Can Elastic Collisions Lead to Right Angles Between Particles?

[itsjorge]

A particle with mass m is at rest. A second one, identical to the last one hits the first one. Show that in the case of a perfectly elastic collision (Q=0) the directions of the two particles make a right angle.

You can't assume that both final velocities will be equal. Here's what I've got from using: 1. the conservation of linear momentum. 2. The conservation of kinetic energy (Q=0). 3. Scalar product between the initial velocities.

v=u₁cosα + u₂ cosβ
u₁ sinα = u₂ sinβ
v²=u₁² + u₂²
cos(α+β)=(u₁² cosα cosβ + u₂² sinα sinβ) / (u₁ u₂)

And hell, I really can't solve this system...

 

[Stephen Tashi]

You haven't defined the variables you used. You didn't say how you established the coordinate system. If you are going to portray a collision in 3-dimensions as 2-dimensional problem, you should say why this is possible.

 

[itsjorge]

OK. That's my bad, I'm sorry. We're in 2 dimensions.
v represents the initial velocity of the particle. (The one hitting).
u represents the final velocity.
α, β represent the angles made with the direction of the particle 1 (The one hitting)
- α for the one that was moving previously and β for the one that was at rest.
They way I got to that system is explained up there.

 

[Stephen Tashi]

Why does your expression derrived from the inner product $(u1 \cos(\alpha), u1 \sin(\alpha)) \cdot (u2\cos(\beta), u2\sin(\beta))$ have the factors $u_1^2 , u_2^2$? Why not the factor $u_1u_2$?

 

[itsjorge]

Agree. You're right. Anyways, is this the way to proceed?

 

[CAF123]

Hi, manipulate your equations from conservation of linear of momentum and kinetic energy (three equations) to eliminate all instances of $v, u_1$ and $u_2$.

 

[Stephen Tashi]

The simplistic way to look at it is that two momentum vectors add to be the vector velocity V and that conservation of energy implies that the lengh of the hypotenuse V squared is the sum of the squares of the lengths of the other two sides. Hence it is a right triangle. Can we put that into algebra?

 

[Stephen Tashi]

Try the law of cosines for triangles.

(By the way, the angle between the two velocity vectors wouldn't be sum of angles $\alpha, \beta$ , it would be a difference of them. The the addition law for $cos(A +B)$ should be consistent with the signs of the cosine terms in the inner product.)

 

[vela]

Another approach is to work with the momentum vectors without breaking them into components. Conservation of momentum gives you
$$\vec{p}_1 + \vec{p}_2 = \vec{p}'_1 + \vec{p}'_2$$ where $\vec{p}_1 = 0$. Square both sides and use the fact that the kinetic energy can be written as $\frac{\vec{p}^2}{2m}$.