# 2D inelastic collision

1. Apr 29, 2014

### mpittma1

1. The problem statement, all variables and given/known data
https://scontent-a-sjc.xx.fbcdn.net/hphotos-prn1/t1.0-9/603671_1409914319284787_3441607163537875560_n.jpg

2. Relevant equations

conservation of total momentum

M1V1+M2V2 = MtotVf

3. The attempt at a solution

I tried using the COTM

as such:

x: M1V1iCOSΘ1+M2V2SINΘ2=MtotVf

y: M1V1iSINΘ1+M2V2COSΘ2=MtotVf

then dividing 1/2:

COTΘ1 + TANΘ2 = 1

Θ2 = TAN-1(1-COTΘ1)

After plugging in the values im getting Θ2 = -10.85°

What am I doing wrong?

2. Apr 29, 2014

### jackarms

I think the main issue is in setting both the x and y equations equal to $M_{tot}V_{f}$, since these are only component equations, so neither will equal the entire momentum. Instead, you'll want to use the components of the total momentum, so $M_{tot}v_{f,x}$ and $M_{tot}v_{f,y}$

3. Apr 29, 2014

### mpittma1

I cant see that working because the ending vector is directly along the x-axis, there is no sin or cos involved with that vector.

4. Apr 29, 2014

### jackarms

No, it works. The angle from the vertical will just be 0 degrees, so you'll have something like:

$P_{x} = P_{tot} \cdot cos(0)$ and
$P_{y} = P_{tot} \cdot sin(0)$

5. Apr 29, 2014

### mpittma1

so the equation becomes TANΘ1+COTΘ2=0

Θ2=COT-1(-TANθ1)

Θ2= -.025°

6. Apr 29, 2014

### jackarms

Well, you can't divide the equations like that.
$$\frac{a + b}{c + d} \neq \frac{a}{c} + \frac{b}{d}$$

7. Apr 29, 2014

### mpittma1

Ok so before dividing I have

x: M1V1iCOSθ1+M2V2iSINθ2=MtotVfCOS(0)

y: M1V1iSINθ1+M2V2iCOSθ2=MtotVfSIN(0)

look legit?

8. Apr 30, 2014

### jackarms

Yes, that looks good. I think the easiest way to solve would be to move the components with $M_{1}v_{1}$ over, and then you could divide the equations.

9. Apr 30, 2014

### Rellek

Hey there!

$$V_{total} = \sqrt{V_{x}^2+V_{y}^2}$$

and

$$\theta = tan^{-1}(\frac{V_{y}}{V_{x}})$$

These seem to be the only two equations that you are forgetting about. As long as you can solve for both component of your resultant velocities, you should be good!