Velocity of Car after Unloading Gravel

[jeeves_17]

1. A 10,000 kg railroad car is rolling at 2.00m/s when a 4000kg load of gravel is suddenly dropped in. What is the car’s speed just after the gravel is loaded?



Relevant Equations

W = 1/2 mv(f)^2 - 1/2 mv(i)^2

3. The Attempt at a Solution [/b]

W = 1/2 (14,000kg)v(f)^2 - 1/2(10,000)(2.0)^2


I'm lost:$

 

[rock.freak667]

The most I can guess is that consider momentum to be conserved..you can find the initial momentum and then you can formulate the final and they should be equal

 

[jeeves_17]

Pi= mi*vi
= (10 000)(2.0)
= 20 000

Pi = Pf = 20 000

Pf = mf*vf
Vf = Pf/mf
= 20 000/14 000
= 1.42857
= 1.43 m/s


?

 

[rock.freak667]

Yes that would seem feasible as if it is heavier it would travel slower

 

[xcvxcvvc]

If you ignore friction and the initial velocity of the gravel then you can use conservation of momentum with ease.
2*10000 = V * (4000 + 10000)

I got 1.43m/s too. Explain the "?" Did you check it with the answer key and it's wrong or are you saying "is this right"? I'd say it's right.

 

[jeeves_17]

? meaning is it right...sorry for the confusion

 

[PnotConserved]

I did this same problem and got 1.43 m/s, but the books answer says .143 m/s. I was getting frustrated until I saw these posts. I guess the key is wrong.