Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 108 km/h sees an obstacle. Her reaction time is 1 s, after which the car brakes with a constant deceleration of 5.5 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A ball is thrown straight up at 5 m/s from the edge of a roof 40 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
A 20 kg block slides down a slope at 20° to the horizontal. The coefficient of kinetic friction is 0.15. 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 15° with a speed of 5 m/s. The coefficient of kinetic friction is 0.2. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is launched at 65° 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 13 m/s and 25° above the horizontal from the top of a 5 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 30 m. The coefficient of static friction between tyres and road is 0.7. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A curve of radius 280 m is banked at 25°, and 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 a car can take the curve without sliding up and out.
A toy launcher has a vertical spring of stiffness 1500 N/m. A 0.2 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 spring of stiffness 800 N/m at the bottom of a 20° slope is compressed by 12 cm against a 2 kg block and released. The coefficient of kinetic friction between block and slope is 0.2. 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.)
A 13 kg projectile moving horizontally at 23 m/s explodes into two fragments that move along the same line. The 9 kg fragment continues forwards at 76 m/s. Find the velocity v′2 of the 4 kg fragment (positive in the original direction).
A 1500 kg car runs into a parked 1300 kg car. The wrecks lock together and skid 20 m before stopping. The coefficient of kinetic friction between tyres and road is 0.7. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A motor delivers a torque of 1920 N·m at 1000 rpm. Find its power output P in kW.
A person stands on a frictionless turntable and holds a 2.5 kg dumbbell in each hand, 0.7 m from the axis. The person and turntable alone have a moment of inertia of 1.5 kg·m² about the axis. They turn at 60 rpm, then pull the dumbbells in to 0.3 m from the axis. Treat the dumbbells as point masses. Find the new speed n2 in rpm.
A 1.1 kW motor running at 1500 rpm drives a pinion with 12 teeth, which meshes with a gear with 24 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.
In a planetary gear set the sun gear has 15 teeth and each planet gear has 15 teeth. The planet carrier is held fixed and the sun is driven at 3000 rpm. Find the speed nr of the ring gear (magnitude).
A simple pendulum has a length of 1.3 m. Use g = 9.81 m/s². Find its period T for small swings.
A body oscillates in simple harmonic motion with period 0.5 s and amplitude 5 cm. What is the shortest time t it needs to move from equilibrium to a displacement of 2.5 cm?
A 5 kg mass on a spring with a stiffness of 12 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency fd of its free vibration in Hz.
A 40 kg mass hangs from spring k₃ = 50 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 15 kN/m and k₂ = 50 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: d = 111.8 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.
Answer: hmax = 41.27 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 = 1.972 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 = 2.819 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: x = 31.24 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: H = 6.538 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 = 14.35 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: vmax = 49.47 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.
Friction points down the slope.
Answer: h = 3.823 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: d = 0.554 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′2 = −96.25 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.
The minus sign means the second fragment flies backwards.
Answer: v1 = 30.94 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 = 201.1 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: n2 = 121.5 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: T2 = 13.59 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: nr = 1000 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.
With nc = 0 the sun drives the ring through the planets, which act as idlers.
The ring turns opposite to the sun.
Answer: T = 2.287 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: t = 0.04167 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: fd = 7.146 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.
Answer: fn = 4.23 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.
ExamProblems · examproblems.com/courses/dynamics