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 1500 m long. Starting from rest, what constant acceleration a does it need?
A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 22 m/s² for 2.5 s, after which the engine stops and the rocket moves freely. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax the rocket reaches.
A 14 kg block slides down a slope at 25° to the horizontal. The coefficient of kinetic friction is 0.2. 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 80 N pulls the front 1 kg block; the 3 kg block is towed behind. The coefficient of kinetic friction is 0.25 for both blocks. Use g = 9.81 m/s². Find the tension T in the connecting string.
A ball is launched from level ground at 20 m/s and must land 25 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.
A ball is launched at 21 m/s and 65° above the horizontal from the top of a 25 m cliff. Ignore air resistance and use g = 9.81 m/s². Find the greatest height H above the ground at the foot of the cliff.
A car takes a flat curve of radius 120 m. The coefficient of static friction between tyres and road is 0.7. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A 0.6 kg ball on a string 1.4 m long is swung in a vertical circle. At the lowest point it moves at 4 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.
A 2 kg ball hangs on a light string of length 3 m. It is pulled aside until the string makes 45° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A spring of stiffness 5000 N/m at the bottom of a 30° slope is compressed by 12 cm against a 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 20 kg projectile moving horizontally at 10 m/s explodes into two fragments that move along the same line. The 6 kg fragment continues forwards at 49 m/s. Find the velocity v′2 of the 14 kg fragment (positive in the original direction).
A 10 g bullet is fired horizontally into a 1.5 kg block hanging from light strings 1.4 m long, and stays in it. The strings swing out to a largest angle of 30° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
How long must a simple pendulum be to have a period of 1.5 s? Use g = 9.81 m/s². Find L.
A 2.6 kg mass on a spring with k = 90 N/m is pulled 8 cm from equilibrium and released. Find its maximum speed vmax.
An object stands 4 cm in front of a thin diverging lens with a focal length of 15 cm. Find the image distance di. Use the sign convention 1/f = 1/do + 1/di (f < 0 for a diverging lens, di < 0 for a virtual image).
Light of wavelength 500 nm passes through a single slit 0.08 mm wide. Find the angle θ2 of dark fringe number 2 from the straight-through direction.
Answer: a = 2.016 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: hmax = 222.9 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.
At the top the velocity is zero.
Answer: a = 2.368 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: T = 60 N
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.
With the same μ on both blocks, friction drops out: T = F·m₁/(m₁ + m₂).
Answer: θ = 18.91 °
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.
The other solution is 90° − θ = 71.09°.
Answer: H = 43.46 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.
Answer: vmax = 28.71 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.
Answer: T = 12.74 N
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.
Towards the centre (up) is positive.
Answer: T = 31.11 N
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). On a circular path the net force towards the centre must equal mv²/r.
Check: T = mg(3 − 2cos θ) = 31.11 N.
Answer: d = 1.251 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′2 = −6.714 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.
The minus sign means the second fragment flies backwards.
Answer: v = 289.7 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.
Answer: L = 0.5591 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: vmax = 0.4707 m/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: di = −3.158 cm
Geometric optics: at a boundary, n₁·sin θ₁ = n₂·sin θ₂ (Snell's law), with angles measured from the normal and n = c/v. Going into a slower medium the ray bends towards the normal. From a denser medium (n₁ > n₂) beyond the critical angle θc = arcsin(n₂/n₁) all light is reflected: total internal reflection. Thin lenses and spherical mirrors obey 1/f = 1/do + 1/di with the magnification m = −di/do. Sign convention: f > 0 for a converging lens and a concave mirror (f = R/2), f < 0 for a diverging lens and a convex mirror; di > 0 for a real image, di < 0 for a virtual one, and m < 0 for an inverted image.
Answer: θ2 = 0.7162 °
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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