Car Crash Analysis: Analyzing Friction & Speed

[Billy12Banana]

Homework Statement

Your consulting agency, with its extensive knowledge of physics, has been subpoenaed to provide expert testimony at an automobile accident case in civil court. The case involves a crash between a Ford Escort car and a tractor-trailer truck.

1. The police department determined that the force required to drag a 130 N (29 lb) car tire across the pavement at a constant velocity is 83 N. Specifications from the truck s manufacturer claim that (for technical reasons) the effective coefficient of friction for truck tires is only 70% that of car tires. What was the coefficient of kinetic friction between the tires and the road for both the car and truck? (Include FBD)

2. After collision, the truck and car skidded at the angles shown in the diagram. The car skidded a distance of 49.8 ft before stopping while the truck skidded 9.5 ft before stopping. (1 ft=.305m) The weight of the car is 13, 600 N and the weight of the truck is 69,700 N. What was the speed of each vehicle just after the collision?

3. The pre-crash angle between the velocities of the truck and car was 90 degrees. What was the speed of each vehicle just before the collision?

4. At this intersection, the truck driver had a flashing yellow light while the car driver had a flashing red light.

a.) The car driver claims to have made a full stop at the light before entering the intersection. He also claims that the truck driver did not see him until after the collision. This would make the truck driver partially liable.
b.) The truck driver claims that the car ran the flashing red light. He also claims to have begun braking in anticipation of a collision; traveling at only 6.7 m/s (15 mph) at the moment of impact.
c.) Police measurements show that the distance for the car from the traffic light to the collision point was 13.0 m(42.5 ft) and the distance for the truck from the traffic light to the collision point was 32.8 ft (1ft=.305 m).
d.) Ford Motor Corporation specifications indicate that the maximum acceleration of a comparably loaded Ford Escort is about 3.0 m/s^2, and the maximum acceleration given for the truck is 1.2 m/s^2.
Who is at fault and why?
​

Homework Equations

f=mu(N)
w=fd
??

The Attempt at a Solution

1. I got .638 and .447 for the coefficients of f for the car and truck respectively.
2. I got 13.88 m/s and 5.09 m/s for v of car and truck respectively.
Don't know how to do 3 and 4.
For 3 I thought it was a conservation of momentum question and tried mv+mv=mv+mv but I really don't know. Please help, thanks! Visual of crash is attached.

 

[jbriggs444]

The Attempt at a Solution

1. I got .638 and .447 for the coefficients of f for the car and truck respectively.

Good.

2. I got 13.88 m/s and 5.09 m/s for v of car and truck respectively.

And I get the same.

Don't know how to do 3 and 4.
For 3 I thought it was a conservation of momentum question and tried mv+mv=mv+mv but I really don't know.

Let's chase down the conservation of momentum argument. You have the velocities and masses of the car and truck after the collision. You also know the angles. Momentum is a vector quantity. You can write down equations for the x and y components separately. Can you do that and show us what you get?

 

[Billy12Banana]

So for #3:
1360v=1360(13.88sin40)+6970(5.09sin10)
v=93.63 m/s for the car
6970v=1360(13.88cos40)+6970(5.09cos10)
v=7.08 m/s for the truck

Is this correct?

 

[haruspex]

1360v=1360(13.88sin40)+6970(5.09sin10)

Yes.

v=93.63 m/s for the car

Far too much. Try that again.
It's always a good idea to do a sanity check, just a mental approximation of the answer: v ~ 14 sin(40)+(697/136) 5 sin(10) ~ 14/√2 + 5 (10/57) [because 57 degrees to a radian, and this a small angle] ~ 10 + 1 = 11.
That might look difficult, but with practice it gets easier.

v=7.08 m/s for the truck

Yes, but I think 7.09 is closer.

 

[Billy12Banana]

13.45 for car.
I'm confused as to how to figure out who was at fault using this information and the information in #4 of the problem.

 

[haruspex]

13.45 for car.
I'm confused as to how to figure out who was at fault using this information and the information in #4 of the problem.

You have found e.g. the speed of the car at impact. The driver claims to have moved off from rest at the light. What does that tell you about the car's acceleration?
Apply the same to the truck.

 

[Billy12Banana]

So since the driver made a full stop, he had to accelerate for 0m/s to 13.45 m.
The truck driver claimed to be braking in anticipation of the collision meaning his acceleration has to be negative.
According to my calculations, the truck's velocity before the collision was 7.09 and 5.09 after. And the truck driver claims to have been traveling at 6.7 m/s at moment of impact. Since 6.7 is between 7.09 and 5.09, it seems like what the truck driver claimed is true.
Is this correct?
In order to utilize parts c and d of #4, do I need to do kinematics?

 

[jbriggs444]

So since the driver made a full stop, he had to accelerate for 0m/s to 13.45 m.

If the car driver had stopped as claimed, he
would have had to accelerate from 0 m/s to 13.45 m/sec over a span of 13.0 meters in order to collide at 13.45 m/sec as observed.

Can you show that this is implausible?

Since 6.7 is between 7.09 and 5.09, it seems like what the truck driver claimed is true.

The truck driver is talking about deceleration prior to impact. You are talking about a reduction in velocity during the impact. Two different things.

Also, there is a difference between a statement being plausible and a statement being true.

 

[Billy12Banana]

Can you also explain why, for these two equations from #3, there is only one "mv" on the left side? Like just 1360v or 6970v on the left? I know I ended up with the right answers but I was confused while doing it. I used mv+mv=mv+mv. I think I just assumed the initial velocity for the other vehicle was zero because I didn't know how else to do it, but why is it zero?
1360v=1360(13.88sin40)+6970(5.09sin10)

6970v=1360(13.88cos40)+6970(5.09cos10)

 

[haruspex]

Can you also explain why, for these two equations from #3, there is only one "mv" on the left side? Like just 1360v or 6970v on the left? I know I ended up with the right answers but I was confused while doing it. I used mv+mv=mv+mv. I think I just assumed the initial velocity for the other vehicle was zero because I didn't know how else to do it, but why is it zero?
1360v=1360(13.88sin40)+6970(5.09sin10)

6970v=1360(13.88cos40)+6970(5.09cos10)

There are two different directions. Each equation relates to motion in one of the two, and in each, one vehicle has no prior velocity.