Comprehensive Study Notes: Collisions in Two Dimensions (FM2)
Welcome to Collisions in 2D!
Hello! If you mastered collisions in one dimension (like marbles hitting head-on), you’re ready for the real world—where things bounce off each other at angles! This chapter takes those core concepts—momentum and impulse—and upgrades them using vectors. Don't worry if vectors seemed tricky before; they are the essential tool that makes 2D collisions manageable. By the end of this section, you'll be able to predict exactly where objects will go after a glancing blow!
1. The Vector Foundation: Momentum and Impulse
In Further Mechanics, motion must be treated as a vector quantity because direction matters just as much as speed. A collision in 2D means we must resolve all motion into two independent directions, usually the x (i) and y (j) axes.
Momentum as a Vector (\(\mathbf{p}\))
Momentum is the product of mass (\(m\)) and velocity (\(\mathbf{v}\)). Since velocity is a vector, momentum is also a vector: $$ \mathbf{p} = m\mathbf{v} $$
If a particle has velocity \(\mathbf{v} = (u_x \mathbf{i} + u_y \mathbf{j})\), its momentum is \(\mathbf{p} = m(u_x \mathbf{i} + u_y \mathbf{j})\).
Impulse as a Vector (\(\mathbf{I}\))
An Impulse is the 'kick' or force applied over a short time during a collision. It is defined as the change in momentum: $$ \mathbf{I} = m\mathbf{v} - m\mathbf{u} $$ Where \(\mathbf{u}\) is the initial velocity and \(\mathbf{v}\) is the final velocity.
- Impulse is measured in Ns (Newton-seconds).
- Since impulse is a vector, we calculate the change in momentum separately for the \(\mathbf{i}\) component and the \(\mathbf{j}\) component.
Quick Review Box: Vectors in 2D
When dealing with collisions, always start by defining your coordinate axes (usually horizontal \(\mathbf{i}\) and vertical \(\mathbf{j}\)) and expressing all initial velocities (\(\mathbf{u}\)) as components.
2. Conservation of Momentum (CoM)
The most powerful tool in collision theory is the Conservation of Momentum. In any collision between two bodies, provided no external forces act (like friction), the total momentum of the system remains constant.
The Principle in Vector Form
For two particles, mass \(m_1\) and \(m_2\), the total momentum before the collision must equal the total momentum after: $$ m_1\mathbf{u}_1 + m_2\mathbf{u}_2 = m_1\mathbf{v}_1 + m_2\mathbf{v}_2 $$
Breaking it Down into Two Equations
Because this is a vector equation, it gives us two separate scalar equations that must hold true simultaneously: one for the horizontal (\(\mathbf{i}\)) components and one for the vertical (\(\mathbf{j}\)) components.
If we write the velocities as \(\mathbf{u} = u_x \mathbf{i} + u_y \mathbf{j}\) and \(\mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j}\):
Equation 1 (Horizontal - \(\mathbf{i}\) direction): $$ m_1 u_{1x} + m_2 u_{2x} = m_1 v_{1x} + m_2 v_{2x} $$
Equation 2 (Vertical - \(\mathbf{j}\) direction): $$ m_1 u_{1y} + m_2 u_{2y} = m_1 v_{1y} + m_2 v_{2y} $$
Key Takeaway: CoM gives you two equations straight away, which is essential for solving for the four unknown final components (\(v_{1x}, v_{1y}, v_{2x}, v_{2y}\)).
3. The Geometry of Impact: Line of Centres
In 1D, objects hit along the same line they travelled on. In 2D, the collision forces only act along the path connecting the two centres of the bodies (assuming smooth spheres).
The Line of Centres (LOC)
The Line of Centres (LOC) (sometimes called the Line of Impact) is the line connecting the centres of the two colliding spheres at the moment of impact.
- The impulsive force acts only along the LOC.
- This means momentum is only transferred between the particles along this line.
The Line of Perpendicularity (LOP)
The line perpendicular to the LOC is called the Line of Perpendicularity (LOP).
- Since there is no force acting along the LOP, the components of velocity perpendicular to the LOC remain unchanged during the collision.
Analogy: The Pool Table
Imagine you hit the cue ball (\(m_1\)) into the 8-ball (\(m_2\)) at an angle. The impulse (the "clack" sound) only pushes the balls apart along the line that connects their centres at the contact point. Any motion the balls had perpendicular to that line before the impact is simply carried on after the impact. This is the central trick in 2D collisions!
4. Newton's Experimental Law (NEL)
We need one more equation beyond the two from CoM. This comes from Newton's Experimental Law (or the Law of Restitution), which relates the relative speeds of separation and approach.
Applying NEL (The Bouncing Rule)
Crucially, NEL only applies to the components of velocity along the Line of Centres (LOC), where the impulse is actually acting.
If we use the symbol \(\parallel\) to denote the velocity component parallel to the LOC: $$ v_{1}^{\parallel} - v_{2}^{\parallel} = -e(u_{1}^{\parallel} - u_{2}^{\parallel}) $$
Where \(e\) is the coefficient of restitution (\(0 \le e \le 1\)):
- If \(e=1\), the collision is perfectly elastic (maximum kinetic energy conserved).
- If \(e=0\), the collision is perfectly inelastic (objects stick together).
- If \(0 < e < 1\), the collision is inelastic (energy is lost).
Did you know? Even if a collision is 'elastic' (E=1), the total kinetic energy is still only conserved if you consider the speeds along the LOC. Since the speeds perpendicular to the LOC are always unchanged (for smooth spheres), the kinetic energy associated with the LOP motion is always conserved.
