2-D Linear Momentum, Collision Problem

Click For Summary
SUMMARY

The discussion focuses on solving a 2-D linear momentum problem involving an inelastic collision between two vehicles, specifically a yellow vehicle (mass 2674 kg) and a red vehicle (mass 1110 kg). Key equations used include momentum conservation (m1v1 + m2v2 = (m1 + m2)vf) and energy conservation principles. The user calculated the deceleration due to friction (a = -8.43 m/s²) and derived post-collision velocities (v1c = 14.14 m/s and v2c = 14.54 m/s). The user encountered difficulties in determining the initial velocities (v1i and v2i) and applying momentum conservation in both x and y directions.

PREREQUISITES
  • Understanding of 2-D linear momentum concepts
  • Familiarity with inelastic collisions and energy conservation equations
  • Knowledge of kinematics and friction effects on motion
  • Ability to apply vector components in momentum calculations
NEXT STEPS
  • Review momentum conservation in both x and y directions during collisions
  • Study the application of kinematics equations to determine initial velocities
  • Learn about the effects of friction on vehicle deceleration in collision scenarios
  • Explore detailed examples of inelastic collision problems in physics textbooks
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and collision problems, as well as professionals in accident reconstruction and forensic analysis.

omicronhuh
Messages
1
Reaction score
0
Good evening! I'm having trouble completing an assignment related to inelastic collisions and momentum.

Homework Statement


Here is the statement:

You are in great shape on this hot July morning. You are working as an intern for the Montreal Police (SPVM) Collision Investigation unit. You assist the investigators in order to determine, as accurately as possible, the circumstances surrounding collisions (initial velocity of the vehicles involved, direction vehicles were heading, etc.) in order to take action against drivers at fault under the Criminal Code (dangerous driving causing death or other charges).

You just get to the scene of a collision involving two vehicles. The Chief Investigating Officer addresses you, “Come here quick, we need you. You know the routine, so get to work.” Your job is to take photos of the vehicles as they are positioned, and take a certain number of measurements, including the length of the skid marks and layout of the debris. You immediately draw the following sketch of the accident and include a few common measurements.Your findings are as follows:
• Numerous debris (broken glass, plastic, etc.) were found at a distance of 12.63 m from the red vehicle
• (2) and 11.87 m from the yellow vehicle (1);
• The yellow vehicle (1) seems to have braked in a straight line;
• The red vehicle (2) veered 6.5º away from its initial course;
• The marks of the impact on the red vehicle (2) show that it was heading 98º away from the yellow
• vehicle (1) at the moment of collision;
• There are skid marks over a distance of 30 m before the debris;
• The posted speed limit on this street is 70 km/h;
• Mass of vehicle 1 (yellow) 2674 kg, mass of vehicle 2 (red) 1110 kg.

I also found out that the friction coefficient of the pavement is mk = 0.86

I'll include the sketch as well:
3ba51d285bfa4b809b172f1729ec3020.png

I'm using the following velocity indicators and ignoring the ones in the sketch.

v1i = initial velocity of the yellow car
v2i = initial velocity of the red car
v1c = velocity of the yellow car just after the collision
v2c = velocity of the red car just after the collision

Homework Equations


This are the relevant equations:
p=mv
m1v1 + m2v2 = (m1 + m2)vf
F = ma
And conservation of energy equations: U1 + K1 + Q11 = U2 + K2 + Q12

The Attempt at a Solution


My attempts at solving this problem are as follows:
I'm starting at point c and comparing it to b to then compare b and a using momentum to solve the problem.

In my first attempt I used energy conservation to try to calculate the velocity of the cars just after the collision.
So K = Q, in this case.

1/2(m1)(v1^c)^2 + 1/2(m2)(v2c)^2 = mk(N1)(L) + mk(N2)(L)
1337v1c^2 + 555v2c^2 = 386,056.11

And I hit a roadblock, I couldn't figure out what other equations to use to get the velocity of either car right after the collision.

So, I calculated the deceleration the cars suffered thanks to the friction. I multiplied N for the friction coefficient and then divided by the mass.

a = -8.43m/s^2

Then with kinematics equations I got the velocity of the cars right after the impact.
v1c^2=-2a(x-x0)
v1c^2 = -2(-8.43)(11.87)
v1c = 14.14 m/s

v2c^2 in the x axis=-2a(x-x0)(cos 6.5)
v2c^2 = -2(-8.43)(12.63)(cos 6.5)
v2c in the x-axis = 14.54 m/s

v2c^2 in the x axis=-2a(x-x0)(sin 6.5)
v2c^2 = -2(-8.43)(12.63)(sin 6.5)
v2c in the y-axis = 4.9 m/s

And with this I try to use the momentum conservation equation:
m1v1i + m2v2i (cos 98) = m1(14.14) + m2(14.54)
2674v1i + 1110(-0.139)v2i = 53,949.76

And is at this point where I'm not sure how to proceed. I've tried using the same deceleration to know v1i with kinematics equations and then using that to know v2i, but the speed that results for the red car turns out to be quite absurd, nearing the 400km/h. I am also not sure I'm using the angles correctly.

Do you guys have any advice for me?
Thanks!
 
Physics news on Phys.org
Looks to me that the vehicles separate (just) after the collision. The red car goes further. So you have two separate energy equations for their subsequent skids. But you effectively did that when you switched to using the kinematics equations.
Not sure how you used kinematics to find "v1i". Are you confusing two different velocities for car 1, the velocity before it started braking and tne velocity just before impact?

You have not used momentum conservation in the y direction during impact.
 
Last edited:

Similar threads

Replies
335
Views
17K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K