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 4 m/s² until it stops. Find the braking distance s.
A stone is dropped from rest into a well. The splash is heard 2 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 box slides across a level floor with an initial speed of 4.5 m/s and comes to rest after 6.9 m. Use g = 9.81 m/s². Find the coefficient of kinetic friction μ.
An 8.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 5 kg mass. The coefficient of kinetic friction between block and table is 0.3. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched from level ground at 70° above the horizontal and lands 95 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 15° above horizontal from the top of a 25 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A block slides from rest down a slope at 25° to the horizontal. The coefficient of kinetic friction is 0.1. Find its speed v after it has slid 15 m along the slope.
A 0.5 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 m long (μ = 0.25) and then runs onto a spring (k = 1500 N/m) on a smooth surface. Find the maximum compression x of the spring.
An 8 g bullet is fired horizontally into a 1 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 23 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 0.4 kg ball hits a wall at 9 m/s and bounces straight back at 14 m/s. The contact lasts 20 ms. Find the average force Favg of the wall on the ball.
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of water, with its top edge 3 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 water, with its top edge 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
A pitot-static tube in a wind tunnel reads a difference of 1300 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.
A hose with an inner diameter of 25 mm siphons water from an open tank. The highest point C of the hose is 2.5 m above the water surface, and the outlet (open to the air) is 4 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 90 g of CO₂ at 0 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of N₂ at 650 kPa and 60 °C, treating it as an ideal gas.
A reversible refrigerator keeps a cold room at 2 °C by removing 3 kW of heat, and rejects heat to the surroundings at 35 °C. Find the rate Q̇H at which it rejects heat.
A heat pump delivers 4 kW of heat to a heating system at 22 °C, taking heat from a source at 10 °C. Its actual COP is 30% of the Carnot COP between these temperatures. Find the electrical power Ẇ it uses.
A string fixed at both ends is 1 m long; waves travel on it at 100 m/s. Find the frequency f5 of its harmonic number 5.
Two thin lenses share an axis, 20 cm apart: lens 1 with f₁ = 8 cm and lens 2 with f₂ = −15 cm. An object stands 50 cm in front of lens 1. Find the position di2 of the final image measured from lens 2 (positive beyond lens 2, real-is-positive convention).
Answer: s = 28.13 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: d = 18.57 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: μ = 0.1496
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 = 40.15 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: v0 = 38.08 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: x = 92.26 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: v = 9.885 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: x = 15.66 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 = 267.7 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: Favg = 460 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 = 51.5 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 = 44.15 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: v = 46.55 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: 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 = 232.2 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 6.573 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: Q̇H = 3.36 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.
Power input: Ẇ = 0.3598 kW.
Answer: Ẇ = 0.5421 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: f5 = 250 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: di2 = −6.168 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.
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