Solving the Mystery of a 4-Way Car Crash

[JeffG]

Homework Statement

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

Homework Equations

Find Velocity

The Attempt at a Solution

I don't know where to start

Thanks for your help

 

[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 P_t

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

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?

 

[JeffG]

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

 

[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

 

[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.

 

[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?

 

[Dewgale]

Not particularly.

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

Theoretically, v_f=v₁+v₂

Therefore, if you disregard time, d_f=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.

 

[JeffG]

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

 

[haruspex]

Since the friction after the crash is unknown, the distance traveled 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 traveling 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.

 

[JeffG]

Since the friction after the crash is unknown, the distance traveled 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 traveling 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.

is it possible you can start me off?

 

[haruspex]

is it possible you can start me off?

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