Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A plane starts from rest and accelerates uniformly at 3.25 m/s² along a runway. Find its speed v after 1700 m.
A tram starts from rest and accelerates at 3 m/s² for 4 s, then runs at constant speed for 7 s, and finally brakes at 2 m/s² until it stops. Find the total distance s between the two stops.
Masses of 7.5 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 ramp at 30° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 9.5 kg mass. The hanging mass moves down. The coefficient of kinetic friction between block and ramp is 0.25. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched at 60° 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 19 m/s and 45° above the horizontal from the top of a 30 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 car travelling at 80 km/h on a level road brakes hard and skids to a stop with its wheels locked. The coefficient of kinetic friction between the tyres and the road is 0.6. Find the stopping distance d.
A 2.5 kg block is released from rest at height 4.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3.5 m long (μ = 0.2) and then runs onto a spring (k = 5000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 15 kg projectile moving horizontally at 13 m/s explodes into two fragments that move along the same line. The 9 kg fragment continues forwards at 32 m/s. Find the velocity v′2 of the 6 kg fragment (positive in the original direction).
A 0.5 kg ball hits a wall at 13 m/s and bounces straight back at 15 m/s. The contact lasts 40 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 1 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 1 m wide sits in the wall of a tank of glycerin, with its top edge 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
Water flows through a horizontal pipe that narrows from 75 mm to 50 mm. In the wide section the pressure is 110 kPa and the velocity is 2.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
A hose with an inner diameter of 20 mm siphons water from an open tank. The highest point C of the hose is 1.5 m above the water surface, and the outlet (open to the air) is 5 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the absolute pressure pC at point C. The atmospheric pressure is 101.3 kPa.
A tank of 20 L contains 420 g of N₂ at 100 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of N₂ goes from 300 kPa, 20 L, 20 °C to 600 kPa and 170 °C. Find the new volume V2.
A heat engine receives heat from a source at 350 °C and rejects heat to the surroundings at 35 °C. Its actual thermal efficiency is 25%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
A heat pump keeps a house at 20 °C while it is 5 °C outside, delivering 9 kW of heat to the house. Find the minimum power input Ẇmin.
An observer moves at 5 m/s towards a stationary source of 440 Hz. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.
An optical fibre has a core with n₁ = 1.52 and a cladding with n₂ = 1.47. Find the largest angle θa to the fibre axis at which light from air can enter the end of the fibre and still be guided by total internal reflection.
Answer: v = 105.1 m/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.
Answer: s = 144 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 = 55.18 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 = 4.003 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: t = 4.944 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 = 39.2 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: d = 41.95 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).
Doubling the speed makes the stopping distance four times longer.
Answer: x = 19.31 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 = −15.5 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: Favg = 350 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 = 117.7 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 = 23.18 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 = 97.3 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: pC = 37.53 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.
Surface and outlet are both open to the air, so the pressure terms cancel.
If this pressure falls to the vapour pressure of the water, the siphon stops working.
Answer: P = 2326 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 15.12 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: ηII = 49.46 %
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 = 0.4605 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′ = 446.4 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: θa = 22.75 °
Geometric optics: at a boundary, n₁·sin θ₁ = n₂·sin θ₂ (Snell's law), with angles measured from the normal and n = c/v. Going into a slower medium the ray bends towards the normal. From a denser medium (n₁ > n₂) beyond the critical angle θc = arcsin(n₂/n₁) all light is reflected: total internal reflection. Thin lenses and spherical mirrors obey 1/f = 1/do + 1/di with the magnification m = −di/do. Sign convention: f > 0 for a converging lens and a concave mirror (f = R/2), f < 0 for a diverging lens and a convex mirror; di > 0 for a real image, di < 0 for a virtual one, and m < 0 for an inverted image.
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