# Homework Help: Conservation of Momentum - Finding Initial Velocity

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1. Nov 16, 2015

### annastewert

1. The problem statement, all variables and given/known data
A 1500-kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2200-kg SUV traveling from east to west. The two cars become embellished due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.75, and the cars slide to a halt at a point 5.39 m west and 6.43 m south of the impact point. How fast was each car traveling just before the collision?

2. Relevant equations
m1v1+m2v2=MV -->Conservation of momentum
Kinematics

3. The attempt at a solution
So I started by using the equation
μmg=ma
and the equation
to find the velocity of the two cars once they have crashed. It came to 11.108m/s

From here I used the conservation of momentum:
1500(Va)+2200(Vb)=(1500+2200)(11.108)
1500Va+2200Vb=41099.6
so then Va=27.4-1.47Vb

Because I have two variables I need a second equation so I was trying to use Kinetic Energy.
½ma(Va)^2+½mb(Vb)^2=½MV^2 +Work Done by Friction

To find force friction I took the normal force of the two cars stuck together
3700⋅9.8=36260
Friction=μNormal
Friction=(0.75)(36260)=27195N
Force⋅distance --> 27195(8.39)=228166.1J

from here we have
750Va^2+1100Vb^2=210172.2+228166.1
Input Va from the previous expression
750(274.-1.47Vb)^2+1100Vb^2=429238.25
cleaning that up a bit
2720.75Vb^2-60417Vb+133831.75=0

I plugged this into a quadratic equation, but I isn't giving me the correct answer... Any ideas where I went wrong?

2. Nov 16, 2015

### Staff: Mentor

Good.

Careful! Remember that momentum is a vector quantity--direction matters.

3. Nov 17, 2015

### annastewert

oooh thank you. I will try that!