# Homework Help: Momentum/Impusle problem [revised]

1. Dec 16, 2007

### harelo

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

So my problem says: "Car A (in a frictionless surface) travels at a velocity and weighs 1100 kg, Car B is parked and weighs 2200 KG, they collide in an inelastic collision (meaning they stick together) and they move a distance of 2.8 M together. The road has a coefficient of friction of 0.4"

2. Relevant equations

Find the velocity of the Car A before impact

3. The attempt at a solution

Last time I posted this question, our professor did not give us a coefficient of friction, which deemed the problem impossible to solve with mathematical values, but now that he realized his mistake, he gave us the problem again (an a re-test) so I just wanted to know if I'm along the right lines.

Since

$$F_{f}$$ = $$U_{k}$$mg,

and

F = $$F_{f}$$,

we can asume that

ma = $$U_{k}$$mg

and the masses would divide out, so

a = $$U_{k}$$g
a = (0.4)(-9.82 $$m/s^{2}$$)
a = -3.982 $$m/s^{2}$$

so then we can use the kinematic equation of

$$V_{f}$$^2 = $$V_{o}$$^2 + 2ad

where $$V_{f}$$ = 0 (this is the only part I'm iffy about), so

$$V_{o}$$ = $$\sqrt{-2ad}$$
$$V_{o}$$ = 4.7 m/s

Is this correct? thank for the help :)

2. Dec 16, 2007

### Feldoh

There will be a vf that is nonzero, unless car A accelerates to 0m/s before it collides with car B

3. Dec 16, 2007

### harelo

Wouldn't the fact that they crashed and there is a force of friction mean they'll eventually come to a rest?

EDIT: but not over that distance, now I see what I did.

So should I approach it some other way? maybe through work?

Last edited: Dec 16, 2007
4. Dec 16, 2007

### Feldoh

Yes eventually but momentum has to be conserved and right after they collide they will still be moving because if the collided them car A had a momentum that was not 0 so then the momentum right after the crash will have to get a nonzero number as well.

Edit: Work backwards, you can use the fact that the two cars went 2.8m after the collision to your advantage.

5. Dec 16, 2007

### harelo

Well since I have a distance, and an acceleration and a mass, I can get work.

W = F$$\Delta$$d
W = (3300)(-3.982)(2.8)
W = -36793.68 J

and since W = KE

W = 1/2mv^2
v = $$\sqrt{2W/m}$$
v = 4.72 m/s

what? ok now I'm lost.

6. Dec 16, 2007

### harelo

Am I missing something in my interpretation of work?

7. Dec 16, 2007

### Feldoh

The acceleration was the value before they collided, that wont be the acceleration after they collide.

8. Dec 16, 2007

### harelo

Hmmmm alright, so I'm guessing kinematic equations won't work... I'm lost, can I get a clue please?

9. Dec 16, 2007

### Staff: Mentor

View the problem as having two parts:
(1) The collision. Hint: What's conserved?
(2) The coming to a stop due to friction (after the collision). What's the acceleration?

Begin by looking at part two and figure out the initial speed of the stuck-together cars immediately after the collision.

10. Dec 16, 2007

### harelo

well I know that Momentum is conserved, however kinetic energy isn't (inelastic collisions, do'h)

and the acceleration would be the the coefficient of friction times acceleration due to gravity (like shown above).

So,

$$m_{1}$$$$v_{1}$$ = ($$m_{1}$$+$$m_{2}$$)$$V^{'}$$

hmmmm... Where can this get me?

Last edited: Dec 16, 2007
11. Dec 16, 2007

### Staff: Mentor

Good. (I see you've already calculated the initial speed of the stuck-together cars immediately after the collision.)

Good. Use conservation of momentum to figure out the initial speed of car A.

12. Dec 16, 2007

### harelo

OHHHHHH how can I be so dumb!!! that speed I Calculated is V'!!! wow I'm dumb!

So now it should be:

$$m_{1}$$$$v_{1}$$ = ($$m_{1}$$+$$m_{2}$$)$$V^{'}$$
$$v_{1}$$ = (($$m_{1}$$+$$m_{2}$$)$$V^{'}$$) / $$m_{1}$$
$$v_{1}$$ = ((3300)(4.72))/1100
$$v_{1}$$ = 14.16 m/s

I think this should be right, is it?

Last edited: Dec 16, 2007
13. Dec 17, 2007

Looks good!