# Components of Velocity (Collisions)

1. May 24, 2012

### FeDeX_LaTeX

1. The problem statement, all variables and given/known data

I solved this problem simply by substituting the initial and final velocities in vector form and applying the principle of conservation of momentum, but there's something I don't understand about this problem.

"Two small smooth spheres A and B have equal radii. The mass of A is 2m kg and the mass of B is m kg. The spheres are moving on a smooth horizontal plane and they collide. Immediately before the collision the velocity of A is (2i – 2j) ms-1 and the velocity of B is (–3ij) ms-1. Immediately after the collision the velocity of A is (i – 3j) ms-1. Find the speed of B immediately after the collision."

I thought that the components of velocity perpendicular to the line of centres of the spheres would be unchanged in the collision, at least, this is what it says in my M4 book. The vertical component of A's velocity before is -2j, but after colliding with B, it's vertical component is -3j. Why is this?

Thanks.

2. Relevant equations

v = eu, conservation of momentum

3. The attempt at a solution

If you simply apply the conversation of momentum principle with the velocities in vector form, you get the correct answer of √2 m/s. But why do the velocities perpendicular to the line of centres change?

2. May 24, 2012

### jegues

I'm not entirely sure what you mean by "velocities perpendicular to the line of centres", but the velocity components in the plane of collision remain will remain unchanged.

3. May 24, 2012

### FeDeX_LaTeX

I just mean this: ---OO---

But plane of collision is the same thing to 'perpendicular to the line of centres'.

So how come in the question this doesn't apply? Is this an error?

4. May 25, 2012

### jegues

To me the plane of collision is the plane at which the two spheres make contact.

The only way I explain how to visualize this would be to imagine two spheres touching each other side by side and at the point where they actually touch place a infinitesimally thin piece of paper in between them.

This upright piece of paper is the plane in which the collision takes place (i.e. where the two spheres make contact).

Since your spheres are rolling around in the xy plane (for simplicity lets assume we are in the plane z = 0) the plane of the collision is going to be perpendicular to the xy plane.

In other words, the plane will stretch out in the positive and negative z direction forever.

Clearly any vectors that were to lie in this plane would cause the spheres to move in either the positive or negative z direction. (They would have to point in one of the two directions)

$$\text{i.e., } \quad \hat{k} \quad \text{ or } \quad -\hat{k}$$

Clearly that is not the case, correct? So I still believe the statement holds in your case.