Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 50 km/h brakes with a constant deceleration of 3 m/s² until it stops. How long does braking take? Find t.
A ball is thrown straight up at 5 m/s from the edge of a roof 30 m above the ground. Ignore air resistance and use g = 9.81 m/s². How long after the throw does it hit the ground? Find t.
A 10 kg block slides down a slope at 45° to the horizontal. The coefficient of kinetic friction is 0.15. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 70 N pulls the front 8 kg block; the 9 kg block is towed behind. The coefficient of kinetic friction is 0.1 for both blocks. Use g = 9.81 m/s². Find the tension T in the connecting string.
A ball is launched from level ground at 20° above the horizontal and lands 95 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is thrown from ground level at 19 m/s and 70° above the horizontal. It lands on the flat roof of a building 9 m high, on its way down. Ignore air resistance and use g = 9.81 m/s². Find the horizontal distance x from the launch point to where it lands.
A curve of radius 220 m is banked at 12°. At what speed v can a car take it with no friction needed at all? Use g = 9.81 m/s².
A 0.8 kg ball on a string 0.5 m long is swung in a vertical circle. At the lowest point it moves at 7.5 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.
A 10 kg ball hangs on a light string of length 0.8 m. It is pulled aside until the string makes 90° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A 2.5 kg block is released from rest at height 4 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 1.5 m long (μ = 0.15) and then runs onto a spring (k = 2000 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.
A 5 kg glider moving at 11 m/s makes a perfectly elastic head-on collision with a 3.5 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).
An 800 kg car runs into a parked 1000 kg car. The wrecks lock together and skid 12.5 m before stopping. The coefficient of kinetic friction between tyres and road is 0.7. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A flywheel with a moment of inertia of 3 kg·m² spins at 800 rpm. A brake applies a constant friction torque of 15 N·m. How long t does it take to stop?
A thin hoop starts from rest and rolls without slipping down a slope inclined at 10° to the horizontal. Use g = 9.81 m/s². Its radius is 0.05 m. Find its angular velocity ω after it has rolled 1.5 m along the slope.
Two parallel shafts are 162 mm apart. They must be connected by a pair of spur gears with a module of 3 mm and a speed reduction of 3.5 : 1. How many teeth N1 must the pinion have?
An open belt connects a 160 mm pulley and a 400 mm pulley whose centres are 800 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
An astronaut on another planet finds that a simple pendulum 0.6 m long has a period of 2.52 s for small swings. Find the gravitational acceleration g on that planet.
A 3.2 kg mass on a spring with k = 360 N/m is pulled 14 cm from equilibrium and released. Find its maximum acceleration amax.
In a free-vibration test of a damped mass–spring system, a displacement peak of 12 mm is followed 4 cycles later by a peak of 2.6 mm. Find the damping ratio ζ.
A 5 kg mass hangs from spring k₃ = 50 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 25 kN/m and k₂ = 5 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: t = 4.63 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: t = 3.035 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.
Up is positive and the ground is h₀ below the starting point.
Only the positive root is physical.
Answer: a = 5.896 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 = 37.06 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.
With the same μ on both blocks, friction drops out: T = F·m₁/(m₁ + m₂).
Answer: v0 = 38.08 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: x = 19.73 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.
Larger root: the ball is descending.
Answer: v = 21.42 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 = 97.85 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: T = 294.3 N
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.
Check: T = mg(3 − 2cos θ) = 294.3 N.
Answer: v = 8.606 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 = 1.941 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 = 29.48 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: t = 16.76 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: ω = 31.97 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: N1 = 24
Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.
Answer: θ = 162.7 °
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: g = 3.73 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 = 15.75 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: ζ = 0.06074
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 = 9.746 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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