Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.
A car driving at 72 km/h brakes with a constant deceleration of 6 m/s² until it stops. Find the braking distance s.
A tram starts from rest and accelerates at 1.5 m/s² for 11 s, then runs at constant speed for 27 s, and finally brakes at 2 m/s² until it stops. Find the total distance s between the two stops.
A 14 kg block slides down a slope at 20° to the horizontal. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 9 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 5.5 kg mass. The table is frictionless. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched at 60° above horizontal from level ground. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A ball is launched at 15° above horizontal from the top of a 20 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
In a fairground rotor, riders stand against the wall of a vertical cylinder of radius 2.1 m. When it spins fast enough the floor drops away and friction holds them up. The coefficient of static friction is 0.3. Use g = 9.81 m/s². Find the smallest angular speed ω in rad/s.
A curve of radius 240 m is banked at 15°, 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 1 kg block slides 18 m down a slope at 40° to the horizontal, with a coefficient of kinetic friction of 0.1. Find the work Wf done against friction.
A spring of stiffness 5000 N/m at the bottom of a 40° slope is compressed by 20 cm against a 0.5 kg block and released. The coefficient of kinetic friction between block and slope is 0.1. How far d up the slope does the block travel from the release point before it stops? (Ignore the short distance over which the spring pushes, i.e. take friction and gravity over the whole distance d.)
A 4 kg glider moving at 9 m/s makes a perfectly elastic head-on collision with a 1 kg glider at rest on a frictionless track. Find the velocity v′1 of the first glider after the collision (positive in its original direction).
A 0.35 kg ball hits a wall at 27 m/s and bounces straight back at 9 m/s. The contact lasts 40 ms. Find the average force Favg of the wall on the ball.
A simple pendulum has a length of 1.6 m. Use g = 9.81 m/s². Find its period T for small swings.
A 0.8 kg mass on a spring with k = 90 N/m is pulled 2 cm from equilibrium and released. Find its maximum acceleration amax.
A string fixed at both ends is 0.75 m long; waves travel on it at 240 m/s. Find the frequency f4 of its harmonic number 4.
A parking sensor at rest sends ultrasound of 30 kHz towards a car that moves away at 15 m/s. Find the frequency f″ of the echo received back at the sensor (speed of sound 343 m/s).
Answer: s = 33.33 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.
Answer: s = 604.3 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.
Answer: a = 0.5897 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 = 3.721 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: t = 3.178 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.
Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.
Starting and landing at the same height: t = 2vy/g.
Answer: t = 2.715 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.
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 = 3.946 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 = 16.34 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: Wf = 13.53 J
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: d = 28.34 m
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′1 = 5.4 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 = 315 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: T = 2.537 s
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: amax = 2.25 m/s²
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: f4 = 640 Hz
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: f″ = 27486 Hz
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|.
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