Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 54 km/h sees an obstacle. Her reaction time is 1.1 s, after which the car brakes with a constant deceleration of 4.5 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 20 m/s² for 2.5 s, after which the engine stops and the rocket moves freely. Ignore air resistance and use g = 9.81 m/s². How long after launch does it hit the ground? Find t.
Masses of 9.5 kg and 4.5 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 5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 5.5 kg mass. The coefficient of kinetic friction between block and table is 0.1. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched from level ground at 45° above the horizontal and lands 55 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 24 m/s and 50° above the horizontal from the top of a 60 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 block slides from rest down a slope at 45° to the horizontal. The coefficient of kinetic friction is 0.1. Find its speed v after it has slid 18 m along the slope.
A 20 kg block starts from rest at the bottom of a 35° slope and is pulled up the slope by a constant force of 215 N parallel to the slope. The coefficient of kinetic friction is 0.25. Find the speed v of the block after it has moved 7 m up the slope.
A 0.5 kg glider moving at 8 m/s makes a perfectly elastic head-on collision with a 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 0.45 kg ball hits a wall at 11 m/s and bounces straight back at 18 m/s. The contact lasts 25 ms. Find the magnitude of the impulse J the wall gives the ball.
A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of oil, with its top edge 2.5 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 150 mm and a throat diameter of 75 mm. The pressure at the inlet is 59 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
Water flows up through a pipe that narrows from 100 mm to 25 mm while rising 3 m. At section 1 the pressure is 400 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 50 L contains 290 g of CO₂ at 30 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of Ar goes from 200 kPa, 2 L, 25 °C to 550 kPa and 150 °C. Find the new volume V2.
A heat engine works between a hot reservoir at 425 °C and a cold reservoir at 35 °C. It takes in 1500 kJ of heat. Find the most work Wmax it could possibly produce.
A heat pump delivers 6 kW of heat to a heating system at 21 °C, taking heat from a source at 0 °C. Its actual COP is 55% of the Carnot COP between these temperatures. Find the electrical power Ẇ it uses.
A string fixed at both ends is 0.5 m long; waves travel on it at 100 m/s. Find the frequency f3 of its harmonic number 3.
Light of wavelength 532 nm falls on two slits 0.15 mm apart. The screen is 1.2 m away. Find the distance y4 from the central bright fringe to bright fringe number 4.
Answer: d = 41.5 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 = 13.82 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.
Solve −h₁ = v₁·t₂ − ½·g·t₂² for the positive root.
Answer: T = 59.91 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: a = 4.671 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: v0 = 23.23 m/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.
Answer: H = 77.23 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: v = 14.99 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).
The block drops a height d·sin θ while friction removes μmg·cos θ·d. The mass cancels.
Answer: v = 6.603 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 = −6.545 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: J = 13.05 N·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 = 196.2 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 = 3.889 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 = 49.56 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: p2 = 338.7 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: P = 332.2 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 1.032 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Wmax = 837.9 kJ
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 = 0.7788 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 = 300 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: y4 = 17.02 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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