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 200 km/h to take off and the usable runway is 1100 m long. Starting from rest, what constant acceleration a does it need?
A tram starts from rest and accelerates at 3 m/s² for 11 s, then runs at constant speed for 13 s, and finally brakes at 3 m/s² until it stops. Find the total distance s between the two stops.
A 6 kg block slides down a frictionless slope at 20° to the horizontal. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 7 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 3 kg mass. The coefficient of kinetic friction between block and table is 0.2. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched at 40° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A ball is launched at 28 m/s and 60° 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.
In a fairground rotor, riders stand against the wall of a vertical cylinder of radius 2.5 m. When it spins fast enough the floor drops away and friction holds them up. The coefficient of static friction is 0.7. Use g = 9.81 m/s². Find the smallest angular speed ω in rad/s.
A curve of radius 120 m is banked at 8°, and the coefficient of static friction between tyres and road is 0.25. Use g = 9.81 m/s². Find the highest speed vmax at which a car can take the curve without sliding up and out.
A toy launcher has a vertical spring of stiffness 1500 N/m. A 0.2 kg ball is placed on it and the spring is compressed by 3 cm, then released. Ignoring air resistance and the spring’s mass, find the height h the ball rises, measured from its position when the spring was compressed.
A 10 kg cart is released from rest at a height of 14 m on a smooth track that leads into a vertical circular loop of radius 5 m. Find the speed v of the cart at the top of the loop.
A 17 kg projectile moving horizontally at 17 m/s explodes into two fragments that move along the same line. The 4 kg fragment continues forwards at 22 m/s. Find the velocity v′2 of the 13 kg fragment (positive in the original direction).
At a crossing a 1800 kg car travelling east at 13 m/s collides with a 1800 kg car travelling north at 18 m/s. The cars lock together. Find the direction θ of the wreck just after the collision, measured north of east.
A 0.8 kg mass on a spring with k = 400 N/m is pulled 11 cm in the positive direction and released from rest at t = 0. Find its displacement x from equilibrium at t = 1.35 s (with sign).
A 0.2 kg mass on a spring with k = 40 N/m is pulled 8 cm from equilibrium and released. Find its maximum acceleration amax.
A small source sends sound equally in all directions. The level is 60 dB at 1 m from it. Find the level β2 at 4 m, ignoring absorption and reflections.
Two thin lenses share an axis, 25 cm apart: lens 1 with f₁ = 8 cm and lens 2 with f₂ = 20 cm. An object stands 25 cm in front of lens 1. Find the position di2 of the final image measured from lens 2 (positive beyond lens 2, real-is-positive convention).
Answer: a = 1.403 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: s = 792 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 = 3.355 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 = 24.72 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.
Answer: x = 25.7 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.
Starting and landing at the same height: t = 2vy/g.
Answer: H = 54.97 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: ω = 2.368 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: vmax = 21.83 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 down the slope.
Answer: h = 0.344 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 = 8.859 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). On a circular path the net force towards the centre must equal mv²/r.
The cart just makes it round when h = 2.5R; then N = 0 at the top.
Answer: v′2 = 15.46 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: θ = 54.16 °
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: x = 3.687 cm
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 = 16 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: β2 = 47.96 dB
Waves: a periodic wave moves one wavelength in one period, so v = fλ with T = 1/f. On a stretched string v = √(T/μ), with T the tension and μ = m/L the mass per metre. Sound in air travels at about 343 m/s at 20 °C, light in vacuum at c = 3.00 × 10⁸ m/s. Loudness is measured on a logarithmic scale: β = 10·log₁₀(I/I₀) dB with I₀ = 10⁻¹² W/m², so 10 times the intensity adds 10 dB. A point source spreads its power P over a sphere: I = P/(4πr²).
Every doubling of the distance lowers the level by about 6 dB.
Answer: di2 = −39.13 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.
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