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A railroad diesel engine coasting at 10 Kmh runs into a stationary flatcar The diesel weighs four times as much as the flatcar. How fast do the cars move when they stick together upon collision?
Wiki User
∙ 15y ago
There are no forces external to the engine and car involved here*, thus this is a case of conservation of momentum. Note, an unknown amount of energy is absorbed by the couplers and other parts of the engine and car, so the conservation of energy equation is not useful here.
Momentum = velocity X mass
Since momentum is conserved, the total velocity X mass before the collision will equal total velocity X mass after the collision.
If we define the mass of the car as M, then the mass of the engine is 4M
Let:
the initial velocity of the engine = Ve1 = 10Kmh
the initial velocity of the car = Vc1 = 0 Kmh
the final velocity = V2 (it is the same for both the car and the engine)
So the initial momentum is: (Ve1) (4M) + (Vc1) (M) = (10) (4M) + (0) (M) = 40M
The final momentum is: (V2) (4M + M) = (V2) (5M) = 5V2M
Setting the initial momentum equal to the final momentum gives:
40M = 5V2M
Doing the algebra gives:
40 = 5V2
8 = V2 So, the answer is the final velocity is 8 Kmh
*We are assuming friction of the wheels on the track is negligible and that the track is level so that gravity can be ignored.
Wiki User
∙ 15y ago
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