5. Case Study: Impact with a Fixed Surface
When a particle hits a fixed, smooth surface (like a wall or a floor), the geometry simplifies dramatically.
We define the axes based on the surface:
- Parallel (LOP): Along the surface.
- Perpendicular (LOC): Normal to the surface.
Rules for Impact with a Fixed, Smooth Surface
1. Motion Parallel to the Surface (LOP)
Since the surface is smooth, there is no tangential friction (no force parallel to the wall).
Therefore, the component of velocity parallel to the surface does not change: $$ v_{\text{parallel}} = u_{\text{parallel}} $$
2. Motion Perpendicular to the Surface (LOC)
The impulse acts perpendicular to the surface. We apply Newton's Experimental Law here. Since the surface is fixed, its speed is zero (\(u_2 = v_2 = 0\)).
For the particle impacting the surface (\(m_1\)): $$ v_{\text{perpendicular}} = -e (u_{\text{perpendicular}}) $$
The perpendicular component reverses direction (due to the negative sign) and is scaled by the coefficient of restitution, \(e\).
Finding the Impulse on the Body
If asked to find the impulse (\(\mathbf{I}\)) exerted on the body by the wall, remember that impulse is the change in momentum: \(\mathbf{I} = m\mathbf{v} - m\mathbf{u}\).
Since the parallel momentum does not change, the impulse is purely perpendicular: $$ \mathbf{I} = m (v_{\text{perp}} - u_{\text{perp}})\mathbf{n} $$ Where \(\mathbf{n}\) is the unit vector perpendicular (normal) to the surface.
6. The Full Problem: Oblique Collision of Two Smooth Spheres
This is the most complex standard problem in this chapter, combining all the concepts above. The key is setting up the coordinate system correctly based on the Line of Centres (LOC).
Don't worry if this seems like a lot of steps—it's highly systematic. If you follow this checklist, you will always find the solution.
Step-by-Step Procedure
Step 1: Define the Collision Axes (LOC and LOP)
Determine the direction of the Line of Centres (LOC) at impact. If the question gives the position vectors of the centres, the LOC is the line connecting those centres.
- Let \(\mathbf{i}^{\prime}\) be the unit vector along the LOC.
- Let \(\mathbf{j}^{\prime}\) be the unit vector along the LOP (perpendicular to LOC).
Step 2: Resolve Initial Velocities
Resolve the initial velocities of both spheres (\(\mathbf{u}_1\) and \(\mathbf{u}_2\)) into components parallel (\(\parallel\)) and perpendicular (\(\perp\)) to the LOC.
$$ \mathbf{u}_1 = u_{1}^{\parallel} \mathbf{i}^{\prime} + u_{1}^{\perp} \mathbf{j}^{\prime} $$ $$ \mathbf{u}_2 = u_{2}^{\parallel} \mathbf{i}^{\prime} + u_{2}^{\perp} \mathbf{j}^{\prime} $$Step 3: Solve for the Perpendicular Components (LOP)
Since the spheres are smooth, there is no impulse force perpendicular to the LOC.
The perpendicular velocity components are unchanged: $$ v_{1}^{\perp} = u_{1}^{\perp} $$ $$ v_{2}^{\perp} = u_{2}^{\perp} $$ (This is 50% of your answer found!)
Step 4: Solve for the Parallel Components (LOC)
Along the Line of Centres (LOC), we have a standard 1D collision. We use CoM and NEL to find the final parallel speeds, \(v_{1}^{\parallel}\) and \(v_{2}^{\parallel}\).
Conservation of Momentum (CoM) along LOC: $$ m_1 u_{1}^{\parallel} + m_2 u_{2}^{\parallel} = m_1 v_{1}^{\parallel} + m_2 v_{2}^{\parallel} \quad (\text{Eq. A}) $$
Newton's Experimental Law (NEL) along LOC: $$ v_{1}^{\parallel} - v_{2}^{\parallel} = -e(u_{1}^{\parallel} - u_{2}^{\parallel}) \quad (\text{Eq. B}) $$
Solve these two simultaneous equations for \(v_{1}^{\parallel}\) and \(v_{2}^{\parallel}\).
Step 5: Recombine to Find Final Velocity Vectors
The final velocity vector for each sphere is the sum of its new parallel component and its unchanged perpendicular component:
$$ \mathbf{v}_1 = v_{1}^{\parallel} \mathbf{i}^{\prime} + v_{1}^{\perp} \mathbf{j}^{\prime} $$ $$ \mathbf{v}_2 = v_{2}^{\parallel} \mathbf{i}^{\prime} + v_{2}^{\perp} \mathbf{j}^{\prime} $$If the question asks for the speed and direction relative to the original axes (\(\mathbf{i}, \mathbf{j}\)), you must convert these final vectors back into the original coordinate system (often using trigonometry).
Common Mistake to Avoid
NEVER apply Newton's Experimental Law (NEL) to the components of velocity that are perpendicular to the line of centres. NEL only applies in the direction of the impulsive forces (the LOC).
Key Takeaway Summary
To solve any 2D collision:
- Use Conservation of Momentum in two directions (x and y) for the entire system.
- Identify the Line of Centres (LOC) and the Line of Perpendicularity (LOP).
- Velocities perpendicular to the LOC (LOP direction) are UNCHANGED.
- Apply Newton's Experimental Law (\(e\)) and CoM only along the LOC to find the two unknown final parallel speeds.