Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 60 km/h brakes with a constant deceleration of 7 m/s² until it stops. Find the braking distance s.
A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 28 m/s² for 2 s, after which the engine stops and the rocket moves freely. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax the rocket reaches.
A 40 kg box is pushed up a ramp at 40° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.15. Use g = 9.81 m/s². Find the pushing force F.
A 1 kg block on a ramp at 20° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 6 kg mass. The hanging mass moves down. The coefficient of kinetic friction between block and ramp is 0.15. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched at 30° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is launched at 40° above horizontal from the top of a 15 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A curve of radius 180 m is banked at 8°. At what speed v can a car take it with no friction needed at all? Use g = 9.81 m/s².
A 1.3 kg ball on a string 1.5 m long is swung in a vertical circle. At the lowest point it moves at 6.5 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.
A toy launcher has a vertical spring of stiffness 1000 N/m. A 0.05 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 1 kg block is released from rest at height 1.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 1.5 m long (μ = 0.25) and then runs onto a spring (k = 500 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 1 kg glider moving at 2 m/s makes a perfectly elastic head-on collision with a 3.5 kg glider at rest on a frictionless track. Find the velocity v′1 of the first glider after the collision (positive in its original direction).
At a crossing a 1400 kg car travelling east at 25 m/s collides with a 1200 kg car travelling north at 8 m/s. The cars lock together. Find the direction θ of the wreck just after the collision, measured north of east.
An 8 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 7 kg and radius 0.4 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the speed v of the block after it has fallen 4.5 m.
A solid sphere rolls without slipping down a slope inclined at 30° to the horizontal. Use g = 9.81 m/s². Find the acceleration a of its centre.
An 80 mm pulley at 1500 rpm drives a 450 mm pulley by a belt with 3 % slip. Find the speed n2 of the driven pulley.
A 3 kW motor at 1450 rpm drives gear A (24 teeth), which meshes with gear B (50 teeth). Gear C (16 teeth) on the same shaft as B drives gear D (32 teeth) on the output shaft. Each gear mesh has an efficiency of 95%. Find the output torque Tout.
A simple pendulum has a length of 1.7 m. Use g = 9.81 m/s². Find its period T for small swings.
A 3.1 kg mass on a spring with k = 410 N/m is pulled 18 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 12 mm is followed one cycle later by a peak of 5.6 mm. Find the damping ratio ζ.
A 20 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 10 kN/m and the lower one k₂ = 100 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
Answer: s = 19.84 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: hmax = 215.8 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.
At the top the velocity is zero.
Answer: F = 297.3 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.
Constant speed: the forces balance.
Answer: a = 7.732 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 = 11.47 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: t = 4.827 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: v = 15.75 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 = 49.37 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: h = 10.19 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: x = 21.01 cm
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 ramp is smooth, so only the strip removes energy.
Answer: v′1 = −1.111 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.
The minus sign means the first glider bounces back.
Answer: θ = 15.34 °
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 = 7.837 m/s
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: a = 3.504 m/s²
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 = 258.7 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: Tout = 74.3 N·m
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: T = 2.616 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: E = 6.642 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 = 0.1204
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 = 3.393 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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