- #1
Negus
- 3
- 0
Homework Statement
You are called to the site of a collision between a small car and a half ton truck on a lonely country road. You are given the following information:
a) On clean dry pavement at a temperature of 20 C, the coefficient of kinetic friction is 0.600 on locked wheels.
b) The truck has left a skid mark that is 25.7 m long
c) The car has left a skid mark that is 20.3 m long
d) The estimated mass of the truck and contents was 1250 kg
e) The estimated mass of the car and contents was 750 kg
f) The truck’s skid mark is due North
g) The car’s skid mark is due East
h) The debris traveled 10.6 m 18.4 degrees E of N
i) The collisions appears to have been completely inelastic
j) The speed limit was 90 km/hr
k) Miraculously no one was seriously hurt
From this data can you work out the speed of the two vehicles?
2. Homework Equations
∑F = mTa
Ffric = μkmg
vf2 = v02 + 2ad
KE = (1/2)mv2 (I do not believe this is required since Ein ≠ Eout)
Momentum has not been covered in the course yet (next chapter), but I know how to use these type of equations:
mtruckvitruck = mdebrisvfsin(θf)
mcarvicar = mdebrisvfcos(θf)
The Attempt at a Solution
In calculating the acceleration on the debris of the collision (which I assumed included the combined vehicle mass), I used the Ffric and ∑F formulas to get
a = Ffric/mtotal
a = μ(mtruck + mcar)g/(mtruck + mcar)
a = μg
With the kinematics equation,
vf2 = v02 + 2ad
and knowing that vf = 0,
0 = v02 + 2ad
-v02 = 2ad
-v02 = 2μgd
v0 = √-2μgd
v0 = √-2(0.6)(-9.8)(10.6)
v0 = 11.16m/s = 40.2km/hr
After getting this value, I used the logic that the initial speed of the debris should be the final speed of the cars at the moment of the collision. Since the distances of the vehicles's skid marks were given, I plugged them into the same kinematics equation.
For the truck, I got 74.4km/hr, and for the car, I got 68.6km/hr. However, I tried solving the problem with the momentum equations just to make sure and I got different results (61.4km/hr for the truck and 32.1km/hr for the car). Not only that, but I was confused whether the skid mark distance implied that final velocities for both vehicles were 0. Any help would be appreciated; clearing any confusion works much better than giving answers.