1. Nov 27, 2007

### 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:\$

2. Nov 27, 2007

### 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

3. Nov 27, 2007

### 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

???

4. Nov 27, 2007

### rock.freak667

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

5. Nov 27, 2007

### 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.

6. Nov 27, 2007

### jeeves_17

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

7. Mar 4, 2009

### 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.