Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 45 km/h sees an obstacle. Her reaction time is 1.4 s, after which the car brakes with a constant deceleration of 5.5 m/s². Find the total stopping distance d from the moment she sees the obstacle.
Ball A is dropped from rest from a height of 30 m. At the same instant ball B is thrown straight up from the ground directly below at 22 m/s. Ignore air resistance and use g = 9.81 m/s². When do the balls meet? Find t.
A 9 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.
An 8 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 0.25. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched from level ground at 26 m/s and must land 30 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.
A ball is launched at 60° above horizontal from the top of a 5 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A 2 kg block slides 16 m down a slope at 30° to the horizontal, with a coefficient of kinetic friction of 0.15. Find the work Wf done against friction.
An 8 kg block starts from rest at the bottom of a 20° slope and is pulled up the slope by a constant force of 65 N parallel to the slope. The coefficient of kinetic friction is 0.15. Find the speed v of the block after it has moved 4 m up the slope.
A 900 kg car moving at 15 m/s runs into a stationary 1000 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?
At a crossing a 1600 kg car travelling east at 8 m/s collides with a 1600 kg car travelling north at 19 m/s. The cars lock together. Find the direction θ of the wreck just after the collision, measured north of east.
A vertical rectangular gate 3 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 glycerin, with its top edge 2 m below the surface. Find the resultant hydrostatic force F on the gate.
Water flows through a horizontal pipe that narrows from 200 mm to 100 mm. In the wide section the pressure is 140 kPa and the velocity is 3 m/s. Ignoring losses, find the pressure p2 in the narrow section.
Water flows up through a pipe that narrows from 150 mm to 75 mm while rising 6 m. At section 1 the pressure is 350 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 10 L contains 410 g of Ar at 0 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of CH₄ at 250 kPa and 200 °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 30 °C. Its actual thermal efficiency is 35%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
A refrigerator keeps its cold space at -25 °C in a kitchen at 32 °C, removing heat from the cold space at 3 kW. Find the minimum power input Ẇmin.
A pipe open at both ends is 0.8 m long. The speed of sound is 343 m/s. Find the frequency f3 of its harmonic number 3.
Light of wavelength 589 nm falls on two slits 0.5 mm apart. The screen is 1.2 m away. Find the spacing Δy between neighbouring bright fringes.
Answer: d = 31.7 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 = 1.364 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, 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 = 39.97 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: θ = 12.9 °
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 other solution is 90° − θ = 77.1°.
Answer: t = 3.149 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: Wf = 40.78 J
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 = 5.205 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: ΔK = 53289 J
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: θ = 67.17 °
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 = 103 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 = 150.6 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: p2 = 72.5 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: p2 = 279.4 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 = 2331 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 1.019 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηII = 60.26 %
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 = 0.6891 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: f3 = 643.1 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: Δy = 1.414 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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