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M Malone
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Homework Statement
"Before the collision, vehicle #1 was traveling east and vehicle #2 was traveling north.
- The driver of vehicle #1 states he was traveling 35 mph as he approached the intersection. He continues to state that vehicle #2 ran the stop sign, pulling out in front of him and causing him to crash.
- The driver of vehicle #2 states that he stopped at the stop sign before pulling out, and did not see vehicle #1 until the moment of impact. (From the stop sign, vehicle #2 would have traveled 30 feet to the point of impact.
- Vehicle #1 skidded on 20 feet of asphalt and 30 feet of grass before coming to rest. The angle of departure for vehicle #1 was 45 degrees north of east.
The weight of vehicle #1 including occupants was 4300 lbs. - Vehicle #2 skidded on 25 feet of asphalt and 35 feet of grass before coming to rest. The angle of departure for vehicle #2 was 35 degrees north of east.
The weight of vehicle #2 including occupants was 3150 lbs.
An acceleration test concluded that a vehicle such as vehicle #2 would have a maximum acceleration of 2.0 m/s^2 at the time of the accident.
Homework Equations
(I'm all over the place, but...)
KE = (1/2)(m)(v^2)
W = Fd(cosθ)
W = ΔKE
F(friction) = (μk)(F(normal)) = (μk)(mg)
KE1 + PE1 + W(external) = KE2 + PE2
J = FΔt = mΔv = Δp
p = mv
The Attempt at a Solution
Originally, I used the fact that the work of friction is equal to the change in kinetic energy, so:
W(f) = ΔKE
(μk)(mg)(d)(cosθ) = (1/2)(m)(v^2)
√[2(μk)(g)(d)(cosθ)] = v
But this doesn't account for the change in surfaces, from asphalt to grass. Also, I'm not even sure if it properly accounts for the two-dimensions of the collision.
I'm aware that this is a 2D elastic collision problem, and understand how to split the x- and y-components of the velocity. However, I don't think I'm approaching this properly.
Any help would be greatly appreciated.
Thank you!