Collisions in two dimensions

Introduction: Moving from One Dimension to Two

Welcome to the world of 2D collisions! So far, you have likely mastered collisions where objects move back and forth along a single straight line. But in the real world—think of a game of pool, a football match, or even subatomic particles bumping into each other—things rarely move in just one direction. Collisions in two dimensions allow us to calculate exactly where objects will go after they hit each other at an angle.

In this chapter, we combine what you know about vectors (from Unit FM1.1) with the laws of momentum and restitution (from Unit FM1.3). Don't worry if it sounds a bit intimidating; the "secret" to 2D collisions is that we simply break one big 2D problem into two easy 1D problems!

1. The Core Secret: Components are Key

The most important thing to remember is that when two objects collide in 2D, we analyze the motion in two specific directions:
1. Parallel to the surface (or the line of impact).
2. Perpendicular to the surface (or the line of impact).

Quick Review: Before we start, remember your basic 1D collision rules:
- Conservation of Momentum: \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \)
- Newton’s Law of Restitution: \( v_1 - v_2 = -e(u_1 - u_2) \)

Analogy: Think of walking on a moving walkway at the airport. If you walk sideways across it, your forward speed (parallel to the walkway) stays the same, even though your sideways position changes. We treat collision components the same way—one direction affects the other very little!

2. Collision with a Smooth Fixed Surface (The Wall)

Imagine a smooth ball hitting a floor or a wall at an angle. To solve this, we resolve the velocity into components parallel to the wall and perpendicular to the wall.

What happens to the components?

1. Parallel to the wall: Since the wall is "smooth," there is no friction. This means there is no force acting parallel to the wall. Therefore, the velocity component parallel to the wall does not change.
\( v_{parallel} = u_{parallel} \)

2. Perpendicular to the wall: This is where the actual "thud" happens. The wall exerts a force on the ball. We apply Newton’s Law of Restitution (\( e \)) here.
\( v_{perpendicular} = -e \times u_{perpendicular} \)

Step-by-Step Process:
1. Draw a diagram showing the initial velocity \( u \) and the angle \( \alpha \) to the wall.
2. Find the components: \( u_{parallel} = u \cos(\alpha) \) and \( u_{perpendicular} = u \sin(\alpha) \) (depending on where the angle is measured!).
3. Keep the parallel component the same.
4. Multiply the perpendicular component by \( -e \).
5. Use Pythagoras (\( a^2 + b^2 = c^2 \)) to find the final speed if needed.

Key Takeaway

In a collision with a smooth wall, only the component of velocity perpendicular to the wall changes. The parallel component stays exactly the same.

3. Oblique Collision of Two Smooth Spheres

This is the "classic" pool ball scenario. When two spheres collide, the force acts along the Line of Centers (the imaginary line connecting the centers of the two spheres at the moment they touch).

The Two Rules of Sphere Collisions:

1. Perpendicular to the Line of Centers: There is no force acting in this direction. Therefore, the velocity components of both spheres perpendicular to the line of centers remain unchanged.
2. Along the Line of Centers: This is treated exactly like a 1D collision. You use:
- Conservation of Momentum.
- Newton’s Law of Restitution (\( e \)).

Common Mistake: Students often try to apply the law of restitution to the total velocity. Don't do this! Only apply \( e \) and momentum conservation to the components along the line of centers.

Did you know? If a sphere hits another identical sphere that is at rest, and the collision is perfectly elastic (\( e = 1 \)), they will always move off at 90 degrees to each other!

4. Impulse in Two Dimensions

Recall from your 1D notes that Impulse (\( I \)) is the change in momentum: \( I = m(v - u) \). In two dimensions, Impulse is a vector.

Because the force only acts perpendicular to the surface (or along the line of centers), the Impulse vector will always act in that same direction. There is zero impulse parallel to a smooth surface.

Example: If a ball hits a horizontal floor, the impulse \( I \) will only have a vertical (\( j \)) component.
\( I = m(v_y - u_y) \)

Quick Review Box

- Smooth Surface: No change in velocity parallel to the surface.
- Restitution (\( e \)): Only applies to the velocity component that is "hitting" the object (perpendicular).
- Momentum: Conserved along the line of impact.
- Impulse: Acts only in the direction of the force (perpendicular to the surface).

5. Problem-Solving Tips for Struggling Students

If you find these problems confusing, follow this checklist every time:
1. Draw a big diagram: Mark the "Line of Impact" clearly.
2. Resolve everything: Turn your initial velocities into \( i \) and \( j \) components relative to that line.
3. Ignore the parallel: Write down immediately that the parallel velocities are the same before and after. This gets you easy marks!
4. Do the 1D math: Focus only on the components along the line of impact to find your unknowns.
5. Re-combine: If the question asks for the final speed, use Pythagoras at the very end.

Encouraging Note: 2D collision problems often look long, but they are very repetitive. Once you solve three or four "sphere vs. wall" problems, you'll realize they all follow the exact same steps!

Summary of Key Terms

Line of Centers: The line passing through the centers of two colliding spheres.
Oblique Impact: A collision where the objects are not moving directly toward each other's centers.
Smooth: A keyword meaning "ignore friction," which implies the velocity parallel to the surface won't change.
Restitution (\( e \)): A value between 0 and 1 that tells us how "bouncy" a collision is.