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

In summary: The collision distorts the cars so the kinetic energy of the collision is not the same as the kinetic energy of the cars before.)The collision is also not head-on because the yellow car stopped and the red car kept going. So the momentum of the red car was not only reduced, it was also turned.So you can not use conservation of momentum for the collision either.So what can you use?You can use the momentum of the red car before the collision and after the collision. You can use the velocity of the yellow car after the collision to help find the velocity of the
  • #1
Gin-Su
4
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
  • #2
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.
 
  • #3
Thank you for steering me in the right direction.
 
  • #4
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:
  • #5
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
  • #6
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.
 
  • #7
Gin-Su said:
I took your advice and worked on it backwards and I think I have the solution.
Good show.
 
  • Like
Likes ffff
  • #8
How would you find the velocities right after the collision?
 
  • #9
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.
 

1. What is a collision investigation problem?

A collision investigation problem is a scientific investigation that involves analyzing the motion and forces involved in a collision between two or more objects. This can include calculating the linear momentum, energy, and other physical quantities to determine the outcome of the collision.

2. How is linear momentum involved in a collision investigation problem?

Linear momentum is a fundamental concept in collision investigation problems. It refers to the product of an object's mass and velocity and is conserved in a closed system. This means that the total linear momentum before a collision is equal to the total linear momentum after the collision.

3. What is the role of energy in a collision investigation problem?

Energy is another important factor in a collision investigation problem. It is also conserved in a closed system, meaning the total energy before a collision is equal to the total energy after the collision. Different types of energy, such as kinetic and potential energy, can be involved in a collision and can be used to calculate the outcome of the collision.

4. How do you approach solving a collision investigation problem?

To solve a collision investigation problem, you first need to identify the objects involved and their initial conditions, such as mass, velocity, and direction of motion. Then, you can use the principles of conservation of linear momentum and energy to analyze the collision and determine the final conditions of the objects.

5. Are there any real-world applications of collision investigation problems?

Collision investigation problems have many real-world applications, such as in car accidents, sports, and industrial accidents. By understanding the physics behind collisions, scientists and engineers can design safer structures and devices and improve safety protocols in various industries.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Advanced Physics Homework Help
Replies
17
Views
4K
  • Introductory Physics Homework Help
2
Replies
62
Views
10K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top