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 25 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 3.5 m/s². How long does it take the police car to catch up? Find t.
A ball is thrown straight up at 11 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 14 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 110 N pulls the front 8 kg block; the 6 kg block is towed behind. The coefficient of kinetic friction is 0.2 for both blocks. Use g = 9.81 m/s². Find the tension T in the connecting string.
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 45° above horizontal from the top of a 5 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A 2.5 kg block slides 17 m down a slope at 35° to the horizontal, with a coefficient of kinetic friction of 0.3. Find the work Wf done against friction.
A 10 kg block is released from rest at height 4 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 2 m long (μ = 0.15) and then runs onto a spring (k = 500 N/m) on a smooth surface. Find the maximum compression x of the spring.
An 18 kg projectile moving horizontally at 11 m/s explodes into two fragments that move along the same line. The 7 kg fragment continues forwards at 52 m/s. Find the velocity v′2 of the 11 kg fragment (positive in the original direction).
At a crossing a 1800 kg car travelling east at 9 m/s collides with a 1800 kg car travelling north at 18 m/s. The cars lock together. Find the speed v′ of the wreck just after the collision.
A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 2.5 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 glycerin, with its top edge 0.5 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 150 mm to 100 mm. In the wide section the pressure is 130 kPa and the velocity is 3 m/s. Ignoring losses, find the pressure p2 in the narrow section.
A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 100 mm and a gauge pressure of 70 kPa. The nozzle exit diameter is 25 mm, and the height difference across the nozzle is negligible. Ignoring losses and air resistance, find the height H the jet rises above the nozzle. Use g = 9.81 m/s².
A tank of 5 L contains 250 g of He at 0 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of Ar at 600 kPa and 90 °C, treating it as an ideal gas.
A reversible (Carnot) heat engine takes in 4200 kJ of heat from a reservoir at 275 °C and rejects 2284 kJ per cycle. Find the temperature TL of the cold reservoir in °C.
A heat pump delivers 14 kW of heat to a heating system at 55 °C, taking heat from a source at 10 °C. Its actual COP is 50% of the Carnot COP between these temperatures. Find the electrical power Ẇ it uses.
A pipe closed at one end is 1.2 m long. The speed of sound is 343 m/s. Find the frequency f7 of its harmonic number 7.
Two slits 0.25 mm apart are lit with monochromatic light. On a screen 1.5 m away, bright fringe number 3 lies 11.39 mm from the central fringe. Find the wavelength λ.
Answer: t = 14.29 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 = 50 m/s.
Answer: hmax = 31.17 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 = 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 = 47.14 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: t = 4.677 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 = 4.963 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 = 102.5 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 = 120.5 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′2 = −15.09 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.
The minus sign means the second fragment flies backwards.
Answer: v′ = 10.06 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 = 183.9 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 = 1.083 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 = 111.7 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: H = 7.164 m
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.
Bernoulli along the jet: at the top the velocity is zero and the pressure is atmospheric again.
Answer: P = 28366 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 7.939 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: TL = 24.94 °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: Ẇ = 3.84 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: f7 = 500.2 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: λ = 632.8 nm
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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