Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 80 km/h brakes with a constant deceleration of 7.5 m/s² until it stops. Find the braking distance s.
A ball is thrown straight up at 20 m/s from the edge of a roof 20 m above the ground. Ignore air resistance and use g = 9.81 m/s². How long after the throw does it hit the ground? Find t.
A box slides across a level floor with an initial speed of 8.5 m/s and comes to rest after 25 m. Use g = 9.81 m/s². Find the coefficient of kinetic friction μ.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 20 N pulls the front 2 kg block; the 5 kg block is towed behind. The coefficient of kinetic friction is 0.25 for both blocks. Use g = 9.81 m/s². Find the acceleration a of the blocks.
A ball is launched at 35° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is thrown from ground level at 37 m/s and 65° above the horizontal. It lands on the flat roof of a building 13 m high, on its way down. Ignore air resistance and use g = 9.81 m/s². Find the horizontal distance x from the launch point to where it lands.
A car takes a flat curve of radius 140 m. The coefficient of static friction between tyres and road is 0.3. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A curve of radius 270 m is banked at 12°, and the coefficient of static friction between tyres and road is 0.1. Use g = 9.81 m/s². Find the highest speed vmax at which a car can take the curve without sliding up and out.
A block slides from rest down a slope at 25° to the horizontal. The coefficient of kinetic friction is 0.25. Find its speed v after it has slid 17 m along the slope.
A spring of stiffness 800 N/m at the bottom of a 40° slope is compressed by 15 cm against a 0.5 kg block and released. The coefficient of kinetic friction between block and slope is 0.2. How far d up the slope does the block travel from the release point before it stops? (Ignore the short distance over which the spring pushes, i.e. take friction and gravity over the whole distance d.)
A 16 kg projectile moving horizontally at 35 m/s explodes into two fragments that move along the same line. The 5 kg fragment continues forwards at 67 m/s. Find the velocity v′2 of the 11 kg fragment (positive in the original direction).
A 0.15 kg ball hits a wall at 16 m/s and bounces straight back at 8 m/s. The contact lasts 20 ms. Find the magnitude of the impulse J the wall gives the ball.
A flywheel with a moment of inertia of 20.5 kg·m² spins at 1800 rpm. A brake applies a constant friction torque of 120 N·m. How many revolutions N does the flywheel make before it stops?
A solid cylinder rolls without slipping down a slope inclined at 30° to the horizontal. Use g = 9.81 m/s². It starts from rest. How long t does it take to roll 3.5 m down the slope?
Two parallel shafts are 60.75 mm apart. They must be connected by a pair of spur gears with a module of 1.5 mm and a speed reduction of 3.5 : 1. How many teeth N1 must the pinion have?
In a compound gear train, gear A (30 teeth) on the input shaft turning at 3000 rpm drives gear B (72 teeth). Gear C (18 teeth) is fixed to the same shaft as B and drives gear D (72 teeth) on the output shaft. Find the output speed nout.
How long must a simple pendulum be to have a period of 1.5 s? Use g = 9.81 m/s². Find L.
A 2.3 kg mass on a spring with k = 480 N/m is pulled 12 cm from equilibrium and released. Find its maximum acceleration amax.
An 8 kg mass on a spring with a stiffness of 8 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency fd of its free vibration in Hz.
A 40 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 100 kN/m and the lower one k₂ = 20 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
Answer: s = 32.92 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 = 4.908 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.
Up is positive and the ground is h₀ below the starting point.
Only the positive root is physical.
Answer: μ = 0.1473
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: a = 0.4046 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: H = 15.09 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.
At the top the vertical velocity is zero: vy² = 2gH.
Answer: x = 100.5 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.
Larger root: the ball is descending.
Answer: vmax = 20.3 m/s
Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.
Answer: vmax = 29.08 m/s
Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.
Friction points down the slope.
Answer: v = 8.086 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: d = 2.305 m
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 = 20.45 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: J = 3.6 N·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: N = 483 rev
Rotation about a fixed axis: the sum of the moments about the axis equals I·α (ΣM = I·α), just as ΣF = m·a for translation. In energy form, a torque T turning through an angle θ (in radians) does work T·θ, which changes the kinetic energy ½·I·ω²; gravity adds or removes m·g·Δh of the centre of mass. Power is P = T·ω.
Answer: t = 1.463 s
Rolling and angular momentum: a body rolling without slipping has v = ω·r, so its kinetic energy is ½·m·v² + ½·I·ω² = ½·m·v²·(1 + I/(m·r²)). Down a slope that gives a = g·sin θ/(1 + I/(m·r²)). When no external moment acts about an axis, the angular momentum about that axis stays constant, I₁·ω₁ = I₂·ω₂, even if kinetic energy is lost.
Answer: N1 = 18
Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.
Answer: nout = 312.5 rpm
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
total ratio = 9.6 : 1
Answer: L = 0.5591 m
Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².
Answer: amax = 25.04 m/s²
Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².
Answer: fd = 4.613 Hz
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Answer: fn = 3.249 Hz
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Springs in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
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