Why Is the Angle 90 Degrees in Elastic Collisions of Equal Mass?

[Abu]

Homework Statement

Prove that in the elastic collision of two objects of identical mass, with one being a target initially at rest, the angle between their final velocity vectors is always 90 degrees.

Homework Equations

m1v1+m2v2 = m1v1'+m2v2'
1/2m1v1^2 +1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2
v1-v2 = v2'-v1'

The Attempt at a Solution

This was my attempt:

Very sloppy diagram ^^^ but it is what I was basing my equations off of.
m1 = m2 = m
Y Direction:
0 = mv1'sinθ1 - mv2'sinθ2
Cancel m's and set equal to each other
v1'sinθ1 = v2'sinθ2
Solve for v1'
v1' = v2'sinθ2/sinθ1
X Direction

mv1 + 0 = mv1'cosθ1 + mv2'cosθ2
Cancel m's
v1 = v1'cosθ1 + v2'cosθ2
Use the v1-v2 = v2'-v1' formula and sub in v1' from Y direction and v1 from x direction
v1'cosθ1 + v2'cosθ2 = v2' - v2sinθ2/sinθ1

So that's as far as I got because from there I didn't know how to simplify it further (I don't know trigonometric identities) and I didn't even think I was on the right track.

I saw from other sources that you can simply use the conservation of kinetic energy equation, cancel m's and the 1/2, and use that as pythagoras.
It was basically: 1/2mv1^2 = 1/2mv1'^2 + 1/2mv2'^2
Cancel the m's and 1/2 and you have
v1^2 = v1'^2 + v2'^2
If that method is correct, can someone explain it to me?

 

[haruspex]

v1^2 = v1'^2 + v2'^2

That is not enough to deduce that the two velocities are at right angles.

So that's as far as I got because from there I didn't know how to simplify it further

You need to bear in mind where you are trying to get to. You want an equation that only involves angles, so use the equations that you have to eliminate all the speeds.

It might help to resolve onto different axes, maybe parallel to the line of centres at collision (so parallel to the departing velocity of the struck ball) and normal to that line. I have not tried it.

 

[neilparker62]

Consider first a 'head on' collison in which the velocity vector and line of centres (presume spherical objects) are parallel:

For elastic collisions Δp = 2μΔv where μ is the reduced mass [ m1 * m2 / (m1 + m2) ] of the colliding objects and Δv is their relative velocity. So in this case we determine:

Δp = 2 m^2 v / (2m) = mv. Thus momentum of moving body will be mv - Δp = 0 and momentum of stationary body will be 0 + Δp = mv.

If instead the line of centres (on collision) is at some angle to the moving object's velocity vector , then when momentum is resolved parallel and tangential to line of centres there will be a complete transfer of the momentum component parallel to line of centres leaving the (initially moving) object with only the component of momentum tangent to line of centres. Thus at right angles to the momentum imparted to the object initially at rest.

 

[Cutter Ketch]

Simplify! In your original equations set all the m’s equal ... and then cancel them. Note that v2 is zero and remove it. Note that momentum must be preserved independently for orthogonal axes and break your vector equations into component equations. Note that without generality you are welcome to choose a coordinate system and choose the coordinate system where one of the initial components is zero and get rid of it.

Now you have three equations written in one initial velocity component and four final velocity components with no other variables it should be much easier to work the problem.

Now how to prove the final velocities are at 90 degrees? Do you know any linear algebra? Do you remember the dot product?

 

[Cutter Ketch]

Without loss of generality, that is

 

[haruspex]

@Cutter Ketch , @neilparker62 , the thread is nearly a year old.

 

[Cutter Ketch]

@Cutter Ketch , @neilparker62 , the thread is nearly a year old.

Ha! Didn’t notice. Wonder what brought it back to the top today?

 

[haruspex]

Ha! Didn’t notice. Wonder what brought it back to the top today?

@neilparker62 seems to be trawling through old threads.

 

[Delta2]

from conservation of momentum we get the vector equation (after simplification of mass)

$$\vec{v_1}=\vec{v_1'}+\vec{v_2'}$$ (1)

from conservation of kinetic energy we get the equation (after simplification of mass)

$${v_1}^2={v_1'}^2+{v_2'}^2$$ (2)

Equation (1) tell us that the vectors $v_1,v_1',v_2'$ form a triangle. Equation (2) tell us that the triangle is orthogonal with $v_1$ being the hypotenuse and $v_1',v_2'$ the sides that are perpendicular to each other.

 

[neilparker62]

@neilparker62 seems to be trawling through old threads.

Yes this is true - please advise if there is any protocol which I am perhaps not observing. Other than being a nerd with a "smart ass" method of solving various problems involving elastic (and perfectly inelastic) collisions.

 

[haruspex]

Yes this is true - please advise if there is any protocol which I am perhaps not observing. Other than being a nerd with a "smart ass" method of solving various problems involving elastic (and perfectly inelastic) collisions.

It's a good idea to show other methods, but it might avoid confusion if you were to state upfront in such posts that it is a comment on an old thread.