Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car speeding at a constant 18 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 3.5 m/s². How far from its starting point does the police car catch up? Find d.
Ball A is dropped from rest from a height of 15 m. At the same instant ball B is thrown straight up from the ground directly below at 19 m/s. Ignore air resistance and use g = 9.81 m/s². At what height y above the ground do they meet?
An 11 kg block slides down a slope at 25° 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.
A 7 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 4.5 kg mass. The coefficient of kinetic friction between block and table is 0.1. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched from level ground at 20° above the horizontal and lands 75 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 35 m/s and 45° above the horizontal from the top of a 45 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.
A 1800 kg car drives up a 5° slope at a constant 72 km/h. Air resistance and rolling resistance together amount to 400 N. Find the power P the wheels must deliver, in kW.
A 2 kg block rests on a rough horizontal table (μ = 0.1). A light string runs from it over a smooth pulley at the edge of the table to a hanging 4 kg block. The system is released from rest. Find the speed v of the blocks after the hanging block has dropped 2 m.
A 1.5 kg glider moving at 9 m/s makes a perfectly elastic head-on collision with a 4 kg glider at rest on a frictionless track. Find the velocity v′2 of the second glider after the collision.
At a crossing a 1600 kg car travelling east at 13 m/s collides with a 1100 kg car travelling north at 13 m/s. The cars lock together. Find the speed v′ of the wreck just after the collision.
A vertical rectangular gate 2 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 1.5 m wide sits in the wall of a tank of oil, with its top edge 1.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
Water flows through a horizontal venturi meter with an inlet diameter of 100 mm and a throat diameter of 60 mm. The pressure at the inlet is 24 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 100 mm while rising 3 m. At section 1 the pressure is 330 kPa and the velocity is 2.25 m/s. Ignoring losses, find the pressure p2 at section 2.
What volume V does 320 g of CH₄ occupy at 300 kPa and 60 °C? Treat it as an ideal gas.
A fixed amount of He goes from 350 kPa, 2 L, 15 °C to 150 kPa and 210 °C. Find the new volume V2.
A heat engine receives heat from a source at 750 °C and rejects heat to the surroundings at 25 °C. Its actual thermal efficiency is 28%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
A Carnot heat engine receives 80 kW of heat from a source at 900 °C and rejects heat to the surroundings at 30 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at -25 °C and also rejects heat to the surroundings at 30 °C. Find the rate Q̇L at which heat is removed from the freezer.
A source of 1200 Hz moves at 15 m/s towards a stationary observer. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.
Light of wavelength 532 nm falls on two slits 0.15 mm apart. The screen is 2.5 m away. Find the distance y4 from the central bright fringe to bright fringe number 4.
Answer: d = 185.1 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 that moment the police car moves at 2v = 36 m/s.
Answer: y = 11.94 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 = 2.812 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 = 29.56 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: v0 = 33.83 m/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.
Answer: H = 76.22 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: P = 38.78 kW
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). Power is force times speed: P = F·v.
Answer: v = 4.985 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′2 = 4.909 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: v′ = 9.349 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 = 110.4 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 = 2.633 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 = 21 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 = 262.6 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 = 184.2 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 7.825 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: ηII = 39.51 %
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 = 267.7 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′ = 1255 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: y4 = 35.47 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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