Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 120 km/h brakes with a constant deceleration of 7.5 m/s² until it stops. How long does braking take? Find t.
A ball is thrown straight up at 14 m/s from the edge of a roof 25 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
A 15 kg sled is pulled across level snow by a rope at 40° above the horizontal with a force of 150 N. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the acceleration a of the sled.
A 3.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 5 kg mass. The table is frictionless. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched at 35° 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 50° 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 400 N/m. A 0.5 kg ball is placed on it and the spring is compressed by 10 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 spring of stiffness 2000 N/m at the bottom of a 15° slope is compressed by 15 cm against a 3 kg block and released. The coefficient of kinetic friction between block and slope is 0.2. How far d up the slope does the block travel from the release point before it stops? (Ignore the short distance over which the spring pushes, i.e. take friction and gravity over the whole distance d.)
A 1 kg glider moving at 11 m/s makes a perfectly elastic head-on collision with a 4.5 kg glider at rest on a frictionless track. Find the velocity v′1 of the first glider after the collision (positive in its original direction).
A 0.6 kg ball hits a wall at 28 m/s and bounces straight back at 8 m/s. The contact lasts 20 ms. Find the average force Favg of the wall on the ball.
A vertical rectangular gate 1 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 m wide sits in the wall of a tank of oil, with its top edge 1 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 19 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
A horizontal venturi meter (inlet 150 mm, throat 75 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 260 mm. The discharge coefficient is 0.98. Find the flow rate Q. Use ρ = 1000 kg/m³ for water, 13 600 kg/m³ for mercury and g = 9.81 m/s².
A tank of 50 L contains 500 g of N₂ at 50 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of CH₄ goes from 400 kPa, 5 L, 20 °C to 100 kPa and 120 °C. Find the new volume V2.
A heat engine receives heat from a source at 350 °C and rejects heat to the surroundings at 10 °C. Its actual thermal efficiency is 22%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
A Carnot heat engine receives 90 kW of heat from a source at 750 °C and rejects heat to the surroundings at 27 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at 0 °C and also rejects heat to the surroundings at 27 °C. Find the rate Q̇L at which heat is removed from the freezer.
A pipe open at both ends is 0.4 m long. The speed of sound is 343 m/s. Find the frequency f3 of its harmonic number 3.
Light of wavelength 600 nm falls on two slits 0.1 mm apart. The screen is 2.5 m away. Find the distance y5 from the central bright fringe to bright fringe number 5.
Answer: t = 4.444 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: hmax = 34.99 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: a = 7.322 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: a = 5.771 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 = 3.976 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 = 52.9 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 = 0.4077 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: d = 1.691 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 = −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.
Solving momentum and energy conservation together gives these two results.
The minus sign means the first glider bounces back.
Answer: Favg = 1080 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 = 104.2 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 = 1.556 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 = 18.68 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: Q = 35.85 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.
Continuity and Bernoulli together give the ideal throat velocity.
Answer: P = 959.2 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 26.82 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: ηII = 40.32 %
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 = 643.4 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: f3 = 1286 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: y5 = 75 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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