# Homework Help: An Energy Question: Semi-Urgent

1. Dec 6, 2009

### tormaplea

1. The problem statement, all variables and given/known data
Two men, of mass 100kg each, stand on a cart of mass 300kg.The cart can roll with negligible friction along a north-south track, and everything is initially at rest. One man runs toward the north and jumps of the cart at 5 m/s, relative to the cart. After he jumps, the second man runs to the south end and jumps off at 5 m/s relative to the cart. Calculate the speed and direction of the cart after both men have jumped off.

2. Relevant equations
Ek=1/2mv^2

3. The attempt at a solution
I don't know how to begin since it is at rest and there for Ek is 0 meaning every other step of this should be 0.

2. Dec 6, 2009

### mgb_phys

Conservation of momentum (mv) might be useful,

3. Dec 6, 2009

### tormaplea

but when i use that I get P1=P2
500kg*0m/s=100kg*5m/s+400kg*v
500=500+400v
0=400v
0/400=error
that is the problem i am dealing with

4. Dec 6, 2009

### MaxL

Tormaplea--you did something weird going between these two lines:

500kg*0m/s=100kg*5m/s+400kg*v
500=500+400v

You didn't multiply the 500 by the 0 m/s. That might be the problem.

5. Dec 6, 2009

### tormaplea

that did solve the first problem i had
0=500+400v
-500/400=v=-1.25
now ive come into another problem
next step is the se3cond person
0=100v+300*-5
1500=100v
15=v
15+(-1.25)=13.75
I do know the answer is .25m/s[N] and i must be missing a step

6. Dec 6, 2009

### MaxL

Your problem is in this line:

0=100v+300*-5

For one thing, it's not clear what you mean by v--is it the v of the jumper, the the v of the cart before the jump, or the v of the cart after the jump? And why is there a zero on the left side? At this point, the momentum of the cart is NOT zero.

I think you need to draw a picture, and maybe try writing this next equation symbolically:

initial_velocity_cart(mass_cart + mass_person) = ...

it always helps me to think of things a little more abstractly.

7. Dec 6, 2009

### MaxL

Oh! And be careful, 5 m/s is the velocity of the second jumper relative to the cart.

8. Dec 6, 2009

### tormaplea

so are you telling me it would be like
-1.25*(100+300)=100*-5+300*v(v of cart)

9. Dec 6, 2009

### MaxL

Almost! The jumper's velocity is 5m/s relative to the cart. That's absolutely not 5 m/s! What is it?

Also, it's not clear if you mean 300*v*v_cart, or rather 300*(v_cart). Keep in mind, you must keep your equations dimensionally consistent (so no adding m*v + m*v^2).

You're doing great!

10. Dec 6, 2009

### tormaplea

I was just stating that v was the velocity of the cart because of post 6, Im trying my best to keep the equations consistent
Also would the velocity of jumper 2 be -6.25?
-1.25+-5???

Last edited: Dec 6, 2009
11. Dec 6, 2009

### MaxL

Yep!

12. Dec 6, 2009

### tormaplea

but how do i turn that number into .25[N]

13. Dec 6, 2009

### MaxL

The answer I got was 5/12 m/s. I'm pretty confident about it.

What's the [N]?

14. Dec 6, 2009

### tormaplea

[N] is north and I also 5/12 but my teacher has already provided us with the answer of .25 [N]. Is there any place you see that we could have messed up on

15. Dec 6, 2009

### MaxL

Ah, I just did it a different way and got .25 to the north.

Okay, so it seems the mistake we were making was in treating the velocity of the jumpers incorrectly. We were saying 5 m/s was the velocity of the jumper relative to the cart before the jump. So if we were standing off the tracks on the ground (According to our incorrect reasoning) we would see the cart and both guys at rest, then one would jump, and he'd be moving 5 m/s relative to us. The cart would also be moving, so the jumper and the cart would be moving away from each other at more than 5 m/s.

The way your teacher apparently WANTS us to think about it is that, after the jump, the jumper and the cart are moving away from one another at 5 m/s. It's a much harder problem, but probably a bit more realistic.

To do it, you'll need FOUR equations--conservation of momentum and a velocity sum equation for each jump.

16. Dec 7, 2009

### tormaplea

does that mean the jumper only jumps at 2.5m/s

17. Dec 7, 2009

### MaxL

It does not. It would if the jumper and the thing they were jumping off of had equal mass (and I'm assuming you meant the first jumper, and 2.5m/s referred to the velocity).

As for the velocity of the cart relative to the ground after the first jumper jumps: You know the initial velocity of the cart, and the initial velocity of the jumper, and the mass of the cart, and the mass of the jumper, and you know the difference between vjumper and vcart immidiately after the jump. You want vjumper and vcart after the jump. 2 equations, 2 unknowns!