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 24 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 6 m/s². How long does it take the police car to catch up? Find t.
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 18 m/s. Ignore air resistance and use g = 9.81 m/s². At what height y above the ground do they meet?
A 16 kg block slides down a frictionless slope at 45° to the horizontal. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 7.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 2.5 kg mass. The coefficient of kinetic friction between block and table is 0.1. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched from level ground at 14 m/s and must land 15 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 55° above horizontal from the top of a 60 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 20° to the horizontal. The coefficient of kinetic friction is 0.15. Find its speed v after it has slid 17 m along the slope.
A 1 kg cart is released from rest at a height of 16 m on a smooth track that leads into a vertical circular loop of radius 5 m. Find the normal force N the track exerts on the cart at the top of the loop.
An 8 g bullet is fired horizontally into a 2 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 16 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
An 8 g bullet is fired horizontally into a 1.5 kg block hanging from light strings 1.9 m long, and stays in it. The strings swing out to a largest angle of 40° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of water, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 3 m wide sits in the wall of a tank of oil, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.
A pitot-static tube in a wind tunnel reads a difference of 650 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.
Water flows up through a pipe that narrows from 75 mm to 40 mm while rising 3 m. At section 1 the pressure is 340 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 at section 2.
What volume V does 160 g of CH₄ occupy at 150 kPa and 50 °C? Treat it as an ideal gas.
A fixed amount of CO₂ goes from 400 kPa, 5 L, 20 °C to 350 kPa and 200 °C. Find the new volume V2.
A heat engine receives heat from a source at 800 °C and rejects heat to the surroundings at 15 °C. Its actual thermal efficiency is 29%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
A heat engine takes in 2700 kJ of heat from a reservoir at 300 °C, produces 864 kJ of work and rejects the rest to a reservoir at 35 °C, per cycle. Find the total entropy generated Sgen per cycle.
An object stands 40 cm in front of a thin converging lens with a focal length of 15 cm. Find the magnification m (negative for an inverted image).
Light of wavelength 589 nm falls on two slits 0.5 mm apart. The screen is 1.2 m away. Find the distance y1 from the central bright fringe to bright fringe number 1.
Answer: t = 8 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.
At that moment the police car moves at 2v = 48 m/s.
Answer: y = 11.59 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 = 6.937 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: a = 1.717 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.
Answer: θ = 24.33 °
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 = 65.67°.
Answer: x = 39 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 = 8.189 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: N = 13.73 N
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). On a circular path the net force towards the centre must equal mv²/r.
The cart just makes it round when h = 2.5R; then N = 0 at the top.
Answer: v = 444.7 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: v = 556.7 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 = 193.1 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 = 103.6 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: v = 32.91 m/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 flow comes to rest at the tip, so its kinetic energy shows up as extra pressure.
Answer: p2 = 301.7 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 = 178.7 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 9.223 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: ηII = 39.65 %
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: Sgen = 1.247 kJ/K
Entropy generation: over a full cycle the working fluid has no net entropy change. The hot reservoir loses QH/TH and the cold reservoir gains QL/TL. Their sum is the entropy generated, which is zero for a reversible cycle and positive for every real one.
Sgen > 0, so the engine is possible but irreversible.
Answer: m = −0.6
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: y1 = 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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