Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A plane needs a speed of 180 km/h to take off and the usable runway is 800 m long. Starting from rest, what constant acceleration a does it need?
Ball A is dropped from rest from a height of 15 m. At the same instant ball B is thrown straight up from the ground directly below at 25 m/s. Ignore air resistance and use g = 9.81 m/s². At what height y above the ground do they meet?
Masses of 6 kg and 4.5 kg hang on either side of a light, frictionless pulley, connected by a light string. The system is released from rest. Use g = 9.81 m/s². Find the tension T in the string.
A block is given a push so that it starts sliding up a ramp at 30° with a speed of 4.5 m/s. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is launched from level ground at 38 m/s and must land 130 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.
A ball is launched at 25° above horizontal from the top of a 30 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A 1700 kg car drives over the top of a hump in the road at 8 m/s. The top of the hump is a circular arc of radius 95 m. Use g = 9.81 m/s². Find the normal force N of the road on the car at the top.
A 2 kg ball on a string 1 m long is swung in a vertical circle. At the lowest point it moves at 7 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.
A car travelling at 30 km/h on a level road brakes hard and skids to a stop with its wheels locked. The coefficient of kinetic friction between the tyres and the road is 0.8. Find the stopping distance d.
A 1 kg cart is released from rest at a height of 13.5 m on a smooth track that leads into a vertical circular loop of radius 3 m. Find the speed v of the cart at the top of the loop.
A 600 kg car moving at 16 m/s runs into a stationary 500 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?
A 0.35 kg ball hits a wall at 26 m/s and bounces straight back at 20 m/s. The contact lasts 50 ms. Find the average force Favg of the wall on the ball.
A uniform slender rod of mass 8 kg and length 1.9 m is pinned at one end and held horizontal. It is released from rest. Use g = 9.81 m/s². Find the vertical force R that the pin exerts on the rod just after release.
A thin hoop rolls without slipping down a slope inclined at 40° to the horizontal. Use g = 9.81 m/s². Find the smallest coefficient of static friction μs that prevents it from slipping.
A belt runs on a 100 mm driving pulley at 720 rpm. The tension is 1200 N on the tight side and 360 N on the slack side. Find the power P transmitted, in kW.
In a compound gear train, gear A (24 teeth) on the input shaft turning at 1440 rpm drives gear B (25 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D (32 teeth) on the output shaft. Find the output speed nout.
An astronaut on another planet finds that a simple pendulum 1.7 m long has a period of 4.25 s for small swings. Find the gravitational acceleration g on that planet.
An 8 g bullet moving at 250 m/s hits and stays in a 2.5 kg block resting on a frictionless surface. The block is attached to a horizontal spring with k = 1200 N/m. Find the amplitude A of the resulting oscillation in cm.
A 12 kg mass is supported by a spring with a stiffness of 10 kN/m and a viscous damper with a damping coefficient of 100 N·s/m. Find the damping ratio ζ.
A 40 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 80 kN/m and the lower one k₂ = 80 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
Answer: a = 1.563 m/s²
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: y = 13.23 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.
Up is positive, the ground is y = 0.
Answer: T = 50.45 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.
The net driving force is the difference of the weights; it moves the total mass.
Answer: d = 1.759 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: θ = 31.01 °
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.
The other solution is 90° − θ = 58.99°.
Answer: t = 4.404 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: N = 15532 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 (down) is positive.
The car feels lighter: the road pushes up less than its weight.
Answer: T = 117.6 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: d = 4.424 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).
Doubling the speed makes the stopping distance four times longer.
Answer: v = 12.13 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). 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: ΔK = 34909 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: Favg = 322 N
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: R = 19.62 N
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.
At the moment of release ω = 0, so there is no centripetal acceleration and the centre accelerates straight down.
R = m·g/4, independent of the length.
Answer: μs = 0.4195
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 = 3.167 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: nout = 691.2 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 = 2.083 : 1
Answer: g = 3.716 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: A = 3.646 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 = 0.1443
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 = 40 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