# Momentum Help

1. Apr 14, 2013

### JeffG

1. The problem statement, all variables and given/known data
The problem is that there are too cars and they crash and we have to gigure out which one was the one that caused the crash?

Also it is a four way stop

here is a pic that would help

2. Relevant equations

Find Velocity

3. The attempt at a solution

I don't know where to start

Thanks for your help

2. Apr 14, 2013

### Dewgale

The only thing I can suggest is look at the formula for the conservation of momentum.

$P_{1}+P_{2}=P_{3}+P_{4}$

P=mv, so let's fill in what we know.

$1500v_{1}+2000v_{2}=P_{3}+P_{4}$

You said the cars combined, so P3=P4. We'll call this Pt

$1500v_{1}+2000v_{2}=3500v_{t}$

That's the farthest I can take you, since we don't know any of the velocities. Could you post the actual wording of the question, and see if there's any information you left out?

3. Apr 14, 2013

### JeffG

This is why it's confusing because all I got was that picture

4. Apr 14, 2013

### JeffG

I just realized that the angle is 30 degrees and the distance they traveled after crash was 11 meters

I made a mistake in the pic

5. Apr 14, 2013

### Dewgale

Well, the only thing I can think of is this:

Since both cars arrive at the same point at the same moment, the time it took them to get there from their different points must have been the same. Therefore,

$\frac{1500d_{1}+2000d_{2}}{t_{o}}=\frac{3500d_{f}}{t_{f}}$

As I was typing this, I saw your edit.

That makes a bit of a difference. If you know the distance they travelled, you can fill that in the formula.

$\frac{1500d_{1}+2000d_{2}}{t_{o}}=\frac{38500}{t_{f}}$

Unfortunately, I'm not sure where to go from there.

6. Apr 14, 2013

### JeffG

Thanks so much for your help, hopefully some on else can help me too

Thanks so much!!

Wait what if one does not stop and one was speeding does that make a difference?

7. Apr 14, 2013

### Dewgale

Not particularly.

I've had an idea, though I have no clue if it's right.

Theoretically, vf=v1+v2

Therefore, if you disregard time, df=d1+d2 should also be correct, though I'm not totally sure.

If it is, then 11*cos(30)= 9.526 m, and 11*sin(30)=5.5 m

$\frac{25289.419}{t_{o}}=\frac{38500}{t_{f}}$

$25289.419t_{f}=38500t_{o}$

t$_{f}\approx1.522t_{o}$

Don't take my word for that, though; I'm not 100% sure it's right.

8. Apr 14, 2013

### JeffG

It looks pretty accurate but Im not even sure we need time for this problem, but again thanks for your help

9. Apr 14, 2013

### haruspex

Since the friction after the crash is unknown, the distance travelled tell us nothing. With information provided, you can determine the ratio of the speeds of the two vehicles, and that's all. E.g., in one extreme, they were both travelling quite slowly but the road surface is an ice rink.
The diagram shows both cars in the middle of the road and the crash in the centre of the junction. OTOH, it gives the width of the road. If we assume each was driving on the right, the 1500kg car had further to go from the stop line to the impact point. If you make some reasonable assumption about the lengths of the vehicles, the ratio of the speeds might be enough to come to a conclusion.

10. Apr 14, 2013

### JeffG

is it possible you can start me off?

11. Apr 14, 2013

### haruspex

Can you calculate the speed ratio? Consider conservation of momentum in the two approach directions and use the known angle after the crash.