Solving Car Crash Scenario: Find v1 of Car A

[Raysi25]

Homework Statement

Car A (m1=1405kg) crashes into Car B (m2=873kg) which is stationary and perpendicular to Car A. Car B is moved a distance of 1.8m out of the way and Car A rolls to a stop 9.6m away at 1.93 m/s. Assuming the coefficient of friction is 0.8, can the velocity of Car A be calculated before impact?

Please see amazing diagram attached for a clear depiction of crash.

Variables:
m1=1405kg
m2=873kg
v1=?
v2=0m/s
µ=0.8
d=1.8m

Homework Equations

W=F*d
F=µmg

The Attempt at a Solution

Force required to move Car B
F=µmg
= 0.8*873*9.8
= 6844N

Work done to move Car B
W=F*d
= 6844 * 1.8
= 12,320J

Not sure where to go from here...can anyone please point me in the right direction?

 

[AlephNumbers]

I am going to assume that is the coefficient of kinetic friction. What we have is an elastic collision. What equations apply to elastic collisions in one dimension? I believe you should be able to find the correct equation in the introductory physics formulay if you do not know it off the top of your head.

 

[CWatters]

Work done to move Car B
W=F*d
= 6844 * 1.8
= 12,320J

Have a think about where that energy came from. Are there any other exchanges of energy involved?

 

[BvU]

Hello Raysi, welome to PF !

You did well in calculating the kinetic energy car B must have had right after getting bumped by car A.
Unlike Aleph, I am quite convinced that a broadside hit like this is anything but an elastic collision,
so all you have is conservation of momentum. And that forces me to assume all the info you have is
in the direction car A was moving originally.

As CW says: the work done to move car B comes from somewhere, probably the bump, right ?
Something similar for car A and for the reasons I describe above, all you can do is add the two momenta
to estimate car A speed beforehand.

---

I am also a bit critical about the problem statement and the amazing diagram: what does it mean
"Car A rolls to a stop 9.6m away at 1.93 m/s":
1) that it has a speed of 1.93 m/s (how acccurately determined !) right after the collision and then gently moves on another 9.6 m while slowing down only minimally
2) that even after maximum braking over 9.6 m (friction coefficient 0.8) car A still has a speed of 1.93 m/s ?

 

[AlephNumbers]

Uh wow. Yeah. Listen to BvU. I'm not sure what I was thinking there. Good luck!

 

[CWatters]

I am also a bit critical about the problem statement...

I agree. No angles given?

 

[Raysi25]

Thanks so much for your help everyone - I'm in over my head at the moment!

To answer some questions:

I think the force on Car A would be -6844N

BvU - The speed after the collision and the stop of car A is assuming that the only resistance is rolling resistance, so brakes are no longer being applied, allowing it to coast for 9.6m.

CWatters - There were no angles in the question, but it was stated that the crash happened at an intersection where the roads were 12m wide. The original image (not the diagram I added) shows Car A stopping in the same lane it started in, giving it a maximum horizontal movement of 6m. Subtracting the width of the car (1.8m) I have since worked out that the maximum angle that Car A can move after the collision is 19.29 degrees (Tan-1(4.2/12)).

So using a conservation of momentum formula I have:

m1v1 + m2v2 = m3v3 + m4v4

If I do that I end up with

1405*v1 + 873*0 = 1405 * 1.93 + 873*v4

so I need to find v1 and v4.

But can I substitute v4 into KE=1/2mv² to get v4?

12,320=0.5*873*v²
v²=28.22
v=5.31m/s

so => 1405*v1 + 873*0 = 1405*1.93 + 873*5.31

v1 = (1405*1.93+873*5.31)/1405
v1 = 5.22 m/s or 18.82km/h

Does that seem like an answer that makes sense? Or is the speed too slow?

 

[haruspex]

Clearly the impact was not in line with B's mass centre, so car B will have spun around. This makes it impossible to know what distance the car's wheels have skidded. Ignoring that, you can calculate the work done on car B, as you have.
That tells you the speed of car B just after collision.

With no angles given, I think you have to assume all the movements are almost in a single line.
There are two options from there.
It's hard to know how to interpret the data regarding A's post-collision movement. Presumably its acceleration is constant. You could take the speed as being immediately after collision, or as its average until it stops. Either way, the stopping distance is of no interest since you've no idea what force is stopping it.
Alternatively, you could assume the collision is completely inelastic, i.e. the two vehicles have the same post-collision speed, and ignore the rolling to a stop data. Once again, though, this is complicated by the off-centre collision. A may have emerged with the greater linear speed.

 

[haruspex]

the force on Car A would be -6844N

No. During the collision it is accelerating car B, so the force would be much greater. Once the collision is over (and we don't know how long it lasted), there is very little force on car A.

 

[Raysi25]

Ah - so I need the force on Car A to find the Work done by Car A, so that I can substitute it into KE=½mv² and get v3.

 

[haruspex]

Ah - so I need the force on Car A to find the Work done by Car A, so that I can substitute it into KE=½mv² and get v3.

No, forces during collision are indeterminable. It's the impulse (momentum) that matters here.