Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 36 km/h brakes with a constant deceleration of 4.5 m/s² until it stops. Find the braking distance s.
A ball is thrown straight up at 7 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 9 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 block is given a push so that it starts sliding up a ramp at 20° with a speed of 3.5 m/s. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is launched at 65° 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 60° above horizontal from the top of a 50 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A 5 kg block slides 15 m down a slope at 40° to the horizontal, with a coefficient of kinetic friction of 0.15. Find the work Wf done against friction.
A 0.5 kg block is released from rest at height 4 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3 m long (μ = 0.2) and then runs onto a spring (k = 4000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 15 g bullet is fired horizontally into a 5 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 4 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 1200 kg car runs into a parked 1500 kg car. The wrecks lock together and skid 14.5 m before stopping. The coefficient of kinetic friction between tyres and road is 0.75. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 2 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 2 m wide sits in the wall of a tank of oil, 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 150 mm to 100 mm. In the wide section the pressure is 390 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
A horizontal venturi meter (inlet 200 mm, throat 100 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 140 mm. The discharge coefficient is 0.97. 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 10 L contains 330 g of Ar at 90 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of CO₂ goes from 200 kPa, 2 L, 30 °C to 200 kPa and 200 °C. Find the new volume V2.
A reversible (Carnot) heat engine takes in 1200 kJ of heat from a reservoir at 625 °C and rejects 378 kJ per cycle. Find the temperature TL of the cold reservoir in °C.
A refrigerator keeps its cold space at 5 °C in a kitchen at 35 °C, removing heat from the cold space at 3 kW. Find the minimum power input Ẇmin.
A pipe closed at one end must have its fundamental (first harmonic) at 132 Hz. The speed of sound is 343 m/s. Find the length L of the pipe.
Light of wavelength 550 nm falls on two slits 0.5 mm apart. The screen is 2.5 m away. Find the spacing Δy between neighbouring bright fringes.
Answer: s = 11.11 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 = 1.95 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 = 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: d = 1.001 m
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.
Both forces point down the slope; the mass cancels.
Answer: t = 1.848 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: t = 7.235 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 = 84.54 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: x = 9.132 cm
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 ramp is smooth, so only the strip removes energy.
Answer: v = 296.2 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: v1 = 32.87 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 = 159.4 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: F = 71.22 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 = 389.5 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: Q = 46.29 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.
Continuity and Bernoulli together give the ideal throat velocity.
Answer: P = 2494 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 3.122 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: TL = 9.767 °C
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 = 0.3236 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: L = 0.6496 m
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 = 2.75 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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