Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 108 km/h brakes with a constant deceleration of 7.5 m/s² until it stops. How long does braking take? Find t.
Ball A is dropped from rest from a height of 20 m. At the same instant ball B is thrown straight up from the ground directly below at 13 m/s. Ignore air resistance and use g = 9.81 m/s². At what height y above the ground do they meet?
A 10 kg block slides down a slope at 40° to the horizontal. The coefficient of kinetic friction is 0.15. 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 115 N pulls the front 5 kg block; the 1 kg block is towed behind. The coefficient of kinetic friction is 0.25 for both blocks. Use g = 9.81 m/s². Find the acceleration a of the blocks.
A ball is launched at 50° 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 20° above horizontal from the top of a 50 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A block slides from rest down a slope at 30° to the horizontal. The coefficient of kinetic friction is 0.2. Find its speed v after it has slid 12 m along the slope.
A spring of stiffness 2000 N/m at the bottom of a 30° slope is compressed by 15 cm against a 2 kg block and released. The coefficient of kinetic friction between block and slope is 0.2. 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 1800 kg car moving at 16 m/s runs into a stationary 500 kg car and the two lock together. What percentage ΔK/K of the original kinetic energy is lost?
A 0.2 kg ball hits a wall at 22 m/s and bounces straight back at 16 m/s. The contact lasts 25 ms. Find the magnitude of the impulse J the wall gives the ball.
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of water, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of seawater, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.
Water flows through a horizontal venturi meter with an inlet diameter of 200 mm and a throat diameter of 150 mm. The pressure at the inlet is 41 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
Water flows up through a pipe that narrows from 200 mm to 75 mm while rising 7 m. At section 1 the pressure is 350 kPa and the velocity is 1.75 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 5 L contains 160 g of CO₂ at 15 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of CO₂ at 1000 kPa and 10 °C, treating it as an ideal gas.
A heat engine works between a hot reservoir at 625 °C and a cold reservoir at 20 °C. Find the highest possible thermal efficiency ηmax.
A heat pump keeps a house at 21 °C while it is -10 °C outside, delivering 12 kW of heat to the house. Find the minimum power input Ẇmin.
A string fixed at both ends must have its fundamental (first harmonic) at 67 Hz; waves travel on it at 80 m/s. Find its length L.
An object is 12 cm in front of a concave mirror with a radius of curvature of 10 cm. Find the magnification m (negative for an inverted image).
Answer: t = 4 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: y = 8.391 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 = 5.179 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 = 16.71 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 = 3.748 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 = 4.315 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: v = 8.772 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: d = 1.703 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: ΔK/K = 21.74 %
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 lost energy goes into deformation, heat and sound.
Answer: J = 7.6 N·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: F = 242.8 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Answer: F = 22.62 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Answer: Q = 193.5 L/s
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
The pipe is horizontal, so only pressure and velocity change.
Answer: p2 = 205.4 kPa
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
Work in Pa, then divide by 1000 to get kPa.
Answer: P = 1742 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 18.69 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηmax = 67.36 %
Second-law limits: no cycle working between reservoirs at absolute temperatures TH and TL can beat the reversible (Carnot) cycle: ηmax = 1 − TL/TH, COPR,max = TL/(TH − TL) and COPHP,max = TH/(TH − TL). Energy is always conserved: QH = W + QL.
Answer: Ẇmin = 1.265 kW
Second-law limits: no cycle working between reservoirs at absolute temperatures TH and TL can beat the reversible (Carnot) cycle: ηmax = 1 − TL/TH, COPR,max = TL/(TH − TL) and COPHP,max = TH/(TH − TL). Energy is always conserved: QH = W + QL.
The least work is needed by a reversible machine, which has the highest COP.
Answer: L = 0.597 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: m = −0.7143
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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