# Momentum Conservation - so hard!

1. Jul 12, 2007

### rum2563

Momentum Conservation --- so hard!

1. The problem statement, all variables and given/known data
Two automobiles collide at an intersection. One car of mass 1.4 X 10^3 kg is traveling at 45km/h ; the other car of mass 1.3 X 10^3 kg is traveling at 39km/h [E]. If the cars have a completely inelastic collision, what is their velocity just after collision?

2. Relevant equations
v1m1 + v2m2 = v(m1 + m2)

3. The attempt at a solution

Given information:
m1 = 1.4 X 10^3 kg
m2 = 1.3 X 10^3 kg
v1 = 45km/h --> 12.5 m/s
v2 = 39 km/h [E] --> 10.83 m/s
vf ' = ?

v1m1 + v2m2 = v(m1 + m2)
(12.5)(1.4 X 10^3) + (10.83)(1.3 X 10^3) = v (1.4 X 10^3 + 1.3 X 10^3 kg)
v = 11.7 m/s --> 42 km/H

The correct answer is 30 km/h.

2. Jul 12, 2007

### Staff: Mentor

Realize that momentum is a vector. Add the momentum vectors to get the total momentum of the cars after they collide and stick together.

3. Jul 12, 2007

### rum2563

I am really confused.

Do I do the following:$$\sqrt{10.8^2 + 12.5^2}$$

I am not sure what to do since there are so many ways to do this, and I am not sure where to begin.

4. Jul 12, 2007

### Staff: Mentor

Not exactly. You added the squares of the velocities; you should have added the squares of the momenta = mass*velocity.

5. Jul 12, 2007

### rum2563

Oohhh. ok.

I did this:

p1= m1v1
= (1.4 X 10 ^3)(12.5)
= 17500 kg.m/s

p2= m2v2
= (1.3 X 10 ^3)(10.8)
= 14040 kg.m/s

$$\sqrt{17500^2 + 14040^2}$$
= 2.24 X 10^4 kg.m/s

Now I think I got it:
2.24 X 10^4 = v (m1 + m2)
v = (2.24 X 10^4)/(1.4 X 10 ^3 + 1.3 X 10 ^3)
v= 8.296296296 m/s
v = 29.86666667 km/h

Therefore, the speed is 30 km/h.

Wow. Thanks for this Doc Al. I appreciate your help very much. You are the best.

Last edited: Jul 12, 2007