An elastic collision in two dimensions

  • Thread starter Woopy
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  • #1
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Homework Statement


http://hypertextbook.com/physics/mechanics/momentum-two-three/an-elastic-collision-in-2d.pdf [Broken]


Homework Equations


P = MV
P= (r,θ)
p'=mv'
m1v1 + m2v2=m1v1' +m2v2'

The Attempt at a Solution


(2.00kg)(13.42 m/s) + (1.00kg)(12.73 m/s) = (2.00kg)(v1')+ (2.00kg)(21.95 m/s)
26.84 kgm/s + 12.73 kgm/s = 2.00kg(v1') + 43.9 kgm/s
39.57 kgm/s = 2.00kg(v1') + 43.9 kgm/s
-4.33 kgm/s = 2.00kg (v1')
-2.165 m/s = v1'
 
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Answers and Replies

  • #2
alphysicist
Homework Helper
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Hi Woopy,

Homework Statement


http://hypertextbook.com/physics/mechanics/momentum-two-three/an-elastic-collision-in-2d.pdf [Broken]


Homework Equations


P = MV
P= (r,θ)
p'=mv'
m1v1 + m2v2=m1v1' +m2v2'

The Attempt at a Solution


(2.00kg)(13.42 m/s) + (1.00kg)(12.73 m/s) = (2.00kg)(v1')+ (2.00kg)(21.95 m/s)

I don't believe this is true. It's not that the magnitudes of the momenta that are conserved; you have to find the components of the momenta.

The total momentum in the x direction is conserved, and separately the total momentum of the y direction is conserved.
 
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  • #3
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How do I know what the Px and Py are before and after? This seems so overwhelming to me
 
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  • #4
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v1
(2.00kg)(13.42 m/s)(cos 63.43)=12.00 kgm/s
(2.00kg)(13.42 m/s)(sin 63.43)=24.00 kgm/s

v2
(1.00kg)(12.73 m/s)(cos 45)=9.00 kgm/s
(1.00kg)(12.73 m/s)(sin 45)=9.00 kgm/s
 
  • #5
Doc Al
Mentor
45,186
1,515

Homework Statement


http://hypertextbook.com/physics/mechanics/momentum-two-three/an-elastic-collision-in-2d.pdf [Broken]
That link doesn't seem to work. Can you describe the problem.
 
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  • #6
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http://hypertextbook.com/physics/mechanics/momentum-two-three/an-elastic-collision-in-2d.pdf [Broken]

im clicking on that and it works for me
 
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  • #7
Doc Al
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http://hypertextbook.com/physics/mechanics/momentum-two-three/an-elastic-collision-in-2d.pdf [Broken]

im clicking on that and it works for me
How strange. When I click the link, it immediately brings me to a this page: http://midwoodscience.org/

But when I cut and paste the URL into a new tab, I see the intended site. I'll take a look.
 
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  • #8
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Theres a coordinate system, with v1 and v1' in the top left, v1 is 2.00kg and going 13.42 m/s, and is at 63.43degrees.

v1' has an unknown velocity and unknown theta. and in the bottom right v2 is 1kg, 12.73 m/s and 45 degrees.

v2' is also below, at 21.95 m/s and 59.93 degrees

I have to find px, py, and k of the 2kg, 1kg and total, both before and after collision
 
  • #9
Doc Al
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How do I know what the Px and Py are before and after? This seems so overwhelming to me
I didn't look over your numbers, but the only unknown in the diagram is the final speed and direction of the red particle. By setting up momentum conservation equations for the x and y directions, you should be able to solve for the components of that particle's final velocity.
 
  • #10
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so like m1v1 + m2v2 = m1v1' + m2v2', and to get the velocity put cos/sin into equation for each term
 
  • #11
Doc Al
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v1
(2.00kg)(13.42 m/s)(cos 63.43)=12.00 kgm/s
(2.00kg)(13.42 m/s)(sin 63.43)=24.00 kgm/s
This is the initial Px and Py for the red particle. But the sign of Py is incorrect.

v2
(1.00kg)(12.73 m/s)(cos 45)=9.00 kgm/s
(1.00kg)(12.73 m/s)(sin 45)=9.00 kgm/s
This is the initial Px and Py for the green particle. But the sign of Px is incorrect.

Compute the final Px & Py of the green particle in a similar manner.

Then set up your two momentum conservation equations and solve for the final Px & Py for the red particle.
 
  • #12
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something tells me that since its in another quadrant, the degrees arent as what they appear?? Like they should be - degrees or something
 
  • #13
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and does "sign" mean sin? I don't understand
 
  • #14
148
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oh, now I think I know what your saying. The sign is incorrect because they are in negative directions.
 
  • #15
Doc Al
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The sign is incorrect because they are in negative directions.
Right.
 
  • #16
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(1.00kg)(21.95 m/s)(cos 59.93)=11.00 kgm/s
(1.00kg)(21.95 m/s)(sin 59.93)= -19.00 kgm/s
 
  • #17
Doc Al
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(1.00kg)(21.95 m/s)(cos 59.93)=11.00 kgm/s
(1.00kg)(21.95 m/s)(sin 59.93)= -19.00 kgm/s
OK. That's the final Px & Py for the green particle.
 
  • #18
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(2.00kg)(13.42 m/s)(cos 63.43) + (1.00kg)(12.73 m/s)(cos 45) = (2.00kg)(v1')(cosθ) + (1.00kg)(21.95 m/s)(cos 59.93)

12.00 kgm/s + 9.00 kgm/s = 2.00kg(v1')(cosθ) + 11.00 kgm/s
21.00 kgm/s = 2.00kg(v1')(cosθ) + 11.00 kgm/s
10.00 kgm/s = 2.00kg(v1')(cosθ)
5.00 m/s = v1'(cosθ)
now what?? theres 2 unknowns
 
  • #19
Doc Al
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Instead of v1'(cosθ), call it v1'x.
 
  • #20
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(2.00kg)(13.42 m/s)(sin 63.43) + (1.00kg)(12.73 m/s)(sin 45) = (2.00kg)(v1')(sinθ) + (1.00kg)(21.95 m/s)(sin 59.93)

-24.00 kgm/s + -9.00 kgm/s = 2.00kg(v1')(sinθ) + 19.00 kgm/s
-33.00 kgm/s = 2.00kg(v1')(sinθ) + 19.00 kgm/s
-52.00 kgm/s = 2.00kg(v1')(sinθ)
-26.00 kgm/s = (v1')(sinθ)
2 unknowns!
 
  • #21
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How do I get the θ though? I don't understand what to do now that I got those 2 unknowns. Also, were my signs correct on the problem above?
 
  • #22
Doc Al
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You have two unknowns, but you also have two equations. (It might be easier for you if you call the components Vx & Vy, instead of Vcosθ & Vsinθ.)
 
  • #23
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the uknown theta isn't important in this equation anyway, is it?
 
  • #24
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and How do I get the K? I know its 1/2 mv2, my guess is that

.5(2.00kg)[(13.42m/s)(cos 63.43)]2
 
  • #25
Doc Al
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the uknown theta isn't important in this equation anyway, is it?
Given Vx & Vy, you can find V and θ if you need to. (Given Vcosθ & Vsinθ, you find V and θ also. Recall that (sinθ)² + (cosθ)² = 1.)
 

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