Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 54 km/h brakes with a constant deceleration of 4.5 m/s² until it stops. How long does braking take? Find t.
A ball is thrown straight up at 17 m/s from the edge of a roof 20 m above the ground. Ignore air resistance and use g = 9.81 m/s². How long after the throw does it hit the ground? Find t.
A 14 kg block slides down a slope at 45° to the horizontal. The coefficient of kinetic friction is 0.2. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 3 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 5.5 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 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 55° above horizontal from the top of a 30 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 45° to the horizontal. The coefficient of kinetic friction is 0.4. Find its speed v after it has slid 8 m along the slope.
A 5 kg block starts from rest at the bottom of a 15° slope and is pulled up the slope by a constant force of 35 N parallel to the slope. The coefficient of kinetic friction is 0.25. Find the speed v of the block after it has moved 8 m up the slope.
A 1.5 kg glider moving at 12 m/s makes a perfectly elastic head-on collision with a 0.5 kg glider at rest on a frictionless track. Find the velocity v′1 of the first glider after the collision (positive in its original direction).
A 1300 kg car runs into a parked 1700 kg car. The wrecks lock together and skid 25 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 1.5 m wide sits in the wall of a tank of water, with its top edge 3 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 2 m wide sits in the wall of a tank of seawater, with its top edge 1 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 100 mm to 50 mm. In the wide section the pressure is 240 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 in the narrow section.
A horizontal venturi meter (inlet 80 mm, throat 40 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 290 mm. The discharge coefficient is 0.97. Find the flow rate Q. Use ρ = 1000 kg/m³ for water, 13 600 kg/m³ for mercury and g = 9.81 m/s².
A tank of 20 L contains 140 g of CH₄ at 60 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of CH₄ at 300 kPa and 190 °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 35 °C. Its actual thermal efficiency is 29%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
A refrigerator keeps its cold space at -10 °C in a kitchen at 25 °C, removing heat from the cold space at 3.5 kW. Find the minimum power input Ẇmin.
Light travels in water (n = 1.33) towards a flat boundary with air (n = 1.00). Find the critical angle θc beyond which it is totally reflected.
Light of wavelength 500 nm falls on two slits 0.4 mm apart. The screen is 3 m away. Find the distance y4 from the central bright fringe to bright fringe number 4.
Answer: t = 3.333 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 = 4.394 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.
Up is positive and the ground is h₀ below the starting point.
Only the positive root is physical.
Answer: a = 5.549 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 = 22.85 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: t = 1.562 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 = 5.741 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 = 8.16 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: v = 5.786 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′1 = 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.
Solving momentum and energy conservation together gives these two results.
Answer: v1 = 36.14 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 = 82.77 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 = 2.481 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 = 228.3 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 = 10.66 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: P = 1209 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 1.25 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηII = 50.53 %
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 = 0.4655 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: θc = 48.75 °
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: y4 = 15 mm
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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