Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car speeding at a constant 17 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 3 m/s². How long does it take the police car to catch up? Find t.
A ball is thrown straight up at 7 m/s from the edge of a roof 10 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
A 15 kg block slides down a slope at 45° to the horizontal. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 5 kg block on a ramp at 30° 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 0.15. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched from level ground at 50° above the horizontal and lands 135 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 60° above horizontal from the top of a 5 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A small ball on a string of length 1.9 m moves in a horizontal circle so that the string makes a constant angle of 50° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the speed v of the ball.
A ball on a string 0.6 m long is swung in a vertical circle. Use g = 9.81 m/s². What is the smallest speed vmin at the top for which the string just stays tight?
A 1800 kg car drives up a 6° slope at a constant 36 km/h. Air resistance and rolling resistance together amount to 500 N. Find the power P the wheels must deliver, in kW.
A 2 kg block rests on a rough horizontal table (μ = 0.3). 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 1.5 m.
A 15 g bullet is fired horizontally into a 2.5 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 13 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 1900 kg car runs into a parked 1700 kg car. The wrecks lock together and skid 11 m before stopping. The coefficient of kinetic friction between tyres and road is 0.75. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A flywheel with a moment of inertia of 28 kg·m² spins at 1900 rpm. A brake applies a constant friction torque of 75 N·m. How many revolutions N does the flywheel make before it stops?
A uniform slender rod of mass 9 kg and length 1.7 m hangs at rest from a frictionless pivot at its top end. A 35 g bullet moving horizontally at 340 m/s hits the rod 0.93 m below the pivot and stays embedded. Find the angular velocity ω of the rod just after the impact.
A belt runs on a 120 mm driving pulley at 960 rpm. The tension is 1000 N on the tight side and 500 N on the slack side. Find the power P transmitted, in kW.
In a compound gear train, gear A (12 teeth) on the input shaft turns at 3000 rpm and drives gear B (80 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 288 rpm. How many teeth ND must gear D have?
An astronaut on another planet finds that a simple pendulum 1.1 m long has a period of 2.21 s for small swings. Find the gravitational acceleration g on that planet.
A body oscillates in simple harmonic motion with period 0.5 s and amplitude 17 cm. What is the shortest time t it needs to move from its maximum displacement of 17 cm to a displacement of 10.2 cm?
An 8 kg mass on a spring with a stiffness of 4 kN/m has a viscous damper giving a damping ratio of 0.45. Find the damped natural frequency fd of its free vibration in Hz.
A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 20 kN/m and k₂ = 50 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
Answer: t = 11.33 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.
At that moment the police car moves at 2v = 34 m/s.
Answer: hmax = 12.5 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 = 5.203 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 = 47.97 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: v0 = 36.67 m/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.
Answer: t = 4.803 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 = 4.125 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.
Tension and gravity together give the inward force: tan θ = v²/(r·g).
Answer: vmin = 2.426 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: P = 23.46 kW
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). Power is force times speed: P = F·v.
Answer: v = 3.209 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 = 267.8 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: v1 = 24.11 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: N = 1176 rev
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: ω = 1.272 rad/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.
The pivot force passes through the pivot, so it has no moment there during the impact.
Answer: P = 3.016 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: ND = 25
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 = 8.891 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: t = 0.07379 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 = 3.178 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 = 4.211 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
ExamProblems · examproblems.com/courses/dynamics