Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 90 km/h brakes with a constant deceleration of 8 m/s² until it stops. How long does braking take? Find t.
A stone is dropped from rest into a well. The splash is heard 3.8 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.
Masses of 5 kg and 0.5 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 2.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 4.5 kg mass. The coefficient of kinetic friction between block and table is 0.2. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched from level ground at 22 m/s and must land 45 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 35° above horizontal from the top of a 50 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 35° to the horizontal. The coefficient of kinetic friction is 0.1. Find its speed v after it has slid 3 m along the slope.
A 4 kg block is released from rest at height 1.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 4 m long (μ = 0.15) 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 2 kg glider moving at 3 m/s makes a perfectly elastic head-on collision with a 5 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 20 g bullet is fired horizontally into a 2 kg block hanging from light strings 1.9 m long, and stays in it. The strings swing out to a largest angle of 40° 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 0.5 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 glycerin, with its top edge 2 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 40 mm. The water level is kept 5.5 m above the orifice and the discharge coefficient is 0.61. Find the flow rate Q. Use g = 9.81 m/s².
A closed water tank has compressed air above the water at a gauge pressure of 40 kPa. Water leaves through an opening with a diameter of 30 mm, 4 m below the water surface, into the open air. Ignoring losses, find the flow rate Q. Use g = 9.81 m/s².
What volume V does 390 g of O₂ occupy at 600 kPa and 90 °C? Treat it as an ideal gas.
Find the density ρ of N₂ at 1000 kPa and 10 °C, treating it as an ideal gas.
A heat engine works between a hot reservoir at 450 °C and a cold reservoir at 30 °C. It takes in 3200 kJ of heat. Find the most work Wmax it could possibly produce.
A heat pump keeps a house at 21 °C while it is 2 °C outside, delivering 11 kW of heat to the house. Find the minimum power input Ẇmin.
A string fixed at both ends is 1 m long; waves travel on it at 120 m/s. Find the frequency f3 of its harmonic number 3.
Light of wavelength 589 nm passes through a single slit 0.1 mm wide. Find the width w of the central bright maximum on a screen 2 m away (between the first dark fringes on either side).
Answer: t = 3.125 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: d = 64.04 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: T = 8.918 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 = 18.92 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: θ = 32.9 °
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 = 57.1°.
Answer: x = 55.87 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 = 5.38 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 = 4.202 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 = −1.286 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.
The minus sign means the first glider bounces back.
Answer: v = 298.3 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 = 42.92 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 = 153 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 = 7.963 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: Q = 8.899 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.
Answer: V = 61.33 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 11.9 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: Wmax = 1859 kJ
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 = 0.7105 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: f3 = 180 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: w = 23.56 mm
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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