Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 45 km/h sees an obstacle. Her reaction time is 0.9 s, after which the car brakes with a constant deceleration of 6 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A tram starts from rest and accelerates at 1 m/s² for 4 s, then runs at constant speed for 5 s, and finally brakes at 4.5 m/s² until it stops. Find the total distance s between the two stops.
A 15 kg sled is pulled across level snow by a rope at 15° above the horizontal with a force of 150 N. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the sled.
A 1 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 2 kg mass. The coefficient of kinetic friction between block and table is 0.2. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched at 35° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A ball is launched at 55° 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.
In a fairground rotor, riders stand against the wall of a vertical cylinder of radius 3.3 m. When it spins fast enough the floor drops away and friction holds them up. The coefficient of static friction is 0.5. Use g = 9.81 m/s². Find the smallest angular speed ω in rad/s.
A ball on a string 0.8 m long is swung in a vertical circle. Use g = 9.81 m/s². What is the smallest speed vmin at the top for which the string just stays tight?
A block slides from rest down a slope at 30° to the horizontal. The coefficient of kinetic friction is 0.15. Find its speed v after it has slid 3 m along the slope.
A 0.5 kg block is released from rest at height 2.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3 m long (μ = 0.2) and then runs onto a spring (k = 1000 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.
A 5 kg glider moving at 8 m/s makes a perfectly elastic head-on collision with a 4.5 kg glider at rest on a frictionless track. Find the velocity v′1 of the first glider after the collision (positive in its original direction).
A 1200 kg car runs into a parked 1400 kg car. The wrecks lock together and skid 4 m before stopping. The coefficient of kinetic friction between tyres and road is 0.75. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A uniform slender rod of mass 8.5 kg and length 2.2 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 person stands on a frictionless turntable and holds a 2.5 kg dumbbell in each hand, 0.9 m from the axis. The person and turntable alone have a moment of inertia of 2 kg·m² about the axis. They turn at 35 rpm, then pull the dumbbells in to 0.3 m from the axis. Treat the dumbbells as point masses. Find the new speed n2 in rpm.
A 5.5 kW motor running at 1500 rpm drives a pinion with 16 teeth, which meshes with a gear with 80 teeth. The gear pair has an efficiency of 96%. Find the torque T1 on the motor shaft.
In a compound gear train, gear A (40 teeth) on the input shaft turning at 1440 rpm drives gear B (48 teeth). Gear C (36 teeth) is fixed to the same shaft as B and drives gear D (48 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.
A 1.9 kg mass on a spring with k = 480 N/m is pulled 17 cm from equilibrium and released. Find its maximum acceleration amax.
A 50 kg mass on a spring with a stiffness of 100 kN/m has a viscous damper giving a damping ratio of 0.5. Find the damped natural frequency fd of its free vibration in Hz.
A 5 kg mass hangs from spring k₃ = 40 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 12 kN/m and k₂ = 20 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: d = 24.27 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: s = 29.78 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 = 7.493 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.
Answer: a = 5.886 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.
Answer: x = 55.17 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.
Starting and landing at the same height: t = 2vy/g.
Answer: t = 4.918 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: ω = 2.438 rad/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: vmin = 2.801 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: v = 4.667 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 = 6.106 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 ramp is smooth, so only the strip removes energy.
Answer: v′1 = 0.4211 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.
Solving momentum and energy conservation together gives these two results.
Answer: v1 = 16.62 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 = 6.689 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: n2 = 86.43 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: T1 = 35.01 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 = 900 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 = 1.6 : 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: amax = 42.95 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: fd = 6.164 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 = 9.49 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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