An elastic collision in two dimensions

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Homework Help Overview

The discussion revolves around an elastic collision in two dimensions, focusing on the conservation of momentum and kinetic energy. Participants are analyzing the problem presented in a linked document, which outlines the parameters of two colliding objects, including their masses, initial velocities, and angles.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the conservation of momentum in both x and y directions, questioning how to determine the momentum components before and after the collision.
  • Some participants express confusion regarding the calculation of momentum components and the implications of angles in different quadrants.
  • There are attempts to set up equations based on the conservation laws, with discussions about the signs of the momentum components and the need to resolve unknowns.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and questioning the correctness of their approaches. Some guidance has been provided regarding the use of momentum conservation equations and the relationship between velocity components and angles. However, there is no explicit consensus on the final approach or solution.

Contextual Notes

Participants are working with a specific homework problem that includes a link to a document, which some find inaccessible. There is a focus on ensuring the correct application of physics principles while navigating the complexities of the problem setup.

Woopy
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Homework Statement


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


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'
 
Last edited by a moderator:
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Hi Woopy,

Woopy said:

Homework Statement


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


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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How do I know what the Px and Py are before and after? This seems so overwhelming to me
 
Last edited:
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
 
Woopy said:

Homework Statement


http://hypertextbook.com/physics/mechanics/momentum-two-three/an-elastic-collision-in-2d.pdf
That link doesn't seem to work. Can you describe the problem.
 
Last edited by a moderator:
http://hypertextbook.com/physics/mechanics/momentum-two-three/an-elastic-collision-in-2d.pdf

im clicking on that and it works for me
 
Last edited by a moderator:
Woopy said:
http://hypertextbook.com/physics/mechanics/momentum-two-three/an-elastic-collision-in-2d.pdf

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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there's 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
 
Woopy said:
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
so like m1v1 + m2v2 = m1v1' + m2v2', and to get the velocity put cos/sin into equation for each term
 
  • #11
Woopy said:
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
something tells me that since its in another quadrant, the degrees arent as what they appear?? Like they should be - degrees or something
 
  • #13
and does "sign" mean sin? I don't understand
 
  • #14
oh, now I think I know what your saying. The sign is incorrect because they are in negative directions.
 
  • #15
Woopy said:
The sign is incorrect because they are in negative directions.
Right.
 
  • #16
(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
Woopy said:
(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
(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?? there's 2 unknowns
 
  • #19
Instead of v1'(cosθ), call it v1'x.
 
  • #20
(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
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
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
the uknown theta isn't important in this equation anyway, is it?
 
  • #24
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
Woopy said:
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.)
 
  • #26
Woopy said:
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
Since 1/2mV^2 = 1/2m(Vx^2 + Vy^2), you can find the contributions to the total KE of each component separately.
 
  • #27
2.00kg object 1.00kg object total
px 12.00kgm/s 5.00kgm/s 9.00kgm/s 11.00kgm/s 21.00kgm/s 16.00kgm/s
py -24.00kgm/s -26.00kgm/s -9.00kgm/s -19.00kgm/s -33.00kgm/s -45.00kgm/s
K .. .. .. .. .. ..
 
  • #28
.5(2.00kg)(13.42)^2(cos 63.43) = 80.55 J
.5(2.00kg)(13.42)^2(sin 63.43) = 161.1 J

.5(1.00kg)(12.73)^2(cos 45) = 57.3 J
.5(1.00kg)(12.73)^2(cos
 
  • #29
man I am really confused right now, those arent right are they
 
  • #30
Doc Al please come back! I'm so close to finished, I've been on these forums for the past 7 hours trying to finish all this homework
 

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