A 40,000 kg railroad car initially traveling at 10 m/s collides inelastically with a 20,000 kg railroad car intially at rest. The cars stick together. What is their final speed?
Use total linear momentum conservation.
40,000*10 = 60,000*Vfinal
Solve for Vfinal
To find the final speed of the two railroad cars after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
The momentum of an object is defined as the product of its mass and velocity. Mathematically, momentum (p) is given by:
p = mass × velocity
Let's assign variables to the relevant quantities:
m1 = mass of the first railroad car (40,000 kg)
v1 = initial velocity of the first railroad car (10 m/s)
m2 = mass of the second railroad car (20,000 kg)
v2 = initial velocity of the second railroad car (0 m/s)
Using the conservation of momentum:
m1v1 + m2v2 = (m1 + m2)vf
In this case, since the two cars stick together, their final velocity (vf) will be the same.
Now we can substitute the values into the equation and solve for vf:
(40,000 kg) × (10 m/s) + (20,000 kg) × (0 m/s) = (40,000 kg + 20,000 kg) × vf
(400,000 kg·m/s) = (60,000 kg) × vf
Dividing both sides by 60,000 kg:
vf = (400,000 kg·m/s) / (60,000 kg)
vf ≈ 6.67 m/s (rounded to two decimal places)
Therefore, the final speed of the two railroad cars, after colliding inelastically and sticking together, is approximately 6.67 m/s.