Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.
A plane needs a speed of 280 km/h to take off and the usable runway is 2300 m long. Starting from rest, what constant acceleration a does it need?
Ball A is dropped from rest from a height of 30 m. At the same instant ball B is thrown straight up from the ground directly below at 25 m/s. Ignore air resistance and use g = 9.81 m/s². At what height y above the ground do they meet?
A 12 kg block slides down a slope at 35° to the horizontal. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 145 N pulls the front 9 kg block; the 10 kg block is towed behind. The coefficient of kinetic friction is 0.3 for both blocks. Use g = 9.81 m/s². Find the acceleration a of the blocks.
A ball is launched from level ground at 30° above the horizontal and lands 35 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 50° above horizontal from the top of a 35 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
In a fairground rotor, riders stand against the wall of a vertical cylinder of radius 3.9 m. When it spins fast enough the floor drops away and friction holds them up. The coefficient of static friction is 0.4. Use g = 9.81 m/s². Find the smallest angular speed ω in rad/s.
A curve of radius 90 m is banked at 30°, and the coefficient of static friction between tyres and road is 0.15. Use g = 9.81 m/s². Find the lowest speed vmin at which a car can take the curve without sliding down the bank.
A block slides from rest down a slope at 15° to the horizontal. The coefficient of kinetic friction is 0.15. Find its speed v after it has slid 18 m along the slope.
An 8 kg block starts from rest at the bottom of a 20° slope and is pulled up the slope by a constant force of 60 N parallel to the slope. The coefficient of kinetic friction is 0.2. Find the speed v of the block after it has moved 9 m up the slope.
A 2 kg glider moving at 8 m/s makes a perfectly elastic head-on collision with a 1 kg glider at rest on a frictionless track. Find the velocity v′2 of the second glider after the collision.
A 0.1 kg ball hits a wall at 25 m/s and bounces straight back at 19 m/s. The contact lasts 40 ms. Find the average force Favg of the wall on the ball.
How long must a simple pendulum be to have a period of 2.1 s? Use g = 9.81 m/s². Find L.
A 2.1 kg mass on a spring with k = 230 N/m is pulled 5 cm from equilibrium and released. Find the total energy E of the oscillation.
A pipe closed at one end must have its harmonic number 7 at 500 Hz. The speed of sound is 343 m/s. Find the length L of the pipe.
Light of wavelength 500 nm passes through a single slit 0.05 mm wide. Find the width w of the central bright maximum on a screen 1 m away (between the first dark fringes on either side).
Answer: a = 1.315 m/s²
Motion with constant acceleration: v = v₀ + a·t, s = v₀·t + ½·a·t² and v² = v₀² + 2·a·s. Pick the equation that contains the three things you know and the one you want, and keep the signs consistent with one positive direction.
Answer: y = 22.94 m
Motion with constant acceleration: v = v₀ + a·t, s = v₀·t + ½·a·t² and v² = v₀² + 2·a·s. Pick the equation that contains the three things you know and the one you want, and keep the signs consistent with one positive direction.
Up is positive, the ground is y = 0.
Answer: a = 4.823 m/s²
Newton’s second law: the net force on a body equals its mass times its acceleration, ΣF = m·a. Draw a free-body diagram, choose a positive direction along the motion, and add every force in that direction: applied forces, friction μ·N against the motion, and the components of gravity.
Down the slope is positive.
Answer: a = 4.689 m/s²
Newton’s second law: the net force on a body equals its mass times its acceleration, ΣF = m·a. Draw a free-body diagram, choose a positive direction along the motion, and add every force in that direction: applied forces, friction μ·N against the motion, and the components of gravity.
Answer: v0 = 19.91 m/s
Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.
Answer: x = 147.5 m
Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.
Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.
Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.
Answer: ω = 2.508 rad/s
Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.
Answer: vmin = 18.63 m/s
Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.
Friction points up the slope.
Answer: v = 6.343 m/s
Work–energy principle: the change in kinetic energy equals the work done by all forces. Writing gravity and springs as potential energy (mgh and ½kx²) turns this into a balance: initial energy = final energy + energy lost to friction (μN·d).
The block drops a height d·sin θ while friction removes μmg·cos θ·d. The mass cancels.
Answer: v = 6.436 m/s
Work–energy principle: the change in kinetic energy equals the work done by all forces. Writing gravity and springs as potential energy (mgh and ½kx²) turns this into a balance: initial energy = final energy + energy lost to friction (μN·d).
Answer: v′2 = 10.67 m/s
Conservation of momentum: when no external force acts during a collision, the total momentum Σm·v before equals the total after. Velocities have a direction, so pick a positive direction and give velocities the other way a minus sign. In an elastic collision kinetic energy is conserved too; in a perfectly inelastic one the bodies stick together.
Solving momentum and energy conservation together gives these two results.
Answer: Favg = 110 N
Conservation of momentum: when no external force acts during a collision, the total momentum Σm·v before equals the total after. Velocities have a direction, so pick a positive direction and give velocities the other way a minus sign. In an elastic collision kinetic energy is conserved too; in a perfectly inelastic one the bodies stick together.
Answer: L = 1.096 m
Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².
Answer: E = 0.2875 J
Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².
Answer: L = 1.2 m
Standing waves and the Doppler effect: a string fixed at both ends, or a pipe open at both ends, fits whole half-wavelengths: L = nλ/2, fn = n·v/(2L), n = 1, 2, 3, … A pipe closed at one end has a node at the closed end and an antinode at the open end: L = nλ/4 with odd n only, fn = n·v/(4L). When source or observer move along the line between them, f′ = f·(v ± vo)/(v ∓ vs): use the upper signs for motion towards each other. Two waves of nearly equal frequency beat at |f₁ − f₂|.
Answer: w = 20 mm
Interference and diffraction: light from two slits a distance d apart reaches a point on the screen with a path difference d·sin θ. Bright fringes appear where d·sin θ = mλ (m = 0, ±1, ±2, …). For a screen at distance L ≫ d, the m-th bright fringe is at y = mλL/d and the fringe spacing is Δy = λL/d. A grating with N lines per millimetre has d = 1/N and sharp maxima at d·sin θ = mλ; since sin θ ≤ 1, the highest order is the largest whole number m ≤ d/λ. A single slit of width a has dark fringes at a·sin θ = mλ, m = ±1, ±2, …
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