Introduction: Moving Beyond One Dimension
Welcome to Further Mechanics 1! In your standard Mathematics course, you mostly dealt with motion along a straight line (one dimension). Now, we’re leveling up. In this chapter, we explore motion where objects move freely in a plane—think of a ship navigating the sea or a plane flying in the air. This requires using vectors to describe position, velocity, and displacement.
Don’t worry if vectors seemed tricky before; we’ll break down how they make two-dimensional mechanics much clearer and easier to handle!
Section 1: Representing Position and Velocity with Vectors
Position, Displacement, Speed, and Velocity
In two dimensions, we always need two pieces of information to locate an object or describe its motion:
- Position (\( \mathbf{r} \)): This is the location of the particle relative to a fixed origin \( O \). Since we are in two dimensions (a plane), this vector points from \( O \) to the particle.
- Displacement (\( \mathbf{s} \)): This is the change in position. If a particle moves from position \( \mathbf{r}_1 \) to \( \mathbf{r}_2 \), the displacement is \( \mathbf{s} = \mathbf{r}_2 - \mathbf{r}_1 \). Displacement is a vector.
- Velocity (\( \mathbf{v} \)): This is the rate of change of position, including both magnitude and direction. Velocity is a vector.
- Speed: This is the magnitude (size) of the velocity vector. Speed is a scalar (it only has size, no direction).
Vector Notation
In Further Mechanics, all positions and velocities are represented using unit vectors \( \mathbf{i} \) and \( \mathbf{j} \), or as column vectors.
Let's say a particle is 3 units to the East (\( \mathbf{i} \) direction) and 4 units North (\( \mathbf{j} \) direction) of the origin:
$$ \text{Position } \mathbf{r} = 3\mathbf{i} + 4\mathbf{j} \quad \text{or} \quad \mathbf{r} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} $$
If its velocity is \( \mathbf{v} = 2\mathbf{i} - 1\mathbf{j} \), this means:
- It is moving at \( 2 \text{ m/s} \) in the positive \( \mathbf{i} \) (horizontal) direction.
- It is moving at \( 1 \text{ m/s} \) in the negative \( \mathbf{j} \) (vertical) direction.
Finding Speed: Speed is the magnitude of the velocity vector. We use Pythagoras' Theorem!
If \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \), then: $$ \text{Speed} = |\mathbf{v}| = \sqrt{a^2 + b^2} $$ Example: If \( \mathbf{v} = 2\mathbf{i} - 1\mathbf{j} \), the speed is \( |\mathbf{v}| = \sqrt{2^2 + (-1)^2} = \sqrt{5} \text{ m/s} \).
Key Takeaway: Vectors are essential
In 2D, always treat Position, Displacement, and Velocity as vectors. Speed is the scalar magnitude of the velocity vector.
Section 2: Motion with Constant Velocity
This chapter focuses only on cases where the velocity is constant. This is a massive simplification! It means the particle is moving in a straight line at a constant speed, and crucially, its acceleration is zero.
The Fundamental Equation
You already know the relationship from 1D mechanics: Displacement = Velocity × Time. In vector form, this is even more powerful:
If a particle starts at position \( \mathbf{r}_0 \) (the initial position vector) and travels at a constant velocity \( \mathbf{v} \), its displacement after time \( t \) is \( \mathbf{v} t \).
Finding Position over Time
The position vector \( \mathbf{r} \) of the particle at time \( t \) is given by:
$$ \mathbf{r} = \mathbf{r}_0 + \mathbf{v} t $$
This equation is used constantly throughout this topic. It tells you exactly where the particle is at any point in time.
Step-by-Step Example: Finding Position
- Identify the initial position vector, \( \mathbf{r}_0 \). (This is the position when \( t=0 \)).
- Identify the constant velocity vector, \( \mathbf{v} \).
- Substitute these into the equation \( \mathbf{r} = \mathbf{r}_0 + \mathbf{v} t \).
- Group the \( \mathbf{i} \) components together and the \( \mathbf{j} \) components together.
Example: A particle starts at \( \mathbf{r}_0 = 5\mathbf{i} - 3\mathbf{j} \) and moves with \( \mathbf{v} = -2\mathbf{i} + 4\mathbf{j} \text{ m/s} \).
Position at time \( t \):
$$
\mathbf{r} = (5\mathbf{i} - 3\mathbf{j}) + (-2\mathbf{i} + 4\mathbf{j}) t
$$
$$
\mathbf{r} = (5 - 2t)\mathbf{i} + (-3 + 4t)\mathbf{j}
$$
The position is split into two independent components: \( x = 5 - 2t \) and \( y = -3 + 4t \). This is why 2D motion is so simple—the horizontal and vertical movements don't interfere with each other!
Common Mistake to Avoid!
Always remember the initial position \( \mathbf{r}_0 \). Students often forget to add this vector, calculating only the displacement \( \mathbf{v} t \) instead of the final position \( \mathbf{r} \).
Key Takeaway: The Master Formula
For constant velocity motion, memorize and master: \( \mathbf{r} = \mathbf{r}_0 + \mathbf{v} t \).
Section 3: Problems Involving Resultant Velocity
Sometimes, an object’s velocity is influenced by its surrounding medium. The classic example is a boat moving in a current, or an aircraft flying in a wind.
The resultant velocity is the actual, observable velocity of the object relative to the ground (or a fixed point).
Combining Velocities (Vector Addition)
If an object \( A \) has velocity \( \mathbf{v}_A \) relative to the medium \( M \) (e.g., water/air), and the medium \( M \) has velocity \( \mathbf{v}_M \) relative to the ground \( G \), then the resultant velocity \( \mathbf{v}_{A G} \) is the vector sum:
$$ \mathbf{v}_{A G} = \mathbf{v}_{A M} + \mathbf{v}_{M G} $$
Analogy: Imagine walking on a moving escalator. Your walking speed (relative to the escalator) combines with the escalator's speed (relative to the ground) to give your overall speed.
Solving Problems: Components vs. Vector Triangles
You can solve resultant velocity problems using two main methods:
- Component Method (Recommended for complex problems):
- Break all given velocities into their \( \mathbf{i} \) (horizontal) and \( \mathbf{j} \) (vertical) components.
- Add the \( \mathbf{i} \) components together.
- Add the \( \mathbf{j} \) components together.
- The resulting vector \( \mathbf{v} = (\sum \mathbf{i})\mathbf{i} + (\sum \mathbf{j})\mathbf{j} \) is the resultant velocity.
- Vector Triangle Method (Good for simple direction/speed questions):
If you are given magnitudes and directions (angles), you can sketch a vector triangle (head-to-tail addition) and use the Sine Rule or Cosine Rule to find the magnitude and direction of the resultant vector.
Did You Know?
The concept of vector resolution (breaking a vector into components) is just trigonometry! If a velocity \( V \) is at an angle \( \theta \) to the positive x-axis, the components are \( V \cos \theta \cdot \mathbf{i} \) and \( V \sin \theta \cdot \mathbf{j} \).
Key Takeaway: Addition is key
Resultant velocity is always the vector sum of the object's velocity and the medium's velocity.
Section 4: Relative Velocity
This is where mechanics gets really interesting. Relative velocity is the velocity of one object as seen by an observer traveling with another object.
Definition and Formula
Let \( A \) and \( B \) be two particles moving with velocities \( \mathbf{v}_A \) and \( \mathbf{v}_B \) relative to the ground.
The velocity of A relative to B, written as \( \mathbf{v}_{A B} \), is the velocity \( A \) *appears* to have if you are standing on \( B \).
The formula is simply the difference between the velocity vectors:
$$ \mathbf{v}_{A B} = \mathbf{v}_A - \mathbf{v}_B $$
Note the order: The first subscript (A) is the particle being observed, and the second subscript (B) is the observer.
Analyzing Relative Motion
The concept of relative velocity is powerful because it simplifies 2D motion down to 1D motion. If you calculate \( \mathbf{v}_{A B} \), you can pretend that observer \( B \) is stationary at the origin, and particle \( A \) is moving directly towards or away from them with velocity \( \mathbf{v}_{A B} \).
Relative Displacement Just like normal position, we define the relative position vector \( \mathbf{r}_{A B} \):
$$ \mathbf{r}_{A B} = \mathbf{r}_A - \mathbf{r}_B $$
This vector tells you the position of \( A \) relative to \( B \). If \( A \) and \( B \) are traveling with constant velocities, the relative position changes over time according to the familiar constant velocity formula:
$$ \mathbf{r}_{A B}(t) = \mathbf{r}_{A B}(0) + \mathbf{v}_{A B} t $$
Where \( \mathbf{r}_{A B}(0) \) is the initial separation vector (the distance between them at \( t=0 \)).
Key Takeaway: The Subtraction Rule
Relative velocity \( \mathbf{v}_{A B} \) is always \( \mathbf{v}_A - \mathbf{v}_B \). This simplifies complex 2D motion into a single line of relative movement.
Section 5: Key Applications of Relative Motion
Relative velocity is essential for solving two standard types of problems: Interception and Closest Approach.
Interception (Collision)
Two objects, \( A \) and \( B \), intercept (or collide) if they are at the same physical location at the same time \( T \). We can solve this in two ways:
- Using Absolute Position:
Set their position vectors equal to each other at time \( T \): $$ \mathbf{r}_A(T) = \mathbf{r}_B(T) $$ Since \( \mathbf{r} = \mathbf{r}_0 + \mathbf{v} t \), we set the \( \mathbf{i} \) components equal and the \( \mathbf{j} \) components equal, resulting in two simultaneous equations to solve for \( T \).
- Using Relative Velocity (often quicker):
For interception to occur, the relative displacement vector \( \mathbf{r}_{A B} \) at time \( T \) must be \( \mathbf{0} \). This means the object \( A \) (as seen by \( B \)) must travel exactly the distance of the initial separation \( \mathbf{r}_{A B}(0) \).
This approach is particularly useful if the question asks what direction one particle must travel to intercept another.
The required condition: \( \mathbf{v}_{A B} \) must be parallel to the initial relative position vector \( \mathbf{r}_{B A}(0) \) (the line connecting B to A).
Closest Approach
When two particles are moving but won't collide, we often need to find the shortest distance between them. This occurs when their relative velocity vector is perpendicular to their relative displacement vector.
We work entirely with the relative displacement vector: $$ \mathbf{r}_{rel} = \mathbf{r}_A - \mathbf{r}_B = \mathbf{R} $$ (where \( \mathbf{R} \) is a function of time \( t \)).
Method 1: Calculus (Minimizing Distance Squared)
This is often the most robust mathematical method. Since the minimum distance \( |\mathbf{R}| \) occurs at the same time \( t \) as the minimum of \( |\mathbf{R}|^2 \), we work with the squared distance to avoid square roots.
Steps:
- Find the relative position vector: \( \mathbf{R} = (a+ct)\mathbf{i} + (b+dt)\mathbf{j} \).
- Calculate the squared distance: \( D^2 = |\mathbf{R}|^2 = (a+ct)^2 + (b+dt)^2 \).
- Differentiate \( D^2 \) with respect to \( t \): \( \frac{d(D^2)}{dt} \).
- Set \( \frac{d(D^2)}{dt} = 0 \) and solve for the time \( t \) at which closest approach occurs.
- Substitute this value of \( t \) back into \( |\mathbf{R}| \) to find the minimum distance.
Method 2: Geometric Approach / Completing the Square
If you prefer algebra over calculus, you can often use the method of completing the square on the expression for \( D^2 \). This helps find the minimum value without differentiation.
Alternatively, the Geometric Approach uses the idea that the closest approach is the perpendicular distance from the observer \( B \)'s initial position to the line of motion of \( A \) (relative to \( B \)).
Steps for Geometric Method:
- Calculate \( \mathbf{v}_{A B} \). This defines the path of relative motion.
- Calculate the initial relative position \( \mathbf{r}_{A B}(0) \).
- The shortest distance is the perpendicular distance from the origin (where \( B \) is imagined to be stationary) to the line passing through \( \mathbf{r}_{A B}(0) \) and running parallel to \( \mathbf{v}_{A B} \).
Note: This often involves finding the angle between \( \mathbf{v}_{A B} \) and \( \mathbf{r}_{A B}(0) \) and using basic trigonometry (SOH CAH TOA).
Quick Review: Relative Motion Checklist
1. Define the Vectors: Identify \( \mathbf{r}_A, \mathbf{v}_A, \mathbf{r}_B, \mathbf{v}_B \).
2. Calculate Relative Velocity: \( \mathbf{v}_{A B} = \mathbf{v}_A - \mathbf{v}_B \).
3. Calculate Relative Position: \( \mathbf{r}_{A B}(t) = (\mathbf{r}_A(0) - \mathbf{r}_B(0)) + \mathbf{v}_{A B} t \).
4. Solve the Problem:
- Interception: Set \( \mathbf{r}_{A B}(T) = \mathbf{0} \).
- Closest Approach: Minimize \( |\mathbf{r}_{A B}(t)|^2 \) using Calculus or Geometry.