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 30 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 5.5 m/s². How long does it take the police car to catch up? Find t.
A stone is dropped from rest into a well. The splash is heard 2.1 s after the stone is released. The speed of sound is 343 m/s; ignore air resistance and use g = 9.81 m/s². Find the depth d of the well.
A 10 kg block slides down a slope at 35° to the horizontal. The coefficient of kinetic friction is 0.1. 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 90 N pulls the front 7 kg block; the 12 kg block is towed behind. The coefficient of kinetic friction is 0.3 for both blocks. Use g = 9.81 m/s². Find the tension T in the connecting string.
A ball is launched at 25° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before 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 0.2 kg ball hangs on a light string of length 3 m. It is pulled aside until the string makes 20° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A 10 kg block starts from rest at the bottom of a 20° slope and is pulled up the slope by a constant force of 60 N parallel to the slope. The coefficient of kinetic friction is 0.15. Find the speed v of the block after it has moved 4 m up the slope.
A 20 kg projectile moving horizontally at 37 m/s explodes into two fragments that move along the same line. The 14 kg fragment continues forwards at 67 m/s. Find the velocity v′2 of the 6 kg fragment (positive in the original direction).
A 10 g bullet is fired horizontally into a 4 kg block hanging from light strings 1.2 m long, and stays in it. The strings swing out to a largest angle of 20° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A vertical rectangular gate 1 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 oil, with its top edge 0.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
A pitot-static tube in a wind tunnel reads a difference of 500 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 50 mm while rising 1 m. At section 1 the pressure is 370 kPa and the velocity is 2.75 m/s. Ignoring losses, find the pressure p2 at section 2.
What volume V does 230 g of CO₂ occupy at 450 kPa and 15 °C? Treat it as an ideal gas.
Find the density ρ of Ar at 900 kPa and 60 °C, treating it as an ideal gas.
A reversible (Carnot) heat engine takes in 3400 kJ of heat from a reservoir at 475 °C and rejects 1400 kJ per cycle. Find the temperature TL of the cold reservoir in °C.
A heat pump keeps a house at 21 °C while it is 7 °C outside, delivering 4 kW of heat to the house. Find the highest possible coefficient of performance COPmax.
A string fixed at both ends is 0.75 m long; waves travel on it at 240 m/s. Find the frequency f2 of its harmonic number 2.
Light of wavelength 600 nm falls at right angles on a grating with 100 lines per millimetre. Find the angle θ2 of the order-2 maximum.
Answer: t = 10.91 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 = 60 m/s.
Answer: d = 20.42 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.
Only the positive root is physical.
Answer: a = 4.823 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 = 56.84 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: x = 70.28 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.
Starting and landing at the same height: t = 2vy/g.
Answer: t = 5.155 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: T = 2.199 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.
Check: T = mg(3 − 2cos θ) = 2.199 N.
Answer: v = 3.177 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 = −33 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 = 477.8 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 = 91.97 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: v = 28.87 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 = 344.8 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: V = 27.82 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 12.98 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: TL = 34.91 °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: COPmax = 21.01
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: f2 = 320 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 = 6.892 °
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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