2D elastic collision

  • Thread starter ambsop
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  • #1
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Homework Statement


Object 1 with velocity v hits object 2, of same mass at velocity 2. Prove that the angle between them after the collision is equal to 90 degrees.


Homework Equations


PE inital = PE final
KE initial = KE final


The Attempt at a Solution



P x component
m1 x v1 + 0 = (m1)(v'1)(costheta1) + (m1)(v'2)(costheta2)

p y comp
0 = (m1)(v'1)(sintheta1) + (m1)(v'2)(sintheta2)

kinetic E equation simplifies to
v1 - v2 = v'2 - v'1

I can't figure out how to combine the three to a point where they simplify.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
vela
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kinetic E equation simplifies to
v1 - v2 = v'2 - v'1
If I recall correctly, this relationship only holds for elastic collisions in one dimension.

Try squaring each of the two equations you have and then add them together. Then use the fact that the collision is elastic to eliminate some of the terms.
 
  • #3
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okay, so i tried that and I'm still stuck. Here's what I did: (' meaning after collision)

V1x = v1'cos1 + v2'cos2
V1y = v1'sin1 + v2'sin2

V^2 = V1x^2 + V2y^2
(v1'cos1)^2 + 2v1'v2'cos1cos2 + (v2'cos2)^2 + (v1'sin1)^2 + 2v1'v2'sin1sin2 + (v2'sin2)^2

From kinetic energy:
V^2 = v1'^2 + v2'^2
(v1' = v1'cos1 + v1'sin1
v2' = v2'cos2 + v2sin2)
(v1'cos1)^2 + 2v1v1cos1sin1 + (v1sin1)^2 + (v2cos2)^2 + 2v2v2cos2sin2 + (v2sin2)^2

Setting the momentum equation to the kinetic energy equation and simplifying leaves me with:

v1v2(cos1cos2 + sin1sin2) = v1v1cos1sin1 + v2v2cos2sin2

using trig identities:
1. sinusinv = 1/2[cos(u-v) - cos(u+v)] and
2. cosucosv = 1/2[cos(u-v) + cos (u+v)] results in the left side of the equation simplifying to:
cos(1-2)v1v2 = v1v1cos1sin1 + v2v2cos2sin2

i don't know what to do with this equation now.
 
  • #4
vela
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You're not squaring the velocities correctly. There is no cross term with the x and y components because they are orthogonal. In other words,
[tex]\vec{v} = v\cos\theta\,\hat{i} + v\sin\theta\,\hat{j},[/tex] so when you square this vector, you get
[tex]v^2 = \vec{v}\cdot\vec{v} = (v \cos\theta)^2 + (v\sin\theta)^2.[/tex] You don't get [itex](v\cos\theta+v\sin \theta)^2[/itex].
 
  • #5
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I'm confused. I think I'm over complicating this problem. So that equation you just gave me, comes from the kinetic energy equation right? But is it v1cos1 or v2cos2? And where does momentum play into this? Do I really need to break it up into the x and y components?
 
  • #6
vela
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What I'm saying is this part is wrong:
From kinetic energy:
V^2 = v1'^2 + v2'^2
(v1' = v1'cos1 + v1'sin1
v2' = v2'cos2 + v2sin2)
If an object is moving 3 m/s in the x-direction and 4 m/s in the y-direction, its speed isn't 8 m/s. It's 5 m/s. You don't simply (scalar) add the individual components to get the speed.

Therefore, the following is wrong
(v1'cos1)^2 + 2v1v1cos1sin1 + (v1sin1)^2 + (v2cos2)^2 + 2v2v2cos2sin2 + (v2sin2)^2
The cross terms shouldn't be there.
 
  • #7
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I now understand why my first approach was wrong, but I don't know how to start this problem.
 
  • #8
vela
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You're almost there. Just fix your mistake.
 

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