Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.
A car driving at 54 km/h brakes with a constant deceleration of 5.5 m/s² until it stops. How long does braking take? Find t.
A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 20 m/s² for 1 s, after which the engine stops and the rocket moves freely. Ignore air resistance and use g = 9.81 m/s². How long after launch does it hit the ground? Find t.
A box slides across a level floor with an initial speed of 2 m/s and comes to rest after 1.4 m. Use g = 9.81 m/s². Find the coefficient of kinetic friction μ.
An 8.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 6 kg mass. The coefficient of kinetic friction between block and table is 0.2. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is kicked from level ground at 26 m/s and 25° above the horizontal towards a wall 11 m away. Ignore air resistance and use g = 9.81 m/s². At what height y does the ball pass the wall?
A ball is thrown from ground level at 34 m/s and 65° above the horizontal. It lands on the flat roof of a building 23 m high, on its way down. Ignore air resistance and use g = 9.81 m/s². Find the horizontal distance x from the launch point to where it lands.
A curve of radius 320 m is banked at 18°. At what speed v can a car take it with no friction needed at all? Use g = 9.81 m/s².
A 1.2 kg ball on a string 0.9 m long is swung in a vertical circle. At the lowest point it moves at 8.5 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 1.5 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 5 kg block rests on a rough horizontal table (μ = 0.4). A light string runs from it over a smooth pulley at the edge of the table to a hanging 3 kg block. The system is released from rest. Find the speed v of the blocks after the hanging block has dropped 2.5 m.
A 5 g bullet is fired horizontally into a 1 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 24 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 0.4 kg ball hits a wall at 20 m/s and bounces straight back at 20 m/s. The contact lasts 40 ms. Find the average force Favg of the wall on the ball.
A 2.5 kg mass on a spring with k = 160 N/m oscillates on a frictionless surface with an amplitude of 18 cm. Find its speed v when it is 15 cm from equilibrium.
A body oscillates in simple harmonic motion with period 0.9 s and amplitude 19 cm. What is the shortest time t it needs to move from its maximum displacement of 19 cm to a displacement of 15.2 cm?
An object stands 8 cm in front of a thin converging lens with a focal length of 12 cm. Find the magnification m (negative for an inverted image).
A grating with 400 lines per millimetre shows the order-3 maximum of a spectral line at 36.9° from the straight-through direction. Find the wavelength λ of the line.
Answer: t = 2.727 s
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: t = 5.528 s
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.
Solve −h₁ = v₁·t₂ − ½·g·t₂² for the positive root.
Answer: μ = 0.1456
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: a = 2.909 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: y = 4.061 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.
The ball is still rising at the wall.
Answer: x = 77.83 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.
Larger root: the ball is descending.
Answer: v = 31.94 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 = 108.1 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: v = 2.476 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 = 436.2 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: Favg = 400 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: v = 0.796 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: t = 0.09217 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: m = 3
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: λ = 500.4 nm
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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