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 18 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 6 m/s². How far from its starting point does the police car catch up? Find d.
A tram starts from rest and accelerates at 3 m/s² for 11 s, then runs at constant speed for 27 s, and finally brakes at 2.5 m/s² until it stops. Find the total distance s between the two stops.
An 11 kg block slides down a slope at 45° to the horizontal. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 60 N pulls the front 9 kg block; the 12 kg block is towed behind. The coefficient of kinetic friction is 0.15 for both blocks. Use g = 9.81 m/s². Find the tension T in the connecting string.
A ball is launched at 40° above horizontal from level ground. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A ball is thrown from ground level at 31 m/s and 40° above the horizontal. It lands on the flat roof of a building 12 m high, on its way down. Ignore air resistance and use g = 9.81 m/s². Find the horizontal distance x from the launch point to where it lands.
A curve of radius 210 m is banked at 12°. At what speed v can a car take it with no friction needed at all? Use g = 9.81 m/s².
A 0.8 kg ball on a light string 0.5 m long is swung in a vertical circle. At the lowest point its speed is 8 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 45° to the horizontal. The coefficient of kinetic friction is 0.1. Find its speed v after it has slid 5 m along the slope.
A 1 kg cart is released from rest at a height of 7 m on a smooth track that leads into a vertical circular loop of radius 2 m. Find the normal force N the track exerts on the cart at the top of the loop.
A 20 kg projectile moving horizontally at 6 m/s explodes into two fragments that move along the same line. The 10 kg fragment continues forwards at 11 m/s. Find the velocity v′2 of the 10 kg fragment (positive in the original direction).
An 8 g bullet is fired horizontally into a 1 kg block hanging from light strings 1.7 m long, and stays in it. The strings swing out to a largest angle of 40° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A flywheel with a moment of inertia of 40 kg·m² spins at 1000 rpm. A brake applies a constant friction torque of 155 N·m. How many revolutions N does the flywheel make before it stops?
A person stands on a frictionless turntable and holds a 6 kg dumbbell in each hand, 0.85 m from the axis. The person and turntable alone have a moment of inertia of 4.5 kg·m² about the axis. They turn at 55 rpm, then pull the dumbbells in to 0.2 m from the axis. Treat the dumbbells as point masses. Find the new speed n2 in rpm.
A 125 mm pulley at 1500 rpm drives a 500 mm pulley by a belt with 2.5 % slip. Find the speed n2 of the driven pulley.
In a compound gear train, gear A (15 teeth) on the input shaft turning at 3000 rpm drives gear B (24 teeth). Gear C (48 teeth) is fixed to the same shaft as B and drives gear D (72 teeth) on the output shaft. Find the output speed nout.
How long must a simple pendulum be to have a period of 2.7 s? Use g = 9.81 m/s². Find L.
A 15 g bullet moving at 200 m/s hits and stays in a 1.5 kg block resting on a frictionless surface. The block is attached to a horizontal spring with k = 800 N/m. Find the amplitude A of the resulting oscillation in cm.
A 5 kg mass is supported by a spring with a stiffness of 12 kN/m and a viscous damper with a damping coefficient of 200 N·s/m. Find the damping ratio ζ.
A 25 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 5 kN/m and the lower one k₂ = 100 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
Answer: d = 108 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 that moment the police car moves at 2v = 36 m/s.
Answer: s = 1290 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 = 6.243 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 = 34.29 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.
With the same μ on both blocks, friction drops out: T = F·m₁/(m₁ + m₂).
Answer: t = 4.98 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.
Starting and landing at the same height: t = 2vy/g.
Answer: x = 79.01 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.
Larger root: the ball is descending.
Answer: v = 20.93 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 = 63.16 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 = 7.901 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: N = 19.62 N
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). On a circular path the net force towards the centre must equal mv²/r.
The cart just makes it round when h = 2.5R; then N = 0 at the top.
Answer: v′2 = 1 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 = 352 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 = 225.2 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: n2 = 145.5 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: n2 = 365.6 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: nout = 1250 rpm
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.
total ratio = 2.4 : 1
Answer: L = 1.811 m
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: A = 8.617 cm
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 = 0.4082
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: keq = 4.762 kN/m
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