What Are the Initial Speeds of Two Cars After a Rear-End Collision?

  • Thread starter Thread starter Nikolan
  • Start date Start date
  • Tags Tags
    Center Momentum
AI Thread Summary
The discussion focuses on calculating the initial speeds of two cars involved in a rear-end collision, where car A is stationary and car B is moving before the impact. The problem involves using the principles of conservation of momentum and the work-energy theorem, particularly the work done by friction as the cars slide to a stop. Participants suggest analyzing each car separately to determine their speeds immediately after the collision. The final calculations involve determining the frictional force and applying it to find the initial speeds based on the distances each car slid. The conversation emphasizes the importance of understanding the relationship between kinetic energy, work done by friction, and momentum conservation in solving the problem.
Nikolan
Messages
5
Reaction score
0

Homework Statement


In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1500 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low μk of 0.21) stops them, at distances dA = 5.3 m and dB = 3.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

(see attachment for image)


Homework Equations



(A = CarA, B = CarB)

PAi + PBi = PAf + PBf
(mA)vAi + (mB)vBi = (mA)vAf + (mB)vBf


The Attempt at a Solution



I've been sitting here looking blankly at this problem for about an hour now. I really have no idea where to start with this, I've tried a few things, but it always ends up giving me two unknown variables. Looking at another thread that was posted on this question, I realize that all the kinetic energy is transferred to thermal in the end, how does that help?

(https://www.physicsforums.com/showthread.php?t=82429)
 

Attachments

  • fig09_60new.gif
    fig09_60new.gif
    7.4 KB · Views: 930
Physics news on Phys.org
Hi Nikolan, welcome to PF

For a) & b) think about one car at a time. After the collision, the work done by friction slowing the car will equal the decrease in kinetic energy of the car.
 
Thank you very much!

I ended up taking what you said and applying it like this:

(Using this solution for both cars A and B)

Fk = (uk)FN to find force of friction.

W of Friction = (Fk)(D) = Kf - Ki = 1/2mvf^2 - 1/2mvi^2

to get:

W = - 1/2mvi^2 => sqrt(((-2)(W))/M) for initial speeds of both cars directly after impact.
I then used Momentum of Cars before = Momentum afterwards
or
Pai + Pbi = Paf + Pbf
mavai + mbmbi = mavaf + mbvbf

and plugging in relevant values to end up with initial speed of car B before collision.Again, thanks for your help! :approve:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top