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 240 km/h to take off and the usable runway is 2000 m long. Starting from rest, what constant acceleration a does it need?
A ball is thrown straight up at 8 m/s from the edge of a roof 5 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
A 7 kg block slides down a slope at 40° 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 block is given a push so that it starts sliding up a ramp at 25° with a speed of 7.5 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 kicked from level ground at 18 m/s and 45° above the horizontal towards a wall 23 m away. Ignore air resistance and use g = 9.81 m/s². At what height y does the ball pass the wall?
A ball is launched at 20° 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 150 m. The coefficient of static friction between tyres and road is 0.6. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A 1.6 kg ball on a light string 0.5 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 20° to the horizontal. The coefficient of kinetic friction is 0.2. Find its speed v after it has slid 9 m along the slope.
A 4 kg block rests on a rough horizontal table (μ = 0.1). 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 m.
A 15 kg projectile moving horizontally at 29 m/s explodes into two fragments that move along the same line. The 6 kg fragment continues forwards at 49 m/s. Find the velocity v′2 of the 9 kg fragment (positive in the original direction).
At a crossing a 1400 kg car travelling east at 22 m/s collides with a 1400 kg car travelling north at 24 m/s. The cars lock together. Find the speed v′ of the wreck just after the collision.
A 13.5 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 15.5 kg and radius 0.3 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the angular acceleration α of the pulley.
A hollow sphere starts from rest and rolls without slipping down a slope inclined at 25° to the horizontal. Use g = 9.81 m/s². Its radius is 0.4 m. Find its angular velocity ω after it has rolled 2.5 m along the slope.
A 4 kW motor running at 960 rpm drives a pinion with 25 teeth, which meshes with a gear with 60 teeth. The gear pair has an efficiency of 95%. Find the torque T1 on the motor shaft.
In a compound gear train, gear A (16 teeth) on the input shaft turning at 1440 rpm drives gear B (40 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D (60 teeth) on the output shaft. Find the output speed nout.
A 1.8 kg mass on a spring with k = 190 N/m is pulled 14 cm in the positive direction and released from rest at t = 0. Find its velocity v at t = 0.8 s (with sign).
A body oscillates in simple harmonic motion with period 2.8 s and amplitude 9 cm. What is the shortest time t it needs to move from equilibrium to a displacement of 6.3 cm?
In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed 2 cycles later by a peak of 3.6 mm. Find the damping ratio ζ.
A 10 kg mass hangs from spring k₃ = 10 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 40 kN/m and k₂ = 10 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.
Answer: a = 1.111 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: hmax = 8.262 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 = 4.427 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: d = 5.586 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: y = 6.983 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.
The ball is already on its way down at the wall.
Answer: x = 44.53 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 = 29.71 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 = 241.5 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 = 5.216 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: v = 2.287 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′2 = 15.67 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′ = 16.28 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: α = 20.77 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.
Answer: ω = 8.817 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.
Rolling without slipping: ω = v/r, and friction does no work at the contact point.
Answer: T1 = 39.79 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: nout = 153.6 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 = 9.375 : 1
Answer: v = −1.343 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: t = 0.3455 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 = 0.08103
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 = 4.594 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.
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
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