Answer:
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum of a system of objects is conserved before and after a collision, provided no external forces act on the system.
Before the collision, the momentum of the first train car is:
p1 = m1v1 = (2000 kg)(20 m/s) = 40000 kg·m/s
where m1 is the mass of the first train car and v1 is its velocity.
The second train car is stationary, so its momentum is:
p2 = m2v2 = (2000 kg)(0 m/s) = 0 kg·m/s
where m2 is the mass of the second train car and v2 is its velocity.
The total momentum before the collision is:
p1 + p2 = 40000 kg·m/s + 0 kg·m/s = 40000 kg·m/s
After the collision, the two train cars stick together and move as a single unit. Let their combined velocity be v:
The total mass of the system after the collision is:
m = m1 + m2 = 2000 kg + 2000 kg = 4000 kg
The total momentum after the collision is:
p = mv = (4000 kg)(v) = 4000v kg·m/s
According to the principle of conservation of momentum, the total momentum before and after the collision must be equal. Therefore:
p1 + p2 = p
Substituting the expressions for p1, p2, and p, we get:
40000 kg·m/s + 0 kg·m/s = 4000v kg·m/s
Solving for v, we get:
v = (40000 kg·m/s) / (4000 kg) = 10 m/s
Therefore, the combined speed of the two train cars after the collision is 10 m/s.
Step-by-step explanation:
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