial = pA_initial + pB_initial
Since Car A is moving and Car B is at rest, the momentum equations become:
pA_initial = mA * vA
pB_initial = mB * vB
We substitute the known values:
p_initial = (40,000 kg * 10 m/s) + (20,000 kg * 0 m/s)
= 400,000 kg·m/s
Now, since the cars stick together and move as one mass after the collision, we can find the final velocity (vf) of the combined system.
The total momentum after the collision is:
p_final = (mA + mB) * vf
Using the principle of conservation of momentum, we can set the initial and final momenta equal to each other:
p_initial = p_final
Therefore, we have:
400,000 kg·m/s = (40,000 kg + 20,000 kg) * vf
Now, we can solve for the final velocity:
vf = 400,000 kg·m/s ÷ 60,000 kg
= 6.67 m/s
Therefore, the final speed of the two cars, when combined and moving together after the collision, is approximately 6.67 m/s.