Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 54 km/h brakes with a constant deceleration of 5 m/s² until it stops. Find the braking distance s.
A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 28 m/s² for 1 s, after which the engine stops and the rocket moves freely. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax the rocket reaches.
Masses of 8 kg and 2.5 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 25° with a speed of 5.5 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 28 m/s and must land 30 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 55° above horizontal from the top of a 40 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A car takes a flat curve of radius 65 m. The coefficient of static friction between tyres and road is 0.5. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A 1.2 kg ball on a light string 1.5 m long is swung in a vertical circle. At the lowest point its speed is 10 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 35° to the horizontal. The coefficient of kinetic friction is 0.4. Find its speed v after it has slid 4 m along the slope.
A 10 kg block is released from rest at height 3 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 2 m long (μ = 0.2) and then runs onto a spring (k = 2500 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 10 g bullet is fired horizontally into a 5 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 6 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
At a crossing a 900 kg car travelling east at 12 m/s collides with a 1300 kg car travelling north at 9 m/s. The cars lock together. Find the direction θ of the wreck just after the collision, measured north of east.
A uniform slender rod of mass 8 kg and length 1 m is pinned at one end and held horizontal. It is released from rest. Use g = 9.81 m/s². Find its angular acceleration α just after release.
A solid sphere of mass 16.5 kg rolls without slipping at 1 m/s. Find its total kinetic energy Ek.
A belt runs on a 125 mm driving pulley at 2880 rpm. The tension is 400 N on the tight side and 160 N on the slack side. Find the power P transmitted, in kW.
In a compound gear train, gear A (16 teeth) on the input shaft turning at 1500 rpm drives gear B (32 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D (80 teeth) on the output shaft. Find the output speed nout.
A simple pendulum has a length of 0.4 m. Use g = 9.81 m/s². Find its period T for small swings.
A 1.3 kg mass on a spring with k = 340 N/m is pulled 3 cm from equilibrium and released. Find its maximum speed vmax.
A 5 kg mass is supported by a spring with a stiffness of 8 kN/m and a viscous damper with a damping coefficient of 240 N·s/m. Find the damping ratio ζ.
A 10 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 40 kN/m and the lower one k₂ = 40 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
Answer: s = 22.5 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: hmax = 53.96 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 the top the velocity is zero.
Answer: T = 37.37 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 = 3.004 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 = 11.02 °
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 = 78.98°.
Answer: x = 41.19 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 = 17.86 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 = 21.14 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 = 4.393 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: x = 45.17 cm
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 ramp is smooth, so only the strip removes energy.
Answer: v = 543.6 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: θ = 47.29 °
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: α = 14.71 rad/s²
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.
Answer: Ek = 11.55 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 = 4.524 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 = 140.6 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 = 10.67 : 1
Answer: T = 1.269 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 = 0.4852 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: ζ = 0.6
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: fn = 7.118 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