Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A plane needs a speed of 220 km/h to take off and the usable runway is 900 m long. Starting from rest, what constant acceleration a does it need?
A tram starts from rest and accelerates at 2 m/s² for 6 s, then runs at constant speed for 17 s, and finally brakes at 3.5 m/s² until it stops. Find the total distance s between the two stops.
Masses of 6.5 kg and 4 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.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 65 N pulls the front 9 kg block; the 7 kg block is towed behind. The coefficient of kinetic friction is 0.15 for both blocks. Use g = 9.81 m/s². Find the tension T in the connecting string.
A ball is launched from level ground at 30 m/s and must land 55 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.
A ball is launched at 15° above horizontal from the top of a 15 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A 0.2 kg ball hangs on a light string of length 0.8 m. It is pulled aside until the string makes 40° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A 2 kg block starts from rest at the bottom of a 30° slope and is pulled up the slope by a constant force of 25 N parallel to the slope. The coefficient of kinetic friction is 0.15. Find the speed v of the block after it has moved 6 m up the slope.
A 4 kg projectile moving horizontally at 31 m/s explodes into two fragments that move along the same line. The 3 kg fragment continues forwards at 63 m/s. Find the velocity v′2 of the 1 kg fragment (positive in the original direction).
A 900 kg car runs into a parked 1000 kg car. The wrecks lock together and skid 18 m before stopping. The coefficient of kinetic friction between tyres and road is 0.6. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 2 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 3 m below the surface. Find the resultant hydrostatic force F on the gate.
Water drains from a large open tank through a sharp-edged orifice with a diameter of 30 mm. The water level is kept 4 m above the orifice and the discharge coefficient is 0.63. Find the flow rate Q. Use g = 9.81 m/s².
Water flows up through a pipe that narrows from 100 mm to 75 mm while rising 7 m. At section 1 the pressure is 350 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 5 L contains 440 g of CO₂ at 70 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of He at 450 kPa and 80 °C, treating it as an ideal gas.
A heat engine receives heat from a source at 750 °C and rejects heat to the surroundings at 35 °C. Its actual thermal efficiency is 35%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
A heat pump delivers 19.5 kW of heat to a heating system at 20 °C, taking heat from a source at -5 °C. Its actual COP is 55% of the Carnot COP between these temperatures. Find the electrical power Ẇ it uses.
An object 2 cm tall stands 50 cm in front of a thin diverging lens with a focal length of 20 cm. Find the image height hi (negative for an inverted image).
A sensor at rest sends ultrasound of 30 kHz towards a car that moves away at 25 m/s and mixes the echo with the emitted signal. Find the beat frequency fb (speed of sound 343 m/s).
Answer: a = 2.075 m/s²
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 = 260.6 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 = 48.58 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 = 28.44 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: θ = 18.42 °
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.
The other solution is 90° − θ = 71.58°.
Answer: x = 74.9 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.
Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.
Answer: T = 2.88 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) = 2.88 N.
Answer: v = 8.709 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 = −65 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: v1 = 30.73 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 = 49.05 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 = 131.3 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: Q = 3.945 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 discharge coefficient accounts for the contraction of the jet and friction at the edge.
Answer: p2 = 279.6 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 = 5705 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 0.6135 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηII = 50.08 %
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 = 3.024 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: hi = 0.5714 cm
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.
Answer: fb = 4076 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|.
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