Unraveling the Mystery of a Two-Car Crash

[loka]

Homework Statement

On May 19, an accident occurred at the corner of two asphalt streets with no stop signs. The speed limit was 40 km/h on both streets. As lead CSI on the case, you must determine whether one or both cars should be ticketed for speeding, which car had right of way at the intersection, and whether one driver or both were at fault for the accident.

Using data tables, you have determined that the mass of car A is 2275 kg and the mass of car B is 1525 kg. You measured the skid marks when the cars were locked together and found that they were 3.705 m long at an angle of 47 degrees east of the northbound street. Witness accounts put both cars 5 m from the intersection at the same time. The coefficient of kinetic friction b/w rubber and asphalt is 0.80. The rules of the road at an intersection w/o stop signs is that if 2 cars get to the intersection at the same time, the right-hand car has right of way.

Please see sketch.

Homework Equations

J = F (t)

p = mv

The Attempt at a Solution

...I'm really at a lost of what to do...

 

[fatra2]

Hi there,

I can't open you image.

Cheers

 

[DorianG]

Think about the motion of the two cars joined together - treat this as a single object which moves in the direction and distance that is stated in the problem. What forces are acting on this object? It is moving with a certain velocity v, then it moves 3.705m, and final velocity is 0 when it's stopped. What can you do to find the initial veloctiy of the object (the two cars)? What principle will help you when you have this value v?

 

[loka]

u=initial velocity
v=final velocity
A=1
B=2

Given:

M₁ = 2275 kg
M₂ = 1525 kg
d = 3.705 m
angle = 47 deg
coeff. of friction = 0.80

----
Q1: Is the final velocity when it stopped or during the 'skidding time'? But one thing is for sure...it's the same for both of them and it's slower than the initial...

so... v₁ = v₂
----

Q2: Does the 5m distance from the intersection have any relevance to the calculations? Or is it just there to tell us that they were at the same distance from the intersection at the same time?
----

Find:
u₁ = ?
u₂ = ?

...once found, I can compare it to the speed limit? ... right?

Random Ideas on how to start it...

A. \sump₁ = \sump₂
m₁u₁ + m₂u₂ = (m₁ + m₂) v
??

B. Forces at an angle? so...and somehow incorporate \mu?
Fn_y = Fg

 

[DorianG]

Q1: Is the final velocity when it stopped or during the 'skidding time'? But one thing is for sure...it's the same for both of them and it's slower than the initial...

Think of this problem in two parts, you're already going along the right lines. Take the second part of the problem first. An object (here two cars together) of mass m₁ and m₂ moves with an initial velocity v. This is what you need to find. For this motion, you know the mass of the object and the distance it takes to slow from v to 0 (ie, it stops). What net force is acting on the object? In what direction? How can you use Newton's Laws to find a, the deceleration on the object, and then find v?

A. LaTeX Code: \\sum p1 = LaTeX Code: \\sum p2
m1u1 + m2u2 = (m1 + m2) v
??

Yes, once you have the combined v, you can then return to conservation of momentum immediately before and after the collision.