Good luck, and let me know if you run into any more problems!

[mpittma1]

Homework Statement

https://scontent-a-sjc.xx.fbcdn.net/hphotos-prn1/t1.0-9/603671_1409914319284787_3441607163537875560_n.jpg

Homework Equations

conservation of total momentum

M₁V₁+M₂V₂ = M_totV_f

The Attempt at a Solution

I tried using the COTM

as such:

x: M₁V_1iCOSΘ₁+M₂V₂SINΘ₂=M_totV_f

y: M₁V_1iSINΘ₁+M₂V₂COSΘ₂=M_totV_f


then dividing 1/2:

COTΘ₁ + TANΘ₂ = 1

Θ₂ = TAN^-1(1-COTΘ₁)

After plugging in the values I am getting Θ₂ = -10.85°

What am I doing wrong?

 

[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 $M_{tot}v_{f,y}$

 

[mpittma1]

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 $M_{tot}v_{f,y}$

I can't see that working because the ending vector is directly along the x-axis, there is no sin or cos involved with that vector.

 

[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
$P_{y} = P_{tot} \cdot sin(0)$

 

[mpittma1]

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

so the equation becomes TANΘ₁+COTΘ₂=0

Θ₂=COT^-1(-TANθ₁)

Θ₂= -.025°

 

[jackarms]

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

 

[mpittma1]

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

Ok so before dividing I have

x: M₁V_1iCOSθ₁+M₂V_2iSINθ₂=M_totV_fCOS(0)

y: M₁V_1iSINθ₁+M₂V_2iCOSθ₂=M_totV_fSIN(0)

look legit?

 

[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.

 

[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!