# Combining Momentum with Friction

1. May 14, 2013

### student34

1. The problem statement, all variables and given/known data

A 1500kg sedan travelling from north to south is struck by a 2200kg truck that is travelling from east to west. The two vehicles become enmeshed due to the impact and slide thereafter. Measurements show the coefficient of friction between the tires and the road is 0.75. The two vehicles slide 5.39m west and 6.43m south of the point of impact. How fast was each vehicle traveling just before the collision?

2. Relevant equations

P1x = P2x
P1y = P2y
P1x: mx*V2x = V2*(mx + my)
P1y: my*V2y = V2*(mx + my)
P1x = mx*Vt
P1y = V2*(my + mx)
P1y = my*Vy
Ffriction(x) = m(s + t)*a
Ffriction(y) = m(s + t)*a
a = g*μ
V2x = (-2*ax*xskid)^(1/2)
V2y = (-2*ay*yskid)^(1/2)

3. The attempt at a solution

mx*V1x = V2y*(mx + my)
V2y = (2*9.8*0.75*6.43)^(1/2) = 9.722...m/s
V2x = (2*9.8*0.75*5.39)^(1/2) = 8.901...m/s

Now I will substitute V2y and V2x into the conservation of momentum equation:

my*V1y = V2y*(mx + my)

Isolating V1y = (9.722...m/s*3700kg)/1500kg = 23.98...m/s which is wrong because the textbook's answers are 12m/s and 21m/s. I get V1x = 14.97...m/s. Both are wrong by 3m/s, but I have no idea why.

Last edited: May 15, 2013
2. May 15, 2013

### SammyS

Staff Emeritus
You can't break the work (done by friction) into x & y components. Work is a scalar quantity, not a vector.

3. May 15, 2013

### student34

I didn't use work or any type of energy equations. I broke acceleration into x and y components.

4. May 15, 2013

### SammyS

Staff Emeritus
The kinematic equation, (vx)2 = 2(ax)x
is equivalent to the work energy theorem (if you include all components).

Since you didn't use work explicitly, let's look at it this way.

For the kinematic equation, (vx)2 = 2(ax)x , you need to use ax, not a. Of course, use ay for to find vy.

The acceleration is in the same direction as the overall displacement. Use that to find its components.

5. May 15, 2013

### student34

I did that in the work in my original post.

6. May 15, 2013

### haruspex

I can't make sense of that. Do you mean mx*V1x = V2x*(mx + my) etc?
As SammyS says, this is not valid. A correct form would be
$v^2 = (-2 a*s_{skid})$ where $v^2 = v^2_x+v^2_y; a^2=a_x^2+a_y^2; x_{skid}^2+y_{skid}^2 = s_{skid}^2$, or in vectors (because a and s are collinear) v.v = -2a.s.

7. May 15, 2013

### student34

Wow, I get it, thank you both so much!!!!

Last edited: May 15, 2013
8. May 15, 2013

### SammyS

Staff Emeritus
That was much clearer than mine.

Thank you haruspex !