# An elastic collision in two dimensions

## 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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alphysicist
Homework Helper
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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How do I know what the Px and Py are before and after? This seems so overwhelming to me

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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

Doc Al
Mentor

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

im clicking on that and it works for me

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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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

Doc Al
Mentor
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.

so like m1v1 + m2v2 = m1v1' + m2v2', and to get the velocity put cos/sin into equation for each term

Doc Al
Mentor
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.

something tells me that since its in another quadrant, the degrees arent as what they appear?? Like they should be - degrees or something

and does "sign" mean sin? I don't understand

oh, now I think I know what your saying. The sign is incorrect because they are in negative directions.

Doc Al
Mentor
The sign is incorrect because they are in negative directions.
Right.

(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

Doc Al
Mentor
(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.

(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

Doc Al
Mentor
Instead of v1'(cosθ), call it v1'x.

(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!

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?

Doc Al
Mentor
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.)

the uknown theta isn't important in this equation anyway, is it?

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

Doc Al
Mentor
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.)