Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 60 km/h sees an obstacle. Her reaction time is 1 s, after which the car brakes with a constant deceleration of 7 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A ball is thrown straight up at 5 m/s from the edge of a roof 5 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 16 kg block slides down a slope at 40° 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.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 90 N pulls the front 3 kg block; the 12 kg block is towed behind. The coefficient of kinetic friction is 0.1 for both blocks. Use g = 9.81 m/s². Find the tension T in the connecting string.
A ball is launched at 60° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is launched at 29 m/s and 30° above the horizontal from the top of a 35 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 block slides from rest down a slope at 35° to the horizontal. The coefficient of kinetic friction is 0.4. Find its speed v after it has slid 4 m along the slope.
A 15 kg block starts from rest at the bottom of a 25° slope and is pulled up the slope by a constant force of 95 N parallel to the slope. The coefficient of kinetic friction is 0.1. Find the speed v of the block after it has moved 4 m up the slope.
A 1300 kg car moving at 6 m/s runs into a stationary 900 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?
A 0.6 kg ball hits a wall at 11 m/s and bounces straight back at 15 m/s. The contact lasts 10 ms. Find the magnitude of the impulse J the wall gives the ball.
A vertical rectangular gate 3 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 1.5 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 venturi meter with an inlet diameter of 200 mm and a throat diameter of 150 mm. The pressure at the inlet is 57 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
Water flows up through a pipe that narrows from 75 mm to 50 mm while rising 3 m. At section 1 the pressure is 340 kPa and the velocity is 3 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 20 L contains 120 g of CO₂ at 95 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of Ar at 100 kPa and 40 °C, treating it as an ideal gas.
A heat engine receives heat from a source at 650 °C and rejects heat to the surroundings at 30 °C. Its actual thermal efficiency is 44%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
A refrigerator keeps its cold space at -20 °C in a kitchen at 35 °C, removing heat from the cold space at 2.5 kW. Find the minimum power input Ẇmin.
An object stands 50 cm in front of a thin diverging lens with a focal length of 25 cm. Find the image distance di. Use the sign convention 1/f = 1/do + 1/di (f < 0 for a diverging lens, di < 0 for a virtual image).
Light of wavelength 500 nm falls at right angles on a grating with 200 lines per millimetre. Find the angle θ2 of the order-2 maximum.
Answer: d = 36.51 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 = 1.641 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 = 4.803 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 = 72 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: H = 22.02 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.
At the top the vertical velocity is zero: vy² = 2gH.
Answer: H = 45.72 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: v = 4.393 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 = 3.223 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 = 9573 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: J = 15.6 N·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 = 128.8 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: Q = 228.2 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 = 292.3 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: P = 417.3 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 1.534 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηII = 65.51 %
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: Ẇmin = 0.5432 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: di = −16.67 cm
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: θ2 = 11.54 °
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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