Collision with two cars (momentum and impulse)

[RioAlvarado]

Homework Statement

[/B]
Two cars are approaching a perpendicular intersection without a stop sign. Car 1 has a mass ## m_1 = 900 kg ## and is heading north at ##v_1 = 25 m/s ##. Car 2 has mass ##m_2 = 700 kg ## and is heading west at an unknown speed ##v_2##. The two cars collide at the intersection, and stick together as a result of the collision. The police report stated that after the collision, the two cars moved together in a direction 40 degrees west of north and stopped after sliding ## d = 35 m ## from the collision point.a. Find the speed of car 2 before the collision. ##v_2 = ##b. What is the speed of the cars immediately after the collision? ## v = ##c. What is the coefficient of kinetic friction between the cars and the pavement? ##μ_k = ##d. What is the magnitude of the impulse exerted on car 1 during the collision? ##J_1 = ##e. If the collision lasted 0.3 seconds, then what is the magnitude of the average force on car 1 during the collision? ##F_{avg} = ##


f. What is the magnitude of the impulse exerted on car 2 during the collision? ##J_2 = ##

Homework Equations

I assume:
##m_1*v_{i1}+m_2*v_{i2} = v_{f1}*m_1 + v_{f2}*m_2) ## eq. 1
##P = mv## eq. 2
##J = \int_{t_i}^{t_f} F dt## eq. 3
##P_f = P_i + J## eq. 4
There may be others, but those are the ones I could think of.

The Attempt at a Solution

I thought to try and use systems of equations to find my speeds, my attempt was:
##20*900 + 800*v_2 =v_{1f}*900 + v_{2f}*800 , tan(40)=\frac{v_{2f}}{v_{1f}}##. However as can be seen I ran into the issue of having two equations and three unknowns. I obtained the ##tan(40)## piece from (what I think to be a correct) extrapolation from the information about both cars moving together in a direction of 40 degrees. Beyond that I am stuck, and obviously unable to continue on to the next parts.My thanks for any and all assistance.

 

[Simon Bridge]

Are you assuming that momentum is conserved in the collision? If so then say so.
Can you justify the assumption: why would momentum be conserved?

You say there are two equations and 3 unknowns - but I only see one equation written down above, and the problem statement only has two unknowns.
I'm guessing your unknowns include v_f1 and v_f2? Then your third equation is the relationship between these two final speeds - spelled out in the problem statement. Read it again more carefully - what do the cars do when they collide? What does that mean about the speed they travel in relation to each other?

Note: Momentum is a vector.
How do you normally account for the different initial directions the cars travel in your equations?

 

[RioAlvarado]

I thank you for your reply, but as it turns out, while I was awaiting responses I continued working on the problem and managed to figure it out. Nevertheless I thank you very much for taking the time to reply and offer assistance.

 

[Simon Bridge]

No worries - for the benifit of people googling to this problem, but who have yet to figure it out, what did you come up with?

 

[RioAlvarado]

Alright, for a. I realized that I needed to consider the alternate angle to the 40, that is 50 degrees. I then said ##tan(50)= \frac{900*25}{700*v_2}## and solved. For speed of the car afterwards I used ##(700+900)*v = \frac{700*v_2}{cos(50)}## and solved that, for the impulse I used the model with says impulse is equal to the change in momentum. So I found the components of the momentum for north and east, used that to find both impulses, which are equal.

 

[Simon Bridge]

I'm afraid your reasoning is not clear - that will start to cost you marks soon if not already.

You didn't really need the alternate (complimentary) angle - you just needed to be careful about how you set up your trig.
If you wanted to treat north as +y and east as +x so you can apply memorized equations, then you get a headache like making sure the correct angle is inside the tangent.
However, you can just add the vectors head-to-tail (sketch it out) to get a right-angle triangle where you know all the angles and one of the sides. The rest is basic geometry.

Alternately, using unit-vector notation, using ##\hat n## for north and ##\hat w## for west:

Conservation of momentum: $$m_1v_1\hat n + m_2v_2\hat w = (m_1+m_2)(\hat w \sin\theta + \hat n\cos\theta)v$$ ... because ##\theta## is taken anticlockwise from the ##\hat n## direction.
A sketch will make it clear.

Group the terms by unit vector and you end up with two equations in two unknowns.
The unknowns are ##v_2## and ##v##.

If you wanted to define east as the +x axis etc, then ##\hat j = \hat n## and ##\hat i = -\hat w## ... will put the equation into a more familiar form.
Whatever you do - get the trig right by using a diagram.

 

[RioAlvarado]

Sorry I couldn't make my solution more clear. I'm not surprised it isn't the best solution, however it did give me precise, and correct answers, and I'm happy with that. If you really think I should elaborate more on my answers for the benefit of those who might be working this problem I could do so tomorrow.