Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 108 km/h brakes with a constant deceleration of 7 m/s² until it stops. How long does braking take? Find t.
A tram starts from rest and accelerates at 2 m/s² for 11 s, then runs at constant speed for 24 s, and finally brakes at 3 m/s² until it stops. Find the total travel time t.
A 12 kg block slides down a slope at 35° to the horizontal. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 3.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 3 kg mass. The coefficient of kinetic friction between block and table is 0.25. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched at 30° 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 15° above horizontal from the top of a 20 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A toy launcher has a vertical spring of stiffness 2000 N/m. A 0.25 kg ball is placed on it and the spring is compressed by 5 cm, then released. Ignoring air resistance and the spring’s mass, find the height h the ball rises, measured from its position when the spring was compressed.
A 10 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 3 kg block. The system is released from rest. Find the speed v of the blocks after the hanging block has dropped 0.5 m.
An 8 kg projectile moving horizontally at 39 m/s explodes into two fragments that move along the same line. The 1 kg fragment continues forwards at 85 m/s. Find the velocity v′2 of the 7 kg fragment (positive in the original direction).
A 0.6 kg ball hits a wall at 14 m/s and bounces straight back at 6 m/s. The contact lasts 5 ms. Find the average force Favg of the wall on the ball.
A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 1 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 4 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 75 mm to 40 mm. In the wide section the pressure is 240 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
A closed water tank has compressed air above the water at a gauge pressure of 70 kPa. Water leaves through an opening with a diameter of 50 mm, 2 m below the water surface, into the open air. Ignoring losses, find the flow rate Q. Use g = 9.81 m/s².
What volume V does 30 g of CO₂ occupy at 700 kPa and 60 °C? Treat it as an ideal gas.
A fixed amount of CO₂ goes from 250 kPa, 10 L, 25 °C to 650 kPa and 120 °C. Find the new volume V2.
A heat engine works between a hot reservoir at 600 °C and a cold reservoir at 25 °C. It takes in 700 kJ of heat. Find the most work Wmax it could possibly produce.
A heat engine takes in 1400 kJ of heat from a reservoir at 600 °C, produces 560 kJ of work and rejects the rest to a reservoir at 15 °C, per cycle. Find the total entropy generated Sgen per cycle.
A string fixed at both ends is 0.65 m long; waves travel on it at 80 m/s. Find the frequency f3 of its harmonic number 3.
Light of wavelength 633 nm falls at right angles on a grating with 300 lines per millimetre. Find the angle θ2 of the order-2 maximum.
Answer: t = 4.286 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 = 42.33 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: a = 3.216 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 = 3.207 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: t = 1.631 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: x = 53.71 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: h = 1.019 m
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 = 1.229 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 = 32.43 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: Favg = 2400 N
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 = 4.519 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 = 236.8 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 = 26.29 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.
Answer: V = 2.697 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 5.072 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Wmax = 461 kJ
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 = 1.312 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: f3 = 184.6 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: θ2 = 22.32 °
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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