Collision Investigation (2-D)

  • #1

Homework Statement


I apologize in advance because this is a long question.
This is a scenario based question where I role play as an intern for the Montreal Police Collision Investigation Unit. Basically I have to determine the initial velocity of vehicles to determine whether or not they are liable under Section 249 (3) of the Criminal Code, meaning if they are going over 30 km/hr over the limit, they face charges.

Now for the data:
  • Debris was found at a distance of 12.63 m from the red vehicle. (I am sure that this is data thrown in to distract)
  • Debris was also found 11.87 m from the yellow vehicle. (I am also sure that this is unneeded.)
  • The yellow vehicle braked in a straight line.
  • The red vehicle veered 6.5º from its initial course.
  • The marks of the impact on the red vehicle show that it was heading 98º away from the yellow vehicle at the moment of collision.
  • Prior the point of collision, there are skid marks over a distance of 30m.
  • The speed limit is 70 km/hr.
  • Mass of vehicle 1 (yellow) 2674 kg
  • Mass of vehicle 2 (red) 1110 kg
The velocity indicators provided were:
Velocity indicators are defined as follows:
  • V1IB: velocity of vehicle 1 before initial braking
  • V1B : velocity of vehicle 1 just before impact
  • V2B : velocity of vehicle 2 just before impact
  • V1A: velocity of vehicle 1 just after impact
  • V2A: velocity of vehicle 2 just after impact
  • (this is for the attached visual of crash)
Friction Data:
Mass of friction block: 9 kg
Ftension = 75.8 N
(I am suppose to be prosecuting the driver of this car and will be presenting my case to my teacher)
(I attached the visual that was given with the worksheet)

Homework Equations


Fk=μk(Fn)
Fnet=ma
Vf^2 = V0^2 + 2ax

The Attempt at a Solution


I have already figured out the coefficient of friction and acceleration using the above equations.
Coefficient of friction: .86
Acceleration: 8.428 m/s^2
I also found the velocity using 70 km/h as my initial and 30 m as x. I ended up getting 70.0036 km/h, which would mean the driver was not speeding but want to make sure that I have the correct math. My teacher also said that I must find the velocity of the car before and after the collision and I am not sure how to approach the problem further.
oCoYtPF.png
 

Answers and Replies

  • #2
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I think you'll have to use vectors to find the east/west and north/south components of the velocity of each car since they are moving nearly perpendicular to each other.

You know the momentum and directions of both cars after the impact, just work your way backward step by step, for both the x and y axes.. you know how far they moved after impact

You don't mention if the 70 kph was for the red or yellow vehicle...

considering the lighter red vehicle hardly veered off course and basically took the yellow vehicle along with it, the red vehicle was going far faster than the yellow one.
 
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  • #3
I am sorry if this sounds like a stupid question I am not great at physics and need reassurance
Since p=mv would I use the equations in the attached photos after determining the angles via force diagrams:
I think you'll have to use vectors to find the east/west and north/south components of the velocity of each car since they are moving nearly perpendicular to each other.

You know the momentum and directions of both cars after the impact, just work your way backward step by step, for both the x and y axes.. you know how far they moved after impact

You don't mention if the 70 kph was for the red or yellow vehicle...

considering the lighter red vehicle hardly veered off course and basically took the yellow vehicle along with it, the red vehicle was going far faster than the yellow one.
img966.png
img965.png
 
  • #4
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From the diagram/description, it seems like only yellow vehicle braked.

You do need the first two statements, it tells you how far away from the initial impact the cars came to a stop, since you have the coefficient of friction, you can find their velocities

I was mistaken about the red vehicle going faster, the written form of the question doesn't match what the diagram (unclearly) shows... From what they say that it veered 6.5 degrees from it's course is NOT what they show.. They show it veered 6.5 degrees from the YELLOW vehicle's initial path.. that changes everything
 
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  • #5
Sorry for double posting the photos don't seem to be coming up clear in the reply so here they are..
m1v1o-m2v2o=m1v1fcosθ1+m2v2fcosθ2
m1v1o-m2v2o=m1v1fsinθ1+m2v2fsinθ2]
 
  • #6
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That looks about right... it's been a long time since I've really had to do this and I'm a little rusty.. I'm still trying to figure out the elastic and inelastic parts of this collision, since the two vehicles moved different distances.... I'll have to get a piece of paper and jot things down to get the brain working
 
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  • #7
haruspex
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I also found the velocity using 70 km/h as my initial and 30 m as x. I ended up getting 70.0036 km/h,
I do not understand that calculation. You are trying to find the initial velocity. There is little point in plugging it in as 70km/h. Besides, your answer implies very nearly constant speed.
Debris was found at a distance of 12.63 m from the red vehicle. (I am sure that this is data thrown in to distract)
Not at all. Take the location of the debris as being where the impact occurred. This tells you how far the cars skidded (we assume) afterwards. You found the acceleration while skidding to a stop, so what were the speeds of the vehicles just after impact?
(I note that different distances are given for the red and yellow vehicles, but the diagram implies that difference merely relates to the length of the yellow vehicle, i.e. they pretty much moved together after impact. The debris will indicate where the front of the yellow vehicle was at impact, whereas the distance from there appears to be measured to the rear of the yellow vehicle, so underestimates the skidding distance. I would take the 12.63m as applying to both.)
 
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  • #8
Can I go ahead and assume the final velocity is 0 to find the initial velocity after the collision? If so I believe I would plug it into Vf^2 = V0^2 + 2ax using the distance 12.63 m.
 
  • #9
haruspex
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Can I go ahead and assume the final velocity is 0 to find the initial velocity after the collision? If so I believe I would plug it into Vf^2 = V0^2 + 2ax using the distance 12.63 m.
Yes.
 
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  • #10
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I'm going to go with @haruspex's suggestion and use 12.63m as the distance for both cars, it sounds reasonable.
Due to the nature of the collision, it looks like no forward energy of the red car was transferred to the yellow car to move it off course.. they slid along each other

OK, so first off, find the energy in each axis to move the vehicles from impact to their final locations

So for the yellow car, which didn't seem to deviate from it's path (you already got this)
frictional force = 2674kg *9.8 m/s2 * .86 = 22536N
deceleration from skidding = (F = MA) A = F/M or 22536N/2674kg = 8.428 m/s2
initial velocity = Vf2 = Vi 2 + 2aΔx
0 = Vi2 + 2*-8.428 * 12.63
Vi = 14.59 m/s...
KE = 1/2 * M*V2
Kinetic energy is .5 * 14.592 * 2674 = 282657 J

This isn't really the crash velocity of the yellow car, the red car had reduced it's speed because it wasn't traveling perpendicular to it.. more later

Now set up a triangle with the hypotenuse = 12.63m and the angle of 6.5*, solve it. I get 1.43m for the short side and 12.55m for the long side
Once again repeat the above , but this time for both X and Y components on the red car
Additional or reduced mass of the car won't change the acceleration due to friction, so it'll also be 8.428m/s2

So in the Y (nearly purely forward direction for the red car, "up" in the diagram) it had
0 = Vi2 + 2*-8.428 * 1.43
Vi(y) = 9.82m/s
KE = 1/2 * M*V2
Kinetic energy in the Y axis is .5* 9.82m/s2*1110kg = 53520 J
Since it wasn't traveling at 90*, it must have been traveling faster than that to get that much KE in purely the Y direction...
Someone check me on this, but here's what I come up with
The red car's SPEED (not velocity) would have been 9.82m/s / sin(98*) = 9.916m/s (really not much different, which makes sense)
Now for the red car's X velocity (will be negative in our coordinates)
V(y) = 9.916 * cos(98) = -1.38 m/s
KE = 1/2 * M*V2
The KE associated with that is .5 * -1.382 * 1110 = -1056 J immediately before the collision.

Now for the KE of the red car immediately after the collision (in the X axis).. Taking 12.55m of X axis skid from our right triangle
0 = Vi^2 * 2ax
Vi = √2*8.426*12.55 = 14.54m/s
And the KE with that is .5* 14.542 * 1110 = 117333 J
So the CHANGE in KE in the X axis is 1056 + 117333= 118389 J of energy imparted to the red car by the yellow car


That 118389 J of energy has to be added back to the yellow car's energy of 282657J
So the yellow car had 401046 J of energy at the point of the collision...
Solving for velocity from KE
KE = .5 V^2 M
401046 = .5 V^2 2674
V = 17.3 m/s

Now we go and figure out the pre-impact skid mark..
Vf2 = Vi 2 + 2aΔx
Vi = 17.32 + 2 * 8.426 * 30
Vi = = 28.36 m/s
And finally convert that to KPH
28.36 * (3600/1000) = 102 kph

Did I mess up somewhere AGAIN? I think that looks a lot more reasonable

Hopefully I caught everything in the edits
 
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  • #11
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Going to edit.. I think I found some problems... ...

Fumbled the KE in a bad way.. fixing now
 
  • #12
haruspex
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Right up to here, except for the wrong units:
Vi = 14.59 m/s2...
Now, remember that this is the speed of both cars in the yellow car's direction just after collision.
The red car continued somewhat in a perpendicular direction. Can you determine that component of the red car's speed after collision?
How does that component of the red car's speed after collision relate to its component in that direction before collision?
 
  • #13
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Currently fixing.. check back in 5 minutes, and I did catch those wrong units too.. hard to see the error in the editor..

My big mistake was the formula for KE was totally goofed..
 
  • #14
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Fixed.. I think... 102 kph for the yellow car before it started to brake.
 
  • #15
haruspex
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0 = Vi2 + 2*-8.428 * 1.43
This doesn't work. The frictional force acts oppositely to the direction of relative motion. It does not operate independently along different co-ordinates.
Because it travelled almost parallel to the yellow car after collision, you can just take its left to right velocity as calculated for the yellow car and see what perpendicular velocity should result in its angle of travel. No need to consider acceleration over again.

In response to a request for elaboration:
If an object is moving 30 degrees E of North, at constant acceleration, and its acceleration in the N direction is -3m/s2, what is its overall acceleration, and what is its acceleration in the E direction?
Let its overall acceleration be a. The N component of that is a cos(30) and the E component is a sin(30), so we have -3 m/s2 = a cos(30), and the easterly acceleration is a sin(30)=-3 m/s2 tan(30).
 
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  • #16
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You can do it two ways I think... The car had an X and Y component of travel from where it started to where it ended up
I could have calculated the total energy over the 12.63m and using the triangle/ratio determined how much momentum it had... I think your method perhaps is simpler (but I didn't think of it) and would come out to the same answer



Here's a diagram of solving the triangle.. I cheated and used autocad but the pythagorean theorem and sin/cosine will do it for you too
PF car crash.jpg
 
  • #17
haruspex
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You can do it two ways I think... The car had an X and Y component of travel from where it started to where it ended up
I could have calculated the total energy over the 12.63m and using the triangle/ratio determined how much momentum it had... I think your method perhaps is simpler (but I didn't think of it) and would come out to the same answer



Here's a diagram of solving the triangle.. I cheated and used autocad but the pythagorean theorem and sin/cosine will do it for you too
View attachment 95652
Sure, but what do you get now for the red car's velocity just after collision in the direction perpendicular to the yellow car's?
 
  • #18
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You're right, and I'm not seeing how the two methods differ, I'd completely accept that the error is in my method... I'd just like to know where
 

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