Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 54 km/h brakes with a constant deceleration of 5.5 m/s² until it stops. How long does braking take? Find t.
Ball A is dropped from rest from a height of 15 m. At the same instant ball B is thrown straight up from the ground directly below at 11 m/s. Ignore air resistance and use g = 9.81 m/s². At what height y above the ground do they meet?
A 12 kg block slides down a frictionless slope at 35° to the horizontal. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A block is given a push so that it starts sliding up a ramp at 30° with a speed of 9 m/s. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is kicked from level ground at 18 m/s and 30° above the horizontal towards a wall 9 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 10 m/s and 60° above the horizontal from the top of a 20 m cliff. Ignore air resistance and use g = 9.81 m/s². Find the greatest height H above the ground at the foot of the cliff.
A car takes a flat curve of radius 120 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 1.3 kg ball on a string 0.6 m long is swung in a vertical circle. At the lowest point it moves at 3 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.
A block slides from rest down a slope at 30° to the horizontal. The coefficient of kinetic friction is 0.35. Find its speed v after it has slid 11 m along the slope.
An 8 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 5 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 13 kg projectile moving horizontally at 12 m/s explodes into two fragments that move along the same line. The 5 kg fragment continues forwards at 24 m/s. Find the velocity v′2 of the 8 kg fragment (positive in the original direction).
At a crossing a 1200 kg car travelling east at 14 m/s collides with a 1600 kg car travelling north at 22 m/s. The cars lock together. Find the speed v′ of the wreck just after the collision.
A uniform slender rod of mass 6 kg and length 1.2 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 of mass 13 kg rolls without slipping at 2.5 m/s. What percentage p of its total kinetic energy is rotational?
A 1.1 kW motor running at 2900 rpm drives a pinion with 24 teeth and a module of 4 mm. Find the tangential force Ft on the pinion teeth.
In a compound gear train, gear A (50 teeth) on the input shaft turns at 960 rpm and drives gear B (80 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 240 rpm. How many teeth ND must gear D have?
An astronaut on another planet finds that a simple pendulum 1.2 m long has a period of 3.57 s for small swings. Find the gravitational acceleration g on that planet.
A 2 kg mass on a spring with k = 180 N/m is pulled 2 cm from equilibrium and released. Find the total energy E of the oscillation.
An 8 kg mass is supported by a spring with a stiffness of 25 kN/m and a viscous damper with a damping coefficient of 360 N·s/m. Find the damping ratio ζ.
A 25 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 = 20 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: t = 2.727 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: y = 5.879 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.
Up is positive, the ground is y = 0.
Answer: a = 5.627 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: d = 5.434 m
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.
Both forces point down the slope; the mass cancels.
Answer: y = 3.561 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: H = 23.82 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.
Answer: vmax = 18.79 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: T = 32.25 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 (up) is positive.
Answer: v = 6.519 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 = 3.364 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 = 4.5 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′ = 13.93 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: p = 50 %
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 share depends only on the shape, not on the mass or the speed.
Answer: Ft = 75.46 N
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 = 30
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: g = 3.717 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 = 0.036 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 = 0.4025
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.
ζ < 1, so the system is underdamped and oscillates as it decays.
Answer: fn = 2.947 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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