Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 54 km/h sees an obstacle. Her reaction time is 1 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 ball is thrown straight up at 11 m/s from the edge of a roof 10 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
A 13 kg block slides down a slope at 20° 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 80 N pulls the front 10 kg block; the 1 kg block is towed behind. The coefficient of kinetic friction is 0.3 for both blocks. Use g = 9.81 m/s². Find the acceleration a of the blocks.
A ball is kicked from level ground at 30 m/s and 70° above the horizontal towards a wall 35 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 27 m/s and 15° 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 150 m. The coefficient of static friction between tyres and road is 0.7. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A 0.8 kg ball on a string 0.7 m long is swung in a vertical circle. At the lowest point it moves at 4 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.
A block slides from rest down a slope at 40° to the horizontal. The coefficient of kinetic friction is 0.1. Find its speed v after it has slid 17 m along the slope.
A 5 kg cart is released from rest at a height of 4 m on a smooth track that leads into a vertical circular loop of radius 1 m. Find the normal force N the track exerts on the cart at the top of the loop.
A 1300 kg car moving at 24 m/s runs into a stationary 2000 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?
A 0.6 kg ball hits a wall at 30 m/s and bounces straight back at 14 m/s. The contact lasts 25 ms. Find the average force Favg of the wall on the ball.
An 8.5 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 15.5 kg and radius 0.2 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the speed v of the block after it has fallen 2 m.
A solid sphere rolls without slipping down a slope inclined at 20° 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?
An 11 kW motor running at 960 rpm drives a pinion with 36 teeth, which meshes with a gear with 50 teeth. The gear pair has an efficiency of 96%. Find the output torque T2 on the gear shaft.
An open belt connects a 200 mm driving pulley at 2880 rpm to a 315 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 1500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
A 3.4 kg mass on a spring with k = 80 N/m is pulled 14 cm in the positive direction and released from rest at t = 0. Find its velocity v at t = 0.1 s (with sign).
A body oscillates in simple harmonic motion with period 1.8 s and amplitude 19 cm. What is the shortest time t it needs to move from its maximum displacement of 19 cm to a displacement of 15.2 cm?
In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 5 cycles later by a peak of 10.4 mm. Find the damping ratio ζ.
A 40 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 30 kN/m and k₂ = 12 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
Answer: d = 33.75 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: hmax = 16.17 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 = 1.972 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 = 4.33 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: y = 39.09 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: v = 37.63 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 = 32.09 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 = 26.13 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: v = 13.74 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: N = 147.2 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: ΔK = 226909 J
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 lost energy goes into deformation, heat and sound.
Answer: Favg = 1056 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: v = 4.531 m/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 = 1.582 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: T2 = 145.9 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: P = 23.39 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: v = −0.3166 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 = 0.1843 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 = 0.04992
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 = 5.157 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
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