Collision Investigation Problem (2-D Linear Momentum, etc.)

Click For Summary
SUMMARY

The discussion focuses on solving a collision investigation problem involving two vehicles, where momentum conservation and energy principles are applied. The key equations include momentum conservation in two dimensions and the work-energy principle, particularly inelastic collisions. The participants emphasize the importance of correctly identifying initial and final velocities, as well as the impact angles, to determine liability under Section 249 (3) of the Criminal Code. The analysis reveals that the yellow vehicle's speed must be calculated to ascertain if it exceeded the speed limit of 70 km/hr.

PREREQUISITES
  • Momentum Conservation in 2-Dimensional Collisions
  • Work-Energy Principle in Inelastic Collisions
  • Newton's Second Law and Frictional Forces
  • Kinematics in One and Two Dimensions
NEXT STEPS
  • Learn to apply the momentum conservation equations in two dimensions for collision analysis.
  • Study the work-energy principle and its application in scenarios involving friction.
  • Explore the calculation of initial velocities using friction data and stopping distances.
  • Review the implications of vehicle dynamics in collision investigations, particularly under legal frameworks.
USEFUL FOR

Students in physics or engineering, collision reconstruction specialists, and law enforcement personnel involved in traffic accident investigations will benefit from this discussion.

Gin-Su
Messages
4
Reaction score
0
My class was given a packet with several scenario questions. I've got the rest of them completed and this is the only one remaining.

So I know that the question involves: Momentum Conservation (2-D): Inelastic collisions, Work-energy: Conservation of energy with friction, Dynamics: Newton's Second Law, frictional force, Kinematics: 1-D and 2-D

Homework Statement


(This is a really long question, I apologize in advance.)

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 into 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
Friction Data:
Mass of friction block: 9 kg
Ftension = 75.8 N
(I apologize in advance for including a picture; however I felt that it would aid in visualizing the scene.)
oCoYtPF.png

Homework Equations



Fkk(Fn)
Fnet=ma
(1/2)v1o2+(1/2)v2o2=(1/2)v1f2+(1/2)v2f2
Pox=Pfx
m1v1o-m2v2o=m1v1fcosθ1+m2v2fcosθ2
Poy=Pfy
m1v1o-m2v2o=m1v1fsinθ1+m2v2fsinθ2]
(Others maybe?)

The Attempt at a Solution



I actually did this initially in a completely different way using Vf2=Vo2+2(a)(Δx), which I now realize is completely wrong. (I ended up getting 95.65 hm/hr for the driver in V1, which doesn't mean anything since I am wrong.)

This was my second attempt:

75.8N=μk(88.2N)
μk=.86

Fnet=ma
kmg=ma
-.86(9.8)=a
-8.428 m/s2=a

This is where I have some trouble, as I am not sure where I would proceed. The angles are throwing me into a bout of confusion as this is the first time I am facing a problem like this.

(1/2)v1o2+(1/2)v2o2=(1/2)v1f2+(1/2)v2f2

So I think I should then use the x and y components.

Pox=Pfx
m1v1o-m2v2o=m1v1fcosθ1+m2v2fcosθ2

2674(v1o)-1110(v2o)=2674(v1f)cos(0)+1110(v2f)cos(6.5)
(I am not 100% sure I should be using 6.5°.)

2674(v1o)-1110(v2o)=2674(v1f)+1110(v2f)cos(6.5)

This is the point at which I am stuck. I am not sure if it's because I am using the wrong equations or if it's because I am overlooking something.

Thank you for taking the time to read this. Any help would be greatly appreciated.
 
Physics news on Phys.org
debris is found all over the place implies that KE is not conserved during the collision
(it is almost perfectly inelastic since the skid marks after collision are almost same length)
but momentum is conserved.
I would run the calculation forward, using the limiting initial velocity for big yellow (100km/hr)
part A: v² formula with skid
part B: inelastic collision; using the 8° impact angle means the red velocity x-component is very small
part C: v² formula with skid ... can the yellow stop in that skid distance? if no, they were over the limit.
 
Thank you for steering me in the right direction.
 
I apologize for the double-posting, but I am currently stuck on part b of what you recommended.

2674(16.307)-1110(vv2i) = (2674)(vf1)cos(0°)+(1110)(vf2)cos(98°)
2674(16.307)-1110(vv2i) = (1110)(vf2)sin(98°)

How could I solve for the multiple variables. I tried solving for one and inputting it into the other equation; however that's not getting me anywhere.
 
Last edited:
Several inconsistencies between diagram and text:
The velocity labels are different (v1A, v1B etc.). The confusion is compounded by then inventing different labels in the working: v1f, v2f. Please define one set of labels, each unique, and stick to them.
the red vehicle ... was heading 98º away from the yellow vehicle
Not according to the picture.
The red vehicle veered 6.5º from its initial course.
Not according to the picture.

Work backwards from the end. Find the two velocities just after collision.
It looks like the yellow car maintained its direction. What does that tell you about the change in the red vehicle's momentum caused by the collision?
 
  • Like
Likes 1 person
haruspex said:
Several inconsistencies between diagram and text:
The velocity labels are different (v1A, v1B etc.). The confusion is compounded by then inventing different labels in the working: v1f, v2f. Please define one set of labels, each unique, and stick to them.

Not according to the picture.

Not according to the picture.

Work backwards from the end. Find the two velocities just after collision.
It looks like the yellow car maintained its direction. What does that tell you about the change in the red vehicle's momentum caused by the collision?

Yeah I apologize for that. I noticed the inconsistency when I was typing it up to. The labels were given to me, so I automatically included it. I apologize for that oversight. The photo was also included by default in the assignment.

I took your advice and worked on it backwards and I think I have the solution. Thank you for your help.
 
Gin-Su said:
I took your advice and worked on it backwards and I think I have the solution.
Good show.
 
  • Like
Likes ffff
How would you find the velocities right after the collision?
 
haruspex said:
Several inconsistencies between diagram and text:
The velocity labels are different (v1A, v1B etc.). The confusion is compounded by then inventing different labels in the working: v1f, v2f. Please define one set of labels, each unique, and stick to them.

Not according to the picture.

Not according to the picture.

Work backwards from the end. Find the two velocities just after collision.
It looks like the yellow car maintained its direction. What does that tell you about the change in the red vehicle's momentum caused by the collision?
How would you find the velocities right after the collision?
 
  • #10
ffff said:
How would you find the velocities right after the collision?
From the friction data, you know the deceleration. From the debris data, you know the distance to come to a stop. From those you can determine the velocity just after collision.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 62 ·
3
Replies
62
Views
13K
Replies
2
Views
3K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K