# Homework Help: Momentum and speed relativity

1. Nov 8, 2015

1. The problem statement, all variables and given/known data
Jennifer (mass 60.0kg) is standing at the left end of a 14.0m long 538.0kg cart that has frictionless wheels and rolls on a frictionless track. Initially both Jennifer and the cart are at rest. Suddenly, Jennifer starts running along the cart at a speed of 5.10m/s relative to the cart. How far will Jennifer have run relative to the ground when she reaches the right end of the cart?

2. Relevant equations
momentum = mv
p is momentum
3. The attempt at a solution
Pi = 0 so

60*5.1 = 538v

v=0.568m/s velocity of cart

14m of cart takes her t(5.668m/s) = 14
t=2.46s

I'm confused about this relativity, do i use velocity of 5.1 when calculating conservation of momentum ? or is is speed relative to ground?

How do i get to a number in meters relative to the ground that she ran?

2. Nov 8, 2015

### J Hann

The initial momentum of Jenifer and cart is zero and will remain zero since there are no external forces.
Now you have the two velocities in that (ground) frame.
Then you can find the velocity of Jenifer to the ground by simple vector addition.

3. Nov 8, 2015

### PeroK

You might ask yourself this: does it matter how fast she runs?

4. Nov 8, 2015

Ok so relative to the ground she is travelling at 5.1m/s - 0.568m/s = 4.523m/s

4.532m/s * 2.46s = 11.148m

This is also an incorrect answer.

I looked over my math and it all looks good, am I missing anything to consider?

5. Nov 8, 2015

I would think so, no?

If she travels slower the cart would also travel slower, so I guess not then?

How could I implement this mathematically I feel a bit stuck here.

6. Nov 8, 2015

### PeroK

First, can you give a good argument why the speed doesn't matter?

In any case a hint is to think more directly about distances (rather than thinking about speeds).

7. Nov 8, 2015

Well since the net momentum is 0 and she increases her speed the speed of the cart would increase to cancel the momentum change

Since we are talking about distances traveled relative to the ground her speed relative to the carts are irrelevant.

Proof: Using her speed of 5.1m/s relative to the cart I find that it takes her 2.46s to reach the end of the cart
Using her speed as 10m/s I find it takes her 1.26s to reach the end of the cart

Using vector addition I find the first case the displacement relative to the ground is 11.148m
Using vector addition I find the second case the displacement relative to the ground is 11.12m

?? So something here must be going wrong

8. Nov 8, 2015

### PeroK

Momentum of the system is 0, relative to the ground. If her speed is $v$ and the cart's speed is $V$ (relative to the ground), then by conservation of momentum:

$mv$ = $MV$ (where $m$ is Jennifer's mass and $M$ is the cart's).

Can you do something with that?

9. Nov 8, 2015

Yeah, I believe i used that here earlier to calculate the speed of the cart with the given speed of jennifer

"
Pi = 0 so

60*5.1 = 538v

v=0.568m/s velocity of cart"

10. Nov 8, 2015

### PeroK

I don't like all these numbers. Numbers obscure what's going on.

Also, if you work in symbols, you would actually see when solving this problem that $v$ does not matter, and that would teach you a lot about the Physics.

Go back to:

$mv = MV$

Forget the $5.1m/s$, which is irrelevant and think about the distances travelled (relative to the ground). Note that $v$ is the speed relative to the ground, not the cart, so you don't know $v$ or $V$.

Hint: let $d$ and $D$ be the distances travelled (relative to the ground) of J and the cart respectively.

11. Nov 8, 2015

Okay

mv=MV

This is conservation of momentum, I know m= 60 M = 538
I only have one equation here trying to poke at two unknowns is there another formula I can pull out somewhere using v=5.1m/s relative to the ground?

12. Nov 8, 2015

### PeroK

Let me help:

$d = vt$ and $D = Vt$

Can you use this to relate $d$ and $D$?

13. Nov 8, 2015

### J Hann

Where did you get 2.46 seconds?
If she runs at 5.1 m/s on a 14 m cart, how long does it take to get to the end of the cart?

14. Nov 8, 2015

I'm not 100% sure but is it d/v = D/V ?

15. Nov 8, 2015

### PeroK

Yes, but we don't know $v$ and $V$, but we do know $m$ and $M$.

16. Nov 8, 2015

There's too many equations being thrown around I'm getting very lost, how could I even relate m to v and M to V throuhg d/v = D/V

17. Nov 8, 2015

### PeroK

Okay. Maybe you want to follow J Hann's suggestion? The problem is that if you just plug numbers into equations, you don't learn much about Physics and, as you've found, one mistake in your calculations and it's very hard to spot.

To give you the next step in any case:

$md = mvt = MVt = MD$

So, not only are the speeds related simply by the relative mass, but the distances travelled (by J and cart) are related in the same way by their relative mass. And that is ... Physics!

Also, this applies to any motion (not just constant velocity): J could walk, speed up, go backwards, go at any variable speed she likes and, at all times:

$md = MD$ (relative to the ground)

18. Nov 8, 2015

Whats confusing me is we started from

Jennifer (mass 60.0kg) is standing at the left end of a 14.0m long 538.0kg cart that has frictionless wheels and rolls on a frictionless track. Initially both Jennifer and the cart are at rest. Suddenly, Jennifer starts running along the cart at a speed of 5.10m/s relative to the cart. How far will Jennifer have run relative to the ground when she reaches the right end of the cart?

And are getting to md = mvt = MVt = MD
and I have no clue where you got these letters from and what they represent and how you kind of just pulled them out of your pocket.

Walk me through it using letters if you can maybe ill understand.

19. Nov 8, 2015

### PeroK

$m$ is Jennifer's mass; $M$ is the mass of the cart; $d$ is the distance travelled by Jennifer (relative to the ground); and $D$ is the distance travelled by the cart (relative to the ground).

20. Nov 8, 2015

Okay, and now im confused on how you figured out that
md = mvt = MVt = MD
how do i know this is true? Where would i be able to pull this formula from?

21. Nov 8, 2015

### PeroK

I've shown it above using conservation of momentum. At all times, Jennifer's speed is (M/m) times the cart's speed, so the distance travelled by Jennifer (at any time) must be (M/m) times the distance travelled by the cart.

22. Nov 8, 2015

yes unfortunatly im not understanding this, thanks for the help though ! :)

23. Nov 8, 2015

### PeroK

Okay, I'm going offline now anyway.

I looked at your numbers in your solution, but I don't see how you got them. You probably need to show how you calculated things. For example:

v=0.568m/s velocity of cart

This isn't correct.

In fact, you probably need to consider velocities relative to the ground even in your solution!

24. Nov 9, 2015

### vela

Staff Emeritus
When you say this is the velocity of the cart, you presumably mean it's the velocity of the cart relative to the ground, right? In other words, you're working in the reference frame where the ground is at rest. If that's the case, then you need to use the velocity of Jennifer relative to the ground on the lefthand side of the equation. That's not equal to 5.1 m/s, is it?

25. Nov 10, 2015

Yeah sorry, I got it eventually, that was a tough question.