Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 90 km/h brakes with a constant deceleration of 6.5 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 35 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
A 9 kg block slides down a slope at 20° to the horizontal. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 2.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 4.5 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 at 60° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A ball is launched at 15° above horizontal from the top of a 55 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A car drives over the top of a hump in the road, which is a circular arc of radius 115 m. Use g = 9.81 m/s². Above what speed vmax do the wheels leave the road at the top?
A 0.2 kg ball on a light string 0.6 m long is swung in a vertical circle. At the lowest point its speed is 7.5 m/s. Ignore air resistance and use g = 9.81 m/s². Find the tension Ttop in the string at the highest point.
A block slides from rest down a slope at 30° to the horizontal. The coefficient of kinetic friction is 0.1. Find its speed v after it has slid 15 m along the slope.
A 3 kg block rests on a rough horizontal table (μ = 0.15). 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 m.
A 1000 kg car moving at 20 m/s runs into a stationary 500 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?
At a crossing a 1300 kg car travelling east at 14 m/s collides with a 1900 kg car travelling north at 8 m/s. The cars lock together. Find the speed v′ of the wreck just after the collision.
A motor delivers a torque of 910 N·m at 1700 rpm. Find its power output P in kW.
A solid sphere of mass 12.5 kg rolls without slipping at 6.5 m/s. Find its total kinetic energy Ek.
A belt runs on an 80 mm driving pulley at 1440 rpm. The tension is 600 N on the tight side and 300 N on the slack side. Find the power P transmitted, in kW.
An open belt connects a 150 mm pulley and a 300 mm pulley whose centres are 800 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
A 0.8 kg mass on a spring with k = 80 N/m oscillates on a frictionless surface with an amplitude of 13 cm. Find its speed v when it is 9 cm from equilibrium.
A 2.4 kg mass is attached to the lower end of an unstretched vertical spring with k = 370 N/m and released from rest. Use g = 9.81 m/s². Find the greatest speed vmax of the mass.
In a free-vibration test of a damped mass–spring system, a displacement peak of 15 mm is followed 2 cycles later by a peak of 7 mm. Find the damping ratio ζ.
A 25 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 = 80 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
Answer: s = 48.08 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 = 40.1 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: a = 0.5897 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 = 18.92 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: x = 141.2 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.
Starting and landing at the same height: t = 2vy/g.
Answer: x = 42.66 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: vmax = 33.59 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: Ttop = 8.94 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.
Kinetic energy is traded for potential energy over a height 2r.
Tension and gravity both point down, towards the centre.
Answer: v = 11.03 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 = 3.488 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: ΔK = 66667 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′ = 7.41 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: P = 162 kW
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: Ek = 369.7 J
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: 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: θ = 169.2 °
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: v = 0.9381 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: vmax = 0.7901 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.
The speed is largest at equilibrium.
Answer: ζ = 0.06054
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 = 3.001 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.
ExamProblems · examproblems.com/courses/dynamics