# Final velocity of both vehicles

1. Aug 11, 2006

### UrbanXrisis

a 2000 kg truck is traveling south, a sports car is traveling west. they become entangled in a collision and leave a skid mark that is 20 meters long in a direction 14 degrees to the west of the direction of travel of the truck. Coefficient of sliding friction between tire and road is 0.6. what are the original velocities of both cars?

I have found three equations with three unknowns and I can solve for all three velocities, but I do not need to know the coefficient of friction and I do not need to know that the the skid mark is 20 meters long.

m1=2000kg
v1=initial velocity of truck
m2=1000kg
v2=initial velocity of car
m3=3000kg
v3=final velocity of both vehicles

equation 1, total momentum of system:

$$m_1v_1+m_2v+2 = m_3 v_3$$

equation 2, momentum in y direction of system:

$$m_1v_1= m_3 v_3 cos(14)$$

equation 3, momentum in x direction of system:

$$m_2v_2= m_3 v_3 sin(14)$$

my question is, if I solved these equations as a matrix, would this give me the original velocities of both cars? I am wondering this because I did not use the coefficient of friction nor did I use the 20 meters long skid mark. I was wondering if this method works. thanks!

2. Aug 12, 2006

### George Jones

Staff Emeritus
How do you know that m2 = 1000 kg? Is this supposed to be included in the question, i.e., is 'a sports car is traveling west' really supposed to be 'a 1000 kg sports car is traveling west'?

I don't understand your first equation. Total momentum is a vector, and an equation for total momentum should be a vector equation, but '2' is a scalar. Also, a vector equation is equivalent to (not independent of) the collection of all component equations.

3. Aug 12, 2006

### Astronuc

Staff Emeritus
As George Jones mentioned, the first equation for total momentum does not look right.

equation 1 $$m_1v_1+m_2v_2 = m_3 v_3$$ with equations 2 & 3,

$$m_1v_1= m_3 v_3 cos(14)$$
$$m_2v_2= m_3 v_3 sin(14)$$

would imply

$$m_3 v_3\,=\,m_3 v_3 cos(14)+m_3 v_3 sin(14)$$, or

1 = cos (14) + sin (14), when the identity is 1 = cos2(x) + sin2(x)

From the length of the skid mark and the coefficient of friction, one would find the initial velocity of the combined mass of the car and truck, as a function of the mass.

This is an inelastic collision and the total kinetic energy is not conserved (some of it goes into the internal energy or deformation of the vehicles). Momentum is conserved.

http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/momentum/2di.html [Broken]

hyperphysics.phy-astr.gsu.edu/hbase/inecol.html
hyperphysics.phy-astr.gsu.edu/Hbase/elacol.html
(unfortunately at the time of this post Hyperphysics is down)

Last edited by a moderator: May 2, 2017
4. Aug 14, 2006

### UrbanXrisis

could you please explain why $$m_1v_1+m_2v_2 = m_3 v_3$$ is not a correct equation?

I have done it using the coefficient of friction, is it correct?

work can be found http://home.earthlink.net/~suburban-xrisis/truck.pdf" [Broken]

Last edited by a moderator: May 2, 2017
5. Aug 15, 2006

### goodboy

momentum is a vector,in your first equation "2"means what?
my solution like this:
P1sin(14)+P2cos(14)=P
P1cos(14)=P2sin(14)
P3=√(2mfs)
there P1 is the momentum of the truck ,
P2 is the momentum of the sport car.
P is the final momentum of both vehicles
m=m1+m2,
f=0.6mg is the force of sliding friction.
s is 20 meters.

Last edited: Aug 15, 2006
6. Aug 15, 2006

### UrbanXrisis

if P1 was the initial momentum of the truck before the collision, then P1sin(14) is not correct since there was no angle before the collision. This also makes P2cos(14) false if we assume that P2 was the initial momentum of the car before the collision

if P1 was the momentum of the truck initially after the collision, then P1 would equal P since the final momentum would be both the car and the truck (since it is an inelastic collision), so that doesnt seem correct either.