Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 100 km/h brakes with a constant deceleration of 6.5 m/s² until it stops. Find the braking distance s.
A ball is thrown straight up at 20 m/s from the edge of a roof 30 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
Masses of 4.5 kg and 2.5 kg hang on either side of a light, frictionless pulley, connected by a light string. The system is released from rest. Use g = 9.81 m/s². Find the tension T in the string.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 80 N pulls the front 9 kg block; the 4 kg block is towed behind. The coefficient of kinetic friction is 0.2 for both blocks. Use g = 9.81 m/s². Find the acceleration a of the blocks.
A ball is kicked from level ground at 40 m/s and 40° above the horizontal towards a wall 96 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 launched at 34 m/s and 50° above the horizontal from the top of a 40 m cliff. Ignore air resistance and use g = 9.81 m/s². At what angle α below the horizontal does it hit the ground?
A block slides from rest down a slope at 35° to the horizontal. The coefficient of kinetic friction is 0.25. Find its speed v after it has slid 9 m along the slope.
A spring of stiffness 5000 N/m at the bottom of a 35° slope is compressed by 20 cm against a 1 kg block and released. The coefficient of kinetic friction between block and slope is 0.15. 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 projectile moving horizontally at 13 m/s explodes into two fragments that move along the same line. The 3 kg fragment continues forwards at 46 m/s. Find the velocity v′2 of the 1 kg fragment (positive in the original direction).
At a crossing a 1500 kg car travelling east at 18 m/s collides with a 1200 kg car travelling north at 21 m/s. The cars lock together. Find the speed v′ of the wreck just after the collision.
A vertical rectangular gate 3 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 m wide sits in the wall of a tank of glycerin, with its top edge 4 m below the surface. Find the depth ycp of the centre of pressure below the surface.
Water flows through a horizontal pipe that narrows from 150 mm to 100 mm. In the wide section the pressure is 130 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
A horizontal venturi meter (inlet 300 mm, throat 150 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 340 mm. The discharge coefficient is 0.96. Find the flow rate Q. Use ρ = 1000 kg/m³ for water, 13 600 kg/m³ for mercury and g = 9.81 m/s².
What volume V does 250 g of N₂ occupy at 750 kPa and 40 °C? Treat it as an ideal gas.
Find the density ρ of CO₂ at 550 kPa and 20 °C, treating it as an ideal gas.
A heat engine works between a hot reservoir at 450 °C and a cold reservoir at 25 °C. Find the highest possible thermal efficiency ηmax.
A refrigerator keeps its cold space at -20 °C in a kitchen at 25 °C, removing heat from the cold space at 4 kW. Find the minimum power input Ẇmin.
A source of 400 Hz moves at 30 m/s towards a stationary observer. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.
An optical fibre has a core with n₁ = 1.62 and a cladding with n₂ = 1.54. Find the largest angle θa to the fibre axis at which light from air can enter the end of the fibre and still be guided by total internal reflection.
Answer: s = 59.35 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: hmax = 50.39 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: T = 31.53 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.
The net driving force is the difference of the weights; it moves the total mass.
Answer: a = 4.192 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 = 32.41 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 already on its way down at the wall.
Answer: α = 60.26 °
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.
Energy or v² = v₀² + 2as in the vertical direction.
Answer: v = 8.07 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 = 14.64 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 = −86 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′ = 13.68 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: F = 386.3 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: ycp = 5.067 m
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.
Pressure grows with depth, so the resultant acts below the centroid.
Answer: p2 = 129.5 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: Q = 160.6 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.
Continuity and Bernoulli together give the ideal throat velocity.
Answer: V = 30.98 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 9.931 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηmax = 58.77 %
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 = 0.711 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: f′ = 438.3 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: θa = 30.18 °
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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