Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 120 km/h brakes with a constant deceleration of 4 m/s² until it stops. Find the braking distance s.
A ball is thrown straight up at 20 m/s from the edge of a roof 20 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 20° to the horizontal. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 7.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 3 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 55° 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 35° above horizontal from the top of a 50 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 3.5 m. When it spins fast enough the floor drops away and friction holds them up. The coefficient of static friction is 0.3. Use g = 9.81 m/s². Find the smallest angular speed ω in rad/s.
A 2 kg cart is released from rest at a height of 3.8 m on a frictionless track and runs through a vertical loop of radius 1 m. Use g = 9.81 m/s². Find the normal force Nbottom of the track on the cart at the bottom of the loop.
A 1 kg block slides 9 m down a slope at 20° to the horizontal, with a coefficient of kinetic friction of 0.2. Find the work Wf done against friction.
A 0.5 kg cart is released from rest at a height of 5.3 m on a smooth track that leads into a vertical circular loop of radius 1.5 m. Find the normal force N the track exerts on the cart at the top of the loop.
A 3.5 kg glider moving at 10 m/s makes a perfectly elastic head-on collision with a 2 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 0.4 kg ball hits a wall at 13 m/s and bounces straight back at 15 m/s. The contact lasts 40 ms. Find the average force Favg of the wall on the ball.
A flywheel with a moment of inertia of 14 kg·m² spins at 2700 rpm. A brake applies a constant friction torque of 115 N·m. How long t does it take to stop?
A hollow sphere rolls without slipping down a slope inclined at 30° to the horizontal. Use g = 9.81 m/s². It starts from rest. How long t does it take to roll 3 m down the slope?
A 4 kW motor running at 960 rpm drives a pinion with 20 teeth and a module of 3 mm. Find the tangential force Ft on the pinion teeth.
An open belt connects a 150 mm driving pulley at 1440 rpm to a 400 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 1000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
An astronaut on another planet finds that a simple pendulum 1 m long has a period of 1.26 s for small swings. Find the gravitational acceleration g on that planet.
A 2.1 kg mass on a spring with k = 160 N/m is pulled 14 cm from equilibrium and released. Find its maximum speed vmax.
A 20 kg mass is supported by a spring with a stiffness of 150 kN/m and a viscous damper with a damping coefficient of 280 N·s/m. Find the damping ratio ζ.
A 50 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 5 kN/m and the lower one k₂ = 60 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
Answer: s = 138.9 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: t = 4.908 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 = 0.5897 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 = 1.401 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 = 64.75 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 = 4.042 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 = 3.057 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: Nbottom = 168.7 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 is positive.
Answer: Wf = 16.59 J
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: N = 10.14 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.
The cart just makes it round when h = 2.5R; then N = 0 at the top.
Answer: v′1 = 2.727 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: Favg = 280 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: t = 34.42 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: t = 1.428 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.
Answer: Ft = 1326 N
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: P = 6.181 kW
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 = 24.87 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: vmax = 1.222 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: ζ = 0.08083
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.
ζ < 1, so the system is underdamped and oscillates as it decays.
Answer: keq = 4.615 kN/m
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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