Components of Velocity (Collisions)

[FeDeX_LaTeX]

Homework Statement

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 (–3i – j) 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.

Homework Equations

v = eu, conservation of momentum

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?

 

[jegues]

Homework Statement

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 (–3i – j) 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.

Homework Equations

v = eu, conservation of momentum

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?

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.

 

[FeDeX_LaTeX]

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.

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?

 

[jegues]

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?

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 let's 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.

 

[asteriatic]

As a scientist, it is important to understand the underlying principles and assumptions behind any problem or solution. In this case, the principle of conservation of momentum is valid only in the direction of the collision. In other words, the total momentum of the system before and after the collision must be equal in the direction of the collision, but not necessarily in the perpendicular direction.

When two spheres collide, there are two components of velocity that need to be considered - the component in the direction of the collision and the component perpendicular to the collision. While the component of velocity in the direction of the collision is conserved, the perpendicular component can change due to the forces acting on the spheres during the collision.

In this problem, the spheres are moving on a smooth horizontal plane, which means there is no friction or external forces acting on them in the horizontal direction. However, there could be forces acting on them in the vertical direction, such as the normal force from the plane. When the spheres collide, these forces can cause a change in the perpendicular component of their velocities.

Additionally, the masses of the spheres also play a role in determining the change in the perpendicular component of their velocities. In this problem, sphere A has a larger mass than sphere B, which means it has a greater inertia and can resist changes in its motion more than sphere B. This can also contribute to the change in the perpendicular component of A's velocity after the collision.

Therefore, it is important to consider all the forces and factors at play when solving collision problems, and not just rely on the principle of conservation of momentum in one direction.