Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car speeding at a constant 22 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 6 m/s². How far from its starting point does the police car catch up? Find d.
A ball is thrown straight up at 20 m/s from the edge of a roof 5 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
A 19 kg block slides down a frictionless slope at 35° to the horizontal. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 2 kg block on a ramp at 25° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 8.5 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 acceleration a of the system.
A ball is launched from level ground at 38 m/s and must land 135 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 60° above horizontal from the top of a 55 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A block slides from rest down a slope at 25° to the horizontal. The coefficient of kinetic friction is 0.3. Find its speed v after it has slid 20 m along the slope.
A 10 kg block is released from rest at height 1.5 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 = 2000 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.
A 5 kg glider moving at 3 m/s makes a perfectly elastic head-on collision with a 4 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 1800 kg car runs into a parked 1600 kg car. The wrecks lock together and skid 14 m before stopping. The coefficient of kinetic friction between tyres and road is 0.8. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A vertical rectangular gate 1.5 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 3 m wide sits in the wall of a tank of oil, with its top edge 2 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 40 mm. In the wide section the pressure is 200 kPa and the velocity is 3 m/s. Ignoring losses, find the pressure p2 in the narrow section.
A hose with an inner diameter of 32 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 2.5 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the flow rate Q.
What volume V does 270 g of N₂ occupy at 500 kPa and 70 °C? Treat it as an ideal gas.
A fixed amount of CO₂ goes from 150 kPa, 2 L, 10 °C to 550 kPa and 250 °C. Find the new volume V2.
A heat engine works between a hot reservoir at 375 °C and a cold reservoir at 40 °C. Find the highest possible thermal efficiency ηmax.
A Carnot heat engine receives 120 kW of heat from a source at 600 °C and rejects heat to the surroundings at 20 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at 0 °C and also rejects heat to the surroundings at 20 °C. Find the rate Q̇L at which heat is removed from the freezer.
Light travels in flint glass (n = 1.62) towards a flat boundary with water (n = 1.33). Find the critical angle θc beyond which it is totally reflected.
Light of wavelength 589 nm falls at right angles on a grating with 1200 lines per millimetre. Find the angle θ1 of the order-1 maximum.
Answer: d = 161.3 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.
At that moment the police car moves at 2v = 44 m/s.
Answer: hmax = 25.39 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: a = 5.627 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.
Down the slope is positive.
Answer: a = 6.982 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: θ = 33.26 °
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 = 56.74°.
Answer: t = 7.061 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.
Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.
Answer: v = 7.691 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 = 3.962 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 ramp is smooth, so only the strip removes energy.
Answer: v′1 = 0.3333 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: v1 = 28 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 = 58.86 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 = 64.75 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 = 148.9 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: Q = 5.633 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.
Surface and outlet are both open to the air, so the pressure terms cancel.
Answer: V = 55 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 1.008 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: ηmax = 51.69 %
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: Q̇L = 1089 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: θc = 55.18 °
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: θ1 = 44.98 °
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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