An elastic collision in two dimensions

[Woopy]

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'

 

[alphysicist]

Hi Woopy,

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.

 

[Woopy]

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

 

[Woopy]

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]

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.

 

[Woopy]

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

im clicking on that and it works for me

 

[Doc Al]

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.

 

[Woopy]

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

 

[Doc Al]

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.

 

[Woopy]

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

 

[Doc Al]

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.

 

[Woopy]

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

 

[Woopy]

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

 

[Woopy]

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

 

[Doc Al]

The sign is incorrect because they are in negative directions.

Right.

 

[Woopy]

(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]

(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.

 

[Woopy]

(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

 

[Doc Al]

Instead of v1'(cosθ), call it v1'x.

 

[Woopy]

(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!

 

[Woopy]

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]

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.)

 

[Woopy]

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

 

[Woopy]

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]

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.)

 

[Doc Al]

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.

 

[Woopy]

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 .. .. .. .. .. ..

 

[Woopy]

.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

 

[Woopy]

man I am really confused right now, those arent right are they

 

[Woopy]

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