Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 54 km/h brakes with a constant deceleration of 6.5 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.
Masses of 10 kg and 5 kg hang on either side of a light, frictionless pulley, connected by a light string. The system is released from rest. Use g = 9.81 m/s². Find the tension T in the string.
A 1 kg block on a ramp at 30° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 1 kg mass. The hanging mass moves down. The coefficient of kinetic friction between block and ramp is 0.1. Use g = 9.81 m/s². Find the tension T in the string.
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 23 m/s and 20° above the horizontal from the top of a 50 m cliff. Ignore air resistance and use g = 9.81 m/s². Find the greatest height H above the ground at the foot of the cliff.
A toy launcher has a vertical spring of stiffness 400 N/m. A 0.15 kg ball is placed on it and the spring is compressed by 8 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 starts from rest at the bottom of a 30° slope and is pulled up the slope by a constant force of 115 N parallel to the slope. The coefficient of kinetic friction is 0.25. Find the speed v of the block after it has moved 2 m up the slope.
A 3.5 kg glider moving at 6 m/s makes a perfectly elastic head-on collision with a 2 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 0.15 kg ball hits a wall at 17 m/s and bounces straight back at 19 m/s. The contact lasts 25 ms. Find the average force Favg of the wall on the ball.
A vertical rectangular gate 3 m wide sits in the wall of a tank of water, with its top edge 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 3 m wide sits in the wall of a tank of seawater, with its top edge 1.5 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 75 mm to 50 mm. In the wide section the pressure is 370 kPa and the velocity is 1 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 110 kPa. Water leaves through an opening with a diameter of 40 mm, 4.5 m below the water surface, into the open air. Ignoring losses, find the jet velocity v. Use g = 9.81 m/s².
A tank of 20 L contains 250 g of N₂ at 90 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of He goes from 250 kPa, 10 L, 20 °C to 750 kPa and 180 °C. Find the new volume V2.
A reversible (Carnot) heat engine takes in 2500 kJ of heat from a reservoir at 650 °C and rejects 807 kJ per cycle. Find the temperature TL of the cold reservoir in °C.
A refrigerator keeps its cold space at -18 °C in a kitchen at 28 °C, removing heat from the cold space at 2.5 kW. Find the minimum power input Ẇmin.
A source of 1200 Hz moves at 30 m/s away from a stationary observer. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.
Light of wavelength 589 nm passes through a single slit 0.15 mm wide. Find the width w of the central bright maximum on a screen 1.5 m away (between the first dark fringes on either side).
Answer: s = 17.31 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: T = 65.4 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.
The net driving force is the difference of the weights; it moves the total mass.
Answer: T = 7.782 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.
Answer: t = 2.243 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: H = 53.15 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.
Answer: h = 0.8699 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 = 4.229 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′1 = 1.636 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.
Answer: Favg = 216 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 = 88.29 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 = 2.939 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 = 368 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 = 17.56 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.
Answer: P = 1347 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 5.153 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: TL = 24.84 °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 = 0.4507 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: f′ = 1103 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: w = 11.78 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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