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 7