AP Physics Help? (581 views)

I was in AP Physics B in school. I sadly don't have my notes from school, but I can certainly be of help for assignments!

What are you working on right now?? We just finished kinematics and I got a 65 on the test :') 

Alright, I actually remember how to do this. I'll try to preserve sig figs as much as I remember to.

Basically, since the two objects are attached and move together, you can assume that this is an inelastic collision—kinetic energy has been lost as some other type of energy. First you want to keep track of your variables:

BEFORE COLLISION
m₁ = mass of the sports car = 920 kg
v₁ = velocity of the sports car --> to be determined
m₂ = mass of the SUV = 2,300 kg
v₂ = velocity of the SUV = 0

Pre-collision momentum of the system = p₁
= m₁v₁ + m₂v₂
= 920v₁ + 0 = 920v₁

AFTER COLLISION
m = mass of the two-car system
v = velocity of the two-car system
s = displacement of the system = 2.8m
μ = coefficient of friction = 0.80

Post-collision momentum = p₂
= mv = 3,220v

Let's find out the relationship between v₁ and v—it'll come handy later. We do this by applying conservation of momentum:

p₁ = p₂
=> 920v₁ = 3,220v
=> v₁ = (3,220/920) = 3.5v — (1) (gee, isn't that 3.5 conveniently simple)

w = weight of the two-car system (needed to find out the work done by friction)
= mg
= (m₁ + m₂)g = 3,220 * 9.8 = 31,556 N

So, friction force in the horizontal direction = F (this is called the braking force, because it also applies to a vehicle for which you've hit the brakes)
Work done by the braking force = W

W = F * s = μ * w * s = 0.80 * 31,556 * 2.8 = 70,685.44 N

You'll need to understand here that the kinetic energy of the two car system (that is after the collision) is used up in braking (because the system comes to rest KE of a resting system is 0 by definition). So you can equate the KE of the system immediately after the collision (when it'll have the most energy) and the W that we calculated (which is the amount of work friction has to do to use up all that KE).

So:

Kinetic Energy = K = ½mv² = 1,610v²

K = W

½mv² = 70,685.44
v = √(70,685.44 / 1,610) = 6.626 m s^-1 — (2)

Remember equation (1), where we found out the relationship between v₁ and v?
I'll reiterate it for you...

v₁ = 3.5v — (1)

Huh. I wonder what would happen if I put (2) and (1) together...

v₁
= 3.5(6.625)
= 23.191 m s^-1

That was more fun than a barrel of physicists. Feel free to ask for clarifications if there's something you need explained or don't understand.
Hopefully you're not doing this homework at the last minute, as I have so often been guilty of doing. >_>