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 200 km/h to take off and the usable runway is 1200 m long. Starting from rest, what constant acceleration a does it need?
A stone is dropped from rest into a well. The splash is heard 4.5 s after the stone is released. The speed of sound is 343 m/s; ignore air resistance and use g = 9.81 m/s². Find the depth d of the well.
A 9 kg block slides down a slope at 30° 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 3.5 kg block on a ramp at 25° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 5 kg mass. The hanging mass moves down. The coefficient of kinetic friction between block and ramp is 0.2. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched from level ground at 24 m/s and must land 35 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 31 m/s and 55° above the horizontal from the top of a 35 m cliff. Ignore air resistance and use g = 9.81 m/s². Find the speed v with which it hits the ground.
A car takes a flat curve of radius 80 m. The coefficient of static friction between tyres and road is 0.3. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A ball on a string 1 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 35° to the horizontal. The coefficient of kinetic friction is 0.2. Find its speed v after it has slid 12 m along the slope.
A 4 kg block is released from rest at height 3.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 1 m long (μ = 0.1) and then runs onto a spring (k = 2500 N/m) on a smooth surface. Find the maximum compression x of the spring.
An 11 kg projectile moving horizontally at 35 m/s explodes into two fragments that move along the same line. The 8 kg fragment continues forwards at 77 m/s. Find the velocity v′2 of the 3 kg fragment (positive in the original direction).
A 0.1 kg ball hits a wall at 23 m/s and bounces straight back at 9 m/s. The contact lasts 50 ms. Find the magnitude of the impulse J the wall gives the ball.
A flywheel with a moment of inertia of 34 kg·m² spins at 1800 rpm. A brake applies a constant friction torque of 85 N·m. How long t does it take to stop?
A uniform slender rod of mass 7 kg and length 0.6 m hangs at rest from a frictionless pivot at its top end. A 45 g bullet moving horizontally at 240 m/s hits the rod 0.39 m below the pivot and stays embedded. Use g = 9.81 m/s². Find the largest angle φ from the vertical through which the rod swings.
A 200 mm pulley at 960 rpm drives a 500 mm pulley by a belt with 1 % slip. Find the speed n2 of the driven pulley.
An 11 kW motor at 1450 rpm drives gear A (20 teeth), which meshes with gear B (32 teeth). Gear C (30 teeth) on the same shaft as B drives gear D (60 teeth) on the output shaft. Each gear mesh has an efficiency of 97%. Find the output torque Tout.
A simple pendulum has a length of 0.2 m. Use g = 9.81 m/s². Find its period T for small swings.
A body oscillates in simple harmonic motion with period 1.8 s and amplitude 8 cm. What is the shortest time t it needs to move from equilibrium to a displacement of 2.4 cm?
A 40 kg mass on a spring with a stiffness of 25 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 5 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 15 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: a = 1.286 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: d = 88.29 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.
Only the positive root is physical.
Answer: a = 2.781 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: a = 3.331 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: θ = 18.3 °
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 = 71.7°.
Answer: v = 40.59 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.
Energy or v² = v₀² + 2as in the vertical direction.
Check with energy: v² = v₀² + 2gh.
Answer: vmax = 15.34 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: vmin = 3.132 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 = 9.822 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 = 32.67 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′2 = −77 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.
The minus sign means the second fragment flies backwards.
Answer: J = 3.2 N·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 = 75.4 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: φ = 60.28 °
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 pivot force passes through the pivot, so it has no moment there during the impact.
Answer: n2 = 380.2 rpm
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: Tout = 218.1 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: T = 0.8971 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 = 0.08729 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: fd = 3.647 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 = 7.434 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.
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