# Mechanics - Collisions

• itsjorge

#### 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...

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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.

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.

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 $u_1u_2$?

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

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##.

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?

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.)

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}##.