Introduction to Vectors and Projectile Motion
Welcome to one of the most exciting parts of Physics! In the previous chapter, Motion and Kinematic Graphs, we looked at objects moving in straight lines. But in the real world, things move in all sorts of directions—think of a football being kicked into the air or a plane flying through a crosswind. To understand these, we need Vectors and Projectile Motion. Don't worry if this seems tricky at first; once you learn how to "split" motion into two directions, it becomes much simpler!
1. Scalars and Vectors
In Physics, we categorize every measurement into two groups:
Scalars: These are quantities that only have magnitude (size). They don't care about direction.
Examples: Distance, speed, mass, time, energy.
Vectors: These have both magnitude and direction. Direction is vital here!
Examples: Displacement, velocity, acceleration, force, momentum.
Vector Notation
Vectors are usually represented by arrows. The length of the arrow shows the magnitude, and the tip shows the direction. In your exams, you might see vectors written in bold or with a small arrow above the letter, but when you are calculating, you focus on their numerical value and their angle.
Quick Review: If you walk 5 meters East and then 5 meters West, your total distance (scalar) is \( 10\text{ m} \), but your displacement (vector) is \( 0\text{ m} \) because you ended up back where you started!
2. Combining and Resolving Vectors
In Unit 1, you need to know how to work with coplanar vectors (vectors acting in the same flat plane). There are two main skills: finding the "total" (resultant) and "splitting" one vector into two parts.
Finding the Resultant Vector
When two forces or velocities act on an object at the same time, we find the resultant. You can do this in two ways:
1. Scale Drawing: You draw the vectors "tip-to-tail" using a ruler and a protractor. The resultant is the line drawn from the very start to the very end.
2. Calculation: If the two vectors are at right angles (\( 90^{\circ} \)) to each other, we use Pythagoras' Theorem and Trigonometry.
If you have a horizontal vector \( V_x \) and a vertical vector \( V_y \):
Magnitude of Resultant \( V = \sqrt{V_x^2 + V_y^2} \)
Angle \( \theta = \tan^{-1}\left(\frac{V_y}{V_x}\right) \)
Resolving a Vector
This is the opposite of finding a resultant. It means taking a single vector acting at an angle and "breaking it down" into two perpendicular components (usually horizontal and vertical). This is the "secret weapon" for solving projectile problems.
If a vector \( V \) acts at an angle \( \theta \) to the horizontal:
Horizontal component: \( V_x = V \cos \theta \)
Vertical component: \( V_y = V \sin \theta \)
Memory Trick: To find the component CO-side (touching) the angle, use CO-sine. The other one is Sine!
Key Takeaway: Resolving vectors allows us to treat complex diagonal motion as two simple, independent straight-line motions.
3. Projectile Motion
A projectile is an object that is thrown or launched into the air and then moves only under the influence of gravity (ignoring air resistance for now). Examples include a launched cannonball or a jumped long-jumper.
The Principle of Independence
This is the most important rule in this chapter: The horizontal and vertical components of a projectile’s motion are completely independent of each other.
This means:
1. What happens horizontally does not affect what happens vertically.
2. The only thing they share is time \( (t) \). The object hits the ground horizontally at the exact same moment it hits the ground vertically.
Breaking Down the Motion
When solving projectile problems, always set up two columns: Horizontal and Vertical.
Horizontal Motion (The Easy Part)
Because there is no horizontal force (ignoring air resistance), there is no acceleration. The object moves at a constant velocity.
Acceleration \( a = 0 \)
Velocity \( v = u \) (constant)
Distance (Range) \( s = u \times t \)
Vertical Motion (The Gravity Part)
Gravity pulls the object down, so there is a constant acceleration. We use the SUVAT equations here (cross-reference: Motion and Kinematics Graphs).
Acceleration \( a = g = 9.81\text{ m s}^{-2} \) (downwards)
We use: \( v = u + at \), \( s = ut + \frac{1}{2}at^2 \), and \( v^2 = u^2 + 2as \).
Did you know? If you drop a bullet and fire another bullet horizontally at the same time from the same height, they will both hit the ground at the exact same time! This is because their vertical acceleration (\( g \)) is identical.
4. Step-by-Step Guide to Projectile Problems
Don't worry if these problems look long; follow these steps every time:
Step 1: Resolve the initial velocity. If the object is launched at an angle \( U \) at angle \( \theta \), find \( U_{\text{horizontal}} = U \cos \theta \) and \( U_{\text{vertical}} = U \sin \theta \).
Step 2: Split your data. Make a table for Horizontal and Vertical values. Remember that vertically, \( a = -9.81\text{ m s}^{-2} \) (if you define upwards as positive).
Step 3: Find the time \( (t) \). Usually, you use the vertical motion to find how long the object is in the air. For example, at the maximum height, the vertical velocity \( v_y = 0 \).
Step 4: Use time to find the horizontal distance. Once you have \( t \), multiply it by the constant horizontal velocity to find the "Range".
Common Mistake to Avoid
A very common error is accidentally using the diagonal initial velocity in a horizontal or vertical equation. Never use the diagonal velocity directly in a SUVAT equation unless the motion is purely along that line. Always resolve it first!
5. Summary of Equations
You will be given an equation list in the exam, but you should be familiar with these from Unit 1:
\( v = u + at \)
\( s = \frac{(u + v)t}{2} \)
\( s = ut + \frac{1}{2}at^2 \)
\( v^2 = u^2 + 2as \)
For projectiles specifically, remember \( g = 9.81\text{ m s}^{-2} \). In calculations, be careful with signs—if you choose upwards to be positive (+), then gravity must be negative (-9.81).
Key Takeaway: Projectile motion is just two linear motion problems happening at the same time. Solve for time in one, and use it in the other!