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 5.5 m/s² until it stops. Find the braking distance s.
A ball is thrown straight up at 10 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.
Masses of 8 kg and 5.5 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 acceleration a of the masses.
A 1.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 4 kg mass. The coefficient of kinetic friction between block and table is 0.1. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is kicked from level ground at 36 m/s and 60° above the horizontal towards a wall 34 m away. Ignore air resistance and use g = 9.81 m/s². At what height y does the ball pass the wall?
A ball is launched at 55° 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.
In a fairground rotor, riders stand against the wall of a vertical cylinder of radius 3.6 m. When it spins fast enough the floor drops away and friction holds them up. The coefficient of static friction is 0.6. Use g = 9.81 m/s². Find the smallest rotation rate n in revolutions per minute.
A 1.5 kg cart is released from rest at a height of 17.1 m on a frictionless track and runs through a vertical loop of radius 4.5 m. Use g = 9.81 m/s². Find the normal force Nbottom of the track on the cart at the bottom of the loop.
A 0.5 kg block slides 3 m down a slope at 35° to the horizontal, with a coefficient of kinetic friction of 0.25. Find the work Wf done against friction.
A 5 kg block rests on a rough horizontal table (μ = 0.25). A light string runs from it over a smooth pulley at the edge of the table to a hanging 2 kg block. The system is released from rest. Find the speed v of the blocks after the hanging block has dropped 2.5 m.
A 0.5 kg glider moving at 9 m/s makes a perfectly elastic head-on collision with a 4 kg glider at rest on a frictionless track. Find the velocity v′2 of the second glider after the collision.
A 2000 kg car runs into a parked 1500 kg car. The wrecks lock together and skid 15 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 shaft transmits 101 kW at 900 rpm. Find the torque T in the shaft.
A person stands on a frictionless turntable and holds a 1.5 kg dumbbell in each hand, 0.7 m from the axis. The person and turntable alone have a moment of inertia of 3 kg·m² about the axis. They turn at 40 rpm, then pull the dumbbells in to 0.35 m from the axis. Treat the dumbbells as point masses. Find the new speed n2 in rpm.
An 80 mm pulley at 1500 rpm drives a 400 mm pulley by a belt with 2.5 % slip. Find the speed n2 of the driven pulley.
In a planetary gear set the sun gear has 20 teeth and each planet gear has 20 teeth. The ring gear is held fixed and the sun is driven at 1440 rpm. Find the speed nc of the planet carrier.
An astronaut on another planet finds that a simple pendulum 1.2 m long has a period of 1.38 s for small swings. Find the gravitational acceleration g on that planet.
A body oscillates in simple harmonic motion with period 3.1 s and amplitude 20 cm. What is the shortest time t it needs to move from its maximum displacement of 20 cm to a displacement of 12 cm?
In a free-vibration test of a damped mass–spring system, a displacement peak of 12 mm is followed 4 cycles later by a peak of 2.6 mm. Find the damping ratio ζ.
A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 15 kN/m and k₂ = 15 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 = 44.89 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 = 3.281 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: a = 1.817 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.
The net driving force is the difference of the weights; it moves the total mass.
Answer: a = 6.867 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: y = 41.39 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.
The ball is still rising at the wall.
Answer: t = 4.391 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: n = 20.35 rpm
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: Nbottom = 126.5 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: Wf = 3.013 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: v = 2.292 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).
Answer: v′2 = 2 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 = 26.85 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: T = 1072 N·m
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: n2 = 53.1 rpm
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.
The kinetic energy increases: the extra energy comes from the work the person does pulling the dumbbells in.
Answer: n2 = 292.5 rpm
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Answer: nc = 360 rpm
Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (ns − nc)/(nr − nc) = −Nr/Ns: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.
Set nr = 0 and solve for the carrier speed.
The carrier turns in the same direction as the sun.
Answer: g = 24.88 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: t = 0.4575 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 = 0.06074
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 = 2.757 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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