Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A plane needs a speed of 260 km/h to take off and the usable runway is 1300 m long. Starting from rest, what constant acceleration a does it need?
A ball is thrown straight up at 5 m/s from the edge of a roof 35 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 20 kg sled is pulled across level snow by a rope at 35° above the horizontal with a force of 140 N. The coefficient of kinetic friction is 0.2. Use g = 9.81 m/s². Find the acceleration a of the sled.
A 3 kg block on a ramp at 35° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 7.5 kg mass. The hanging mass moves down. The coefficient of kinetic friction between block and ramp is 0.25. Use g = 9.81 m/s². Find the tension T in the string.
A ball is kicked from level ground at 38 m/s and 70° above the horizontal towards a wall 66 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 60° above horizontal from the top of a 10 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A small ball on a string of length 1.7 m moves in a horizontal circle so that the string makes a constant angle of 35° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the time T for one revolution.
A 1.5 kg cart is released from rest at a height of 19.6 m on a frictionless track and runs through a vertical loop of radius 7 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 0.5 kg block slides 6 m down a slope at 25° to the horizontal, with a coefficient of kinetic friction of 0.2. Find the work Wf done against friction.
A spring of stiffness 1000 N/m at the bottom of a 40° slope is compressed by 10 cm against a 1 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.)
An 8 g bullet is fired horizontally into a 4.5 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 23 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 20 g bullet is fired horizontally into a 1.5 kg block hanging from light strings 0.6 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 uniform slender rod of mass 6 kg and length 0.6 m is pinned at one end and held horizontal. It is released from rest. Use g = 9.81 m/s². Find the acceleration atip of the free end just after release.
A thin hoop starts from rest and rolls without slipping down a slope inclined at 35° to the horizontal. Use g = 9.81 m/s². Find its speed v after it has rolled 5 m along the slope.
A belt runs on an 80 mm driving pulley at 720 rpm. The tension is 1200 N on the tight side and 600 N on the slack side. Find the power P transmitted, in kW.
In a compound gear train, gear A (24 teeth) on the input shaft turning at 2900 rpm drives gear B (50 teeth). Gear C (48 teeth) is fixed to the same shaft as B and drives gear D (80 teeth) on the output shaft. Find the output speed nout.
An astronaut on another planet finds that a simple pendulum 1.5 m long has a period of 3.99 s for small swings. Find the gravitational acceleration g on that planet.
A 2.7 kg mass on a spring with k = 360 N/m is pulled 4 cm from equilibrium and released. Find the total energy E of the oscillation.
In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed one cycle later by a peak of 31.1 mm. Find the damping ratio ζ.
A 50 kg mass hangs from spring k₃ = 12 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 10 kN/m and k₂ = 10 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.
Answer: a = 2.006 m/s²
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 = 3.229 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 = 4.575 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: T = 37.38 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: y = 54.84 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 already on its way down at the wall.
Answer: x = 30.36 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: T = 2.367 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.
Tension and gravity together give the inward force: tan θ = v²/(r·g).
Answer: Ntop = 8.829 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 = 5.335 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: d = 0.5841 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 = 1197 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 = 126.1 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: atip = 14.71 m/s²
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.
At the moment of release ω = 0, so there is no centripetal acceleration and the centre accelerates straight down.
atip = 1.5·g for any length: the free end falls faster than a dropped stone.
Answer: v = 5.304 m/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.
Rolling without slipping: ω = v/r, and friction does no work at the contact point.
Answer: P = 1.81 kW
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: nout = 835.2 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 = 3.472 : 1
Answer: g = 3.72 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: E = 0.288 J
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 = 0.04002
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 = 1.949 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.
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
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