Chapter 4: Motion in a Plane
When an object moves in two dimensions (a plane), we need vectors to fully describe its position, velocity, and acceleration. This chapter extends kinematics to 2D motion.
Scalars and Vectors
- Scalar: magnitude only (mass, temperature, speed, time)
- Vector: magnitude AND direction (displacement, velocity, acceleration, force)
Vectors are represented as arrows; their length shows magnitude, and the arrow shows direction.
Vector Operations
Addition (Triangle/Parallelogram Law):
The resultant of two vectors A and B is found by placing them head-to-tail; the vector from the tail of A to the head of B is A + B.
Magnitude of resultant: |R| = √(A2 + B2 + 2AB cos(theta))
where theta is the angle between A and B.
- Resolution of Vectors:
- Any vector V at angle theta to the x-axis has:
- Vx = V cos(theta) (horizontal component)
- Vy = V sin(theta) (vertical component)
Projectile Motion
A projectile is an object launched into the air and moving under gravity alone. The horizontal and vertical motions are independent.
Key formulas
- Key results (launch angle theta, initial speed u):
- Time of flight: T = 2u sin(theta)/g
- Maximum height: H = u2 sin2(theta)/(2g)
- Range: R = u2 sin(2 theta)/g
- Maximum range at theta = 45°: Rmax = u2/g
The trajectory of a projectile is a parabola.
Circular Motion
Uniform circular motion (UCM): object moves in a circle at constant speed. Speed is constant but velocity changes direction → acceleration exists.
- Angular displacement (theta): angle swept, in radians
- Angular velocity (omega) = d(theta)/dt = v/r, unit: rad/s
- Period T = 2pi/omega = 2pi r/v
- Frequency f = 1/T
- Centripetal acceleration: ac = v2/r = omega2 r, directed toward centre
- Centripetal force: Fc = m v2/r = m omega2 r
The centripetal acceleration is always directed toward the centre of the circle.
---
Find the resultant of two forces 3 N and 4 N acting at right angles.
|R| = √(32 + 42) = √(9 + 16) = √(25) = 5 N, at angle arctan(4/3) to the 3 N force.
A ball is kicked at 30 m/s at 30° to the horizontal (g = 10 m/s2). Find the range.
R = u2 sin(2 theta)/g = 900 × sin(60°)/10 = 900 × 0.866/10 = 77.9 m
Find the maximum height in Example 2.
H = u2 sin2(theta)/(2g) = 900 × 0.25/(20) = 225/20 = 11.25 m
A particle moves in a circle of radius 2 m at speed 4 m/s. Find centripetal acceleration.
ac = v2/r = 16/2 = 8 m/s2 toward the centre.
A vector of magnitude 10 N acts at 60° to the x-axis. Find its components.
Fx = 10 cos(60°) = 10 × 0.5 = 5 N; Fy = 10 sin(60°) = 10 × 0.866 = 8.66 N
An aircraft flies 100 km north then 100 km east. Find the resultant displacement.
|R| = √(1002 + 1002) = 100 √(2) = 141.4 km, at 45° north of east (northeast).
At what angle should a ball be thrown to achieve maximum range?
Maximum range is achieved at theta = 45°, giving Rmax = u2/g.
---
Common mistakes
- Treating projectile motion as a single equation — always separate horizontal and vertical components.
- Confusing angular velocity (omega, rad/s) with linear velocity (v, m/s): v = omega × r.
- Thinking centripetal force is a new "type" of force — it is simply the net force directed toward the centre, which may be tension, gravity, friction, or a combination.
Summary
Vectors require both magnitude and direction. Projectile motion combines horizontal uniform motion with vertical free fall, producing a parabolic path. Uniform circular motion involves constant speed but changing velocity direction, resulting in centripetal acceleration v2/r directed toward the centre.