Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 108 km/h brakes with a constant deceleration of 5.5 m/s² until it stops. Find the braking distance s.
A stone is dropped from rest into a well. The splash is heard 2.8 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.
A 9 kg block slides down a slope at 40° to the horizontal. The coefficient of kinetic friction is 0.2. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 6 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 0.2. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched from level ground at 15° above the horizontal and lands 85 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 32 m/s and 20° above the horizontal from the top of a 10 m cliff. Ignore air resistance and use g = 9.81 m/s². Find the speed v with which it hits the ground.
A car takes a flat curve of radius 25 m. The coefficient of static friction between tyres and road is 0.6. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A 5 kg cart is released from rest at a height of 40 m on a frictionless track and runs through a vertical loop of radius 10 m. Use g = 9.81 m/s². Find the normal force Ntop of the track on the cart at the top of the loop.
A toy launcher has a vertical spring of stiffness 400 N/m. A 0.2 kg ball is placed on it and the spring is compressed by 10 cm, then released. Ignoring air resistance and the spring’s mass, find the height h the ball rises, measured from its position when the spring was compressed.
A spring of stiffness 5000 N/m at the bottom of a 25° slope is compressed by 12 cm against a 3 kg block and released. The coefficient of kinetic friction between block and slope is 0.3. 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 700 kg car moving at 21 m/s runs into a stationary 1100 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?
A 10 g bullet is fired horizontally into a 5 kg block hanging from light strings 0.7 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 flywheel with a moment of inertia of 32.5 kg·m² spins at 1700 rpm. A brake applies a constant friction torque of 115 N·m. How many revolutions N does the flywheel make before it stops?
A solid sphere rolls without slipping down a slope inclined at 15° to the horizontal. Use g = 9.81 m/s². Find the smallest coefficient of static friction μs that prevents it from slipping.
Two parallel shafts are 105 mm apart. They must be connected by a pair of spur gears with a module of 2 mm and a speed reduction of 2.5 : 1. How many teeth N2 must the gear have?
In a compound gear train, gear A (36 teeth) on the input shaft turns at 1440 rpm and drives gear B (80 teeth). Gear C (20 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 324 rpm. How many teeth ND must gear D have?
How long must a simple pendulum be to have a period of 2.8 s? Use g = 9.81 m/s². Find L.
A 0.5 kg mass on a spring with k = 130 N/m is pulled 17 cm from equilibrium and released. Find its maximum speed vmax.
In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed 4 cycles later by a peak of 11 mm. Find the damping ratio ζ.
A 5 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 25 kN/m and k₂ = 5 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
Answer: s = 81.82 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 = 35.65 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: a = 4.803 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: T = 32.11 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 = 40.84 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: v = 34.93 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.
Energy or v² = v₀² + 2as in the vertical direction.
Check with energy: v² = v₀² + 2gh.
Answer: vmax = 12.13 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: Ntop = 147.2 N
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.
Towards the centre is positive.
Answer: h = 1.019 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: d = 1.761 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: ΔK = 94325 J
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.
The lost energy goes into deformation, heat and sound.
Answer: v = 342.7 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: N = 712.7 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: μs = 0.07656
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: N2 = 75
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: ND = 40
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.
Answer: L = 1.948 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: vmax = 2.741 m/s
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: ζ = 0.03989
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 = 12.33 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
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