Colision Investigation Problem

  • #1

Homework Statement



You are starting a job for Montreal Police Colision Investigators. Your boss tells you to file a report on an accident on Notre-Dame Street. There has been an accident involving 2 cars.

Your findings are as follows:

  • Numerous debris were found at a distance of 12.25 m from the car
  • The two cars stuck together and there are skid marks over 12.25 m
  • There are also skid marks over a distance of 30 m before your debries
  • The posted speed limit is 70 km/h
  • Other data: m1=2,674 kg and m2=1,100kg
  • From a friction block: Fv= 88.2N and Fh = 75.8N
  • The second car was at rest

Write a report for your superior on wether the car at fault should be charged with speeding (over 70 km/h)

Homework Equations



μk = Fk/Fn

The Attempt at a Solution



The coeficent of kinetic friction is 0.86 (μk = Fv/Fh)
We have:
Mass of both cars
initial velocity of second car (0)
final velocity of first car (0)
distance of skid marks

We need 2 know:

Initial velocity of 1st car.

How can we do this with only mass, distance of skid marks, and coeffiecent of friction?
 

Answers and Replies

  • #2
CWatters
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Hint: Work backwards. Energy = force * distance. You know the friction force and you know the distance so you can calculate the energy the cars had just after impact. You know their mass so you can calculate their velocity just after the impact.
 
  • #3
Hmmm I haven't thought about finding energy.
Since Energy = net force * distance I will need to find net force and I already have distance.
And Fnet = Ff + Fa

Ff = (uk)(Fn)
Ff = 75.8 N

But how do I find Fa?
 
  • #4
CWatters
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I think you can assume that the only force decelerating the cars after the impact is friction Ff. What else is there?
 
  • #5
You're right the only force would be friction force.
So,
Ff = uk * Fn
Ff = .86 * 88.2 N
Ff = 75.9 N
With friction force i can find energy.
E = F * d
E = 75.9 N * 12.25 m
E = 929.8 J
So now that i know the energy my next step is to find velocity.
KE = (1/2)mv^2
KE = E
929.8 J = (.5)(2674+1100)(v^2)
.49 = (v^2)
.7 meters per second = v
But that is the velocity after the colision. What step do i need to take next to get the velocity prior to the first car breaking?
 
  • #6
CWatters
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Continue to work backwards step by step.

I haven't checked your working but once you have the correct velocity of both cars after the impact your next step is to work out the velocity of the moving car just before the impact. Hint: Conservation of Momentum.
 
  • #7
Solved

We finished the problem. Thank you for your help! Working backwards did the trick.

https://www.dropbox.com/s/3a3ivx9mscn3qxj/2014-03-10%2010.57.32.jpg
https://www.dropbox.com/s/905gzbunimv5fdt/2014-03-10%2010.58.33.jpg
 
  • #8
Solved

We finished the problem. Thank you for your help! Working backwards did the trick.

https://www.dropbox.com/s/3a3ivx9mscn3qxj/2014-03-10%2010.57.32.jpg
https://www.dropbox.com/s/905gzbunimv5fdt/2014-03-10%2010.58.33.jpg
Do you have any working links for this solution? Thank you
 
  • #9
berkeman
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Do you have any working links for this solution? Thank you
Welcome to PhysicsForums.

This thread is from 2014, so it's not likely the poster is still active on the forums (their last activity was from near the end of 2014).

If you have a similar schoolwork problem, start a new thread and fill out the Template with as much information as you can. Show as much work as you can, and you will get good help here. :smile:
 

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