Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 100 km/h brakes with a constant deceleration of 5 m/s² until it stops. Find the braking distance s.
A ball is thrown straight up at 16 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 15 kg block slides down a slope at 45° to the horizontal. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A block is given a push so that it starts sliding up a ramp at 20° with a speed of 4 m/s. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is launched at 25° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A ball is launched at 50° above horizontal from the top of a 20 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A 2 kg block slides 4 m down a slope at 25° to the horizontal, with a coefficient of kinetic friction of 0.15. Find the work Wf done against friction.
A 2.5 kg block is released from rest at height 4 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3.5 m long (μ = 0.1) and then runs onto a spring (k = 2000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 10 g bullet is fired horizontally into a 3.5 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 21 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 15 g bullet is fired horizontally into a 1 kg block hanging from light strings 0.6 m long, and stays in it. The strings swing out to a largest angle of 15° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A vertical rectangular gate 1 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 1.5 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.
A pitot-static tube in a wind tunnel reads a difference of 950 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.
A horizontal venturi meter (inlet 200 mm, throat 100 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 240 mm. The discharge coefficient is 0.96. Find the flow rate Q. Use ρ = 1000 kg/m³ for water, 13 600 kg/m³ for mercury and g = 9.81 m/s².
What volume V does 230 g of He occupy at 550 kPa and 15 °C? Treat it as an ideal gas.
Find the density ρ of He at 500 kPa and 20 °C, treating it as an ideal gas.
A heat engine works between a hot reservoir at 800 °C and a cold reservoir at 35 °C. Find the highest possible thermal efficiency ηmax.
A heat engine takes in 1200 kJ of heat from a reservoir at 600 °C, produces 636 kJ of work and rejects the rest to a reservoir at 20 °C, per cycle. Find the total entropy generated Sgen per cycle.
A small source sends sound equally in all directions. The level is 80 dB at 1 m from it. Find the level β2 at 8 m, ignoring absorption and reflections.
Two slits 0.1 mm apart are lit with monochromatic light. On a screen 1.2 m away, bright fringe number 4 lies 28.27 mm from the central fringe. Find the wavelength λ.
Answer: s = 77.16 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: hmax = 18.05 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 = 6.243 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: d = 1.87 m
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.
Both forces point down the slope; the mass cancels.
Answer: x = 112.8 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.
Starting and landing at the same height: t = 2vy/g.
Answer: t = 4.761 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: Wf = 10.67 J
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: x = 29.92 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 = 712.5 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.
Answer: v = 42.86 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 = 40.47 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 = 197 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 = 39.79 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: Q = 59.98 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.
Continuity and Bernoulli together give the ideal throat velocity.
Answer: V = 250.3 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 0.8212 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηmax = 71.29 %
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: Sgen = 0.5496 kJ/K
Entropy generation: over a full cycle the working fluid has no net entropy change. The hot reservoir loses QH/TH and the cold reservoir gains QL/TL. Their sum is the entropy generated, which is zero for a reversible cycle and positive for every real one.
Sgen > 0, so the engine is possible but irreversible.
Answer: β2 = 61.94 dB
Waves: a periodic wave moves one wavelength in one period, so v = fλ with T = 1/f. On a stretched string v = √(T/μ), with T the tension and μ = m/L the mass per metre. Sound in air travels at about 343 m/s at 20 °C, light in vacuum at c = 3.00 × 10⁸ m/s. Loudness is measured on a logarithmic scale: β = 10·log₁₀(I/I₀) dB with I₀ = 10⁻¹² W/m², so 10 times the intensity adds 10 dB. A point source spreads its power P over a sphere: I = P/(4πr²).
Every doubling of the distance lowers the level by about 6 dB.
Answer: λ = 589 nm
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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