Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 108 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 19 m/s from the edge of a roof 20 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 40° 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.
An 8 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 6 kg mass. The coefficient of kinetic friction between block and table is 0.1. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched from level ground at 35° above the horizontal and lands 25 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 25° above horizontal from the top of a 40 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A small ball on a string of length 0.7 m moves in a horizontal circle so that the string makes a constant angle of 25° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the time T for one revolution.
A curve of radius 220 m is banked at 18°, and the coefficient of static friction between tyres and road is 0.1. Use g = 9.81 m/s². Find the highest speed vmax at which a car can take the curve without sliding up and out.
A car travelling at 60 km/h on a level road brakes hard and skids to a stop with its wheels locked. The coefficient of kinetic friction between the tyres and the road is 0.5. Find the stopping distance d.
A 1 kg block is released from rest at height 3 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 1.5 m long (μ = 0.1) and then runs onto a spring (k = 1000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 7 kg projectile moving horizontally at 40 m/s explodes into two fragments that move along the same line. The 6 kg fragment continues forwards at 60 m/s. Find the velocity v′2 of the 1 kg fragment (positive in the original direction).
At a crossing a 1600 kg car travelling east at 25 m/s collides with a 1800 kg car travelling north at 18 m/s. The cars lock together. Find the speed v′ of the wreck just after the collision.
A uniform slender rod of mass 20 kg and length 2.9 m is pinned at one end and held horizontal. It is released from rest. Use g = 9.81 m/s². Find the acceleration atip of the free end just after release.
A person stands on a frictionless turntable and holds a 1.5 kg dumbbell in each hand, 0.65 m from the axis. The person and turntable alone have a moment of inertia of 4 kg·m² about the axis. They turn at 60 rpm, then pull the dumbbells in to 0.2 m from the axis. Treat the dumbbells as point masses. Find the new speed n2 in rpm.
A belt runs on a 160 mm driving pulley at 1440 rpm. The tension is 400 N on the tight side and 100 N on the slack side. Find the power P transmitted, in kW.
In a compound gear train, gear A (24 teeth) on the input shaft turns at 1500 rpm and drives gear B (32 teeth). Gear C (36 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 562.5 rpm. How many teeth ND must gear D have?
An astronaut on another planet finds that a simple pendulum 0.8 m long has a period of 2.92 s for small swings. Find the gravitational acceleration g on that planet.
A 2.2 kg mass on a spring with k = 130 N/m is pulled 14 cm from equilibrium and released. Find the total energy E of the oscillation.
A 15 kg mass on a spring with a stiffness of 100 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 40 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 12 kN/m and the lower one k₂ = 50 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
Answer: s = 112.5 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 = 38.4 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 = 4.051 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 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.
Answer: v0 = 16.16 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: t = 3.735 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: T = 1.598 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.
Tension and gravity together give the inward force: tan θ = v²/(r·g).
Answer: vmax = 30.79 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.
Friction points down the slope.
Answer: d = 28.32 m
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).
Doubling the speed makes the stopping distance four times longer.
Answer: x = 23.65 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 = −80 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: v′ = 15.14 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: atip = 14.71 m/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.
atip = 1.5·g for any length: the free end falls faster than a dropped stone.
Answer: n2 = 76.71 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: P = 3.619 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: ND = 72
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: g = 3.704 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: E = 1.274 J
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 = 11.91 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 = 2.476 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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