An elastic collision in two dimensions

In summary: I feel like I'm so close yet so far away..I just don't understand the sign thing and then the px and py thing, and then the directions...I give upHere's the summary:In summary, the problem involves an elastic collision between a 2.00kg object and a 1.00kg object in a 2-dimensional coordinate system. The initial velocities of the two objects are given, and the goal is to find the final velocities and kinetic energy of both objects. Using the conservation of momentum equations and setting up separate equations for the x and y directions, the initial and final values for the x and y components of the momenta are found. The final velocities of both objects are then determined by solving for the
  • #1
Woopy
149
0

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'
 
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  • #2
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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  • #3
How do I know what the Px and Py are before and after? This seems so overwhelming to me
 
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  • #4
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
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.
 
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  • #6
http://hypertextbook.com/physics/mechanics/momentum-two-three/an-elastic-collision-in-2d.pdf

im clicking on that and it works for me
 
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  • #7
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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  • #8
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
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
 
  • #31
Note that when you calculate the KE, you are already given the speeds for the particles before the collision and the speed of the green particle after the collision. So for them there is no need to use components, just use KE = 1/2mV^2, where V is the given speed of the particle.

For the speed of the red particle after the collision, you'll have to figure that out. Using momentum equations, you can solve for Vx & Vy for that particle. Then you can calculate its KE either by first finding the total speed (V^2 = Vx^2 + Vy^2) or just by using the components of its velocity:
KE = 1/2mV^2 = 1/2mVx^2 + 1/2mVy^2.
 

What is an elastic collision in two dimensions?

An elastic collision in two dimensions is a type of collision between two objects where both the total kinetic energy and the total momentum are conserved. This means that after the collision, the objects will have the same combined mass and velocity as they did before the collision.

How is an elastic collision in two dimensions different from an inelastic collision?

In an inelastic collision, some of the kinetic energy is lost and converted into other forms of energy, such as heat or sound. In an elastic collision, all of the kinetic energy is conserved and there is no loss of energy.

What factors affect the outcome of an elastic collision in two dimensions?

The mass and velocity of the objects involved in the collision are the main factors that affect the outcome. The angle at which the objects collide and any external forces acting on the objects can also impact the outcome.

Can an elastic collision in two dimensions occur in real life?

Yes, elastic collisions occur in real life, although they may be difficult to observe. Some examples include collisions between billiard balls, collisions between atoms or molecules, and collisions between subatomic particles.

How is an elastic collision in two dimensions calculated?

The equations used to calculate an elastic collision in two dimensions are the conservation of momentum equation and the conservation of kinetic energy equation. These equations take into account the masses and velocities of the objects before and after the collision to determine the final velocities of the objects.

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