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 7.5 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 32 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 25 kg box is pushed up a ramp at 25° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.2. Use g = 9.81 m/s². Find the pushing force F.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 75 N pulls the front 6 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 tension T in the connecting string.
A ball is launched at 25° 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 50° above horizontal from the top of a 30 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A block slides from rest down a slope at 35° to the horizontal. The coefficient of kinetic friction is 0.3. Find its speed v after it has slid 5 m along the slope.
A spring of stiffness 800 N/m at the bottom of a 20° slope is compressed by 12 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 5 kg glider moving at 7 m/s makes a perfectly elastic head-on collision with a 2 kg glider at rest on a frictionless track. Find the velocity v′2 of the second glider after the collision.
A 1100 kg car runs into a parked 1800 kg car. The wrecks lock together and skid 12 m before stopping. The coefficient of kinetic friction between tyres and road is 0.5. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A vertical rectangular gate 1 m wide sits in the wall of a tank of water, with its top edge 0.5 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 water, with its top edge 3 m below the surface. Find the depth ycp of the centre of pressure below the surface.
Water drains from a large open tank through a sharp-edged orifice with a diameter of 30 mm. The water level is kept 3 m above the orifice and the discharge coefficient is 0.64. Find the flow rate Q. Use g = 9.81 m/s².
Water flows up through a pipe that narrows from 75 mm to 50 mm while rising 7 m. At section 1 the pressure is 220 kPa and the velocity is 3 m/s. Ignoring losses, find the pressure p2 at section 2.
What volume V does 340 g of CO₂ occupy at 100 kPa and 65 °C? Treat it as an ideal gas.
Find the density ρ of CO₂ at 350 kPa and 40 °C, treating it as an ideal gas.
A heat engine receives heat from a source at 450 °C and rejects heat to the surroundings at 15 °C. Its actual thermal efficiency is 36%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
A Carnot heat engine receives 50 kW of heat from a source at 450 °C and rejects heat to the surroundings at 25 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at -5 °C and also rejects heat to the surroundings at 25 °C. Find the rate Q̇L at which heat is removed from the freezer.
An observer moves at 5 m/s away from a stationary source of 800 Hz. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.
An optical fibre has a core with n₁ = 1.5 and a cladding with n₂ = 1.42. 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: t = 3.704 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: hmax = 426.2 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: F = 148.1 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.
Constant speed: the forces balance.
Answer: T = 30 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: t = 2.068 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: x = 144.4 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.
Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.
Answer: v = 5.671 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 = 1.216 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 = 10 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.
Solving momentum and energy conservation together gives these two results.
Answer: v1 = 28.6 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 = 18.39 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 = 4.083 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: Q = 3.471 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 discharge coefficient accounts for the contraction of the jet and friction at the edge.
Answer: p2 = 133 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: V = 217.2 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 5.916 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηII = 59.85 %
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: Q̇L = 262.7 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.
Answer: f′ = 788.3 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 = 28.9 °
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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