2D inelastic collision

  • Thread starter mpittma1
  • Start date
  • #1
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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 = MtotVf

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?
 

Answers and Replies

  • #2
284
41
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
55
0
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 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
284
41
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
55
0
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°
 
  • #6
284
41
Well, you can't divide the equations like that.
$$\frac{a + b}{c + d} \neq \frac{a}{c} + \frac{b}{d}$$
 
  • #7
55
0
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: M1V1iCOSθ1+M2V2iSINθ2=MtotVfCOS(0)

y: M1V1iSINθ1+M2V2iCOSθ2=MtotVfSIN(0)

look legit?
 
  • #8
284
41
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
107
13
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!
 

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