Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A plane starts from rest and accelerates uniformly at 1.75 m/s² along a runway. Find its speed v after 1300 m.
A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 34 m/s² for 2.5 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.
A 3 kg block slides down a slope at 20° 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.
A 3 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 1 kg mass. The table is frictionless. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched from level ground at 75° above the horizontal and lands 35 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 50° above horizontal from the top of a 25 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A car drives over the top of a hump in the road, which is a circular arc of radius 65 m. Use g = 9.81 m/s². Above what speed vmax do the wheels leave the road at the top?
A 0.9 kg ball on a string 1.3 m long is swung in a vertical circle. At the lowest point it moves at 5.5 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.
A toy launcher has a vertical spring of stiffness 1000 N/m. A 0.5 kg ball is placed on it and the spring is compressed by 5 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 20 kg block starts from rest at the bottom of a 20° slope and is pulled up the slope by a constant force of 135 N parallel to the slope. The coefficient of kinetic friction is 0.15. Find the speed v of the block after it has moved 2 m up the slope.
A 1800 kg car moving at 20 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 0.6 kg ball hits a wall at 12 m/s and bounces straight back at 9 m/s. The contact lasts 25 ms. Find the average force Favg of the wall on the ball.
A 7 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 4 kg and radius 0.1 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the speed v of the block after it has fallen 4 m.
A hollow sphere rolls without slipping down a slope inclined at 10° to the horizontal. Use g = 9.81 m/s². Find the smallest coefficient of static friction μs that prevents it from slipping.
A 3 kW motor running at 2900 rpm drives a pinion with 24 teeth, which meshes with a gear with 36 teeth. The gear pair has an efficiency of 97%. Find the torque T1 on the motor shaft.
In a planetary gear set the sun gear has 12 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).
How long must a simple pendulum be to have a period of 2.6 s? Use g = 9.81 m/s². Find L.
A 2.4 kg mass is attached to the lower end of an unstretched vertical spring with k = 310 N/m and released from rest. Use g = 9.81 m/s². Find the greatest speed vmax of the mass.
A 12 kg mass on a spring with a stiffness of 10 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 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 10 kN/m and k₂ = 40 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: v = 67.45 m/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.
Answer: hmax = 474.5 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.
At the top the velocity is zero.
Answer: a = 2.433 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 = 7.357 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 = 26.2 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 = 6.14 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: vmax = 25.25 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: T = 29.77 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: h = 0.2548 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 = 2.837 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: ΔK = 78261 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 = 504 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: v = 7.813 m/s
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: μs = 0.07053
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: T1 = 9.879 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 = 857.1 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: L = 1.68 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: vmax = 0.8632 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.
The speed is largest at equilibrium.
Answer: fd = 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.
Answer: fn = 3.559 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.
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