# Question Involving Conservation of Momentum

1. Mar 10, 2014

### Jay520

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 embeshed 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

vxf2 = vx02 + 2aΔx
vyf2 = vy02 + 2aΔy

3. The attempt at a solution

The question asks to find the velocity of both cars before the impact. I figure this could be solved by applying conservation of momentum. Since friction would be so small compared to the force from the collision, we could set the initial momentum equal to the momentum after the collision. This would allow me to find initial velocities for both cars.

But before I do this, I would need to find the velocity for both cars after collision. I found this by using the two equations above. I know the acceleration (due to friction) after the collision, and I know the displacement after the collision, so I should be able to find velocity immediately after collision. First, I find acceleration.

F = mgμ = ma
∴ a = gμ

Using the above equations:

vx = √(2aΔx) = √(2gμΔx) = √(2*9.8*0.75*5.39) = 8.901 m/s
vy = √(2aΔx) = √(2gμΔy) = √(2*9.8*0.75*6.43) = 9.72 m/s

These are the components for the velocity of the two cars after the collision. Using conservation of momentum, I can find the velocity of the cars before the collision. Note that the sedan is the only car with momentum initially in the y direction, and the SUV is the only car with momentum initially in the x direction.

(subscript a is for SUV; subscript b is for the sedan)

mava = (ma+mb)vx
∴va = 14.97 m/s

mbvb = (ma+mb)vy
∴vb = 23.98 m/s

Unfortunately, I am told that these answers are wrong. The answers from the back of my textbook are 12 m/s and 21 m/s. I even tried using the work energy theorem (to calculate the velocities after the collision), and I still get the same answer. Anyone see what I'm doing wrong, or if there is a problem with the textbook perhaps? Maybe something wrong with my calculator?

Last edited: Mar 10, 2014
2. Mar 10, 2014

### Jay520

I figured out the proper way to do this problem. Does anyone know why the above method does not work?

3. Mar 10, 2014

### Simon Bridge

Yep - by the two step method you used:

Step 1: find the relationship between the velocity immediately after the collision and the stopping distance.
lets be a bit more careful with notation: - but putting x and y as the components of the displacement after the collision and d as the total displacement this time, then you did:
\begin{align}v^2 & =2ad\\ \implies v_x^2+v_y^2 & = 2\mu g x + 2\mu g y\end{align}
... but: $d=\sqrt{ x^2+y^2}$

4. Jan 29, 2017

### Anneke

What is the proper way to do this?

5. Jan 29, 2017

### haruspex

Try Simon's method. Note that he is saying equation (2) is wrong. See if you can write it correctly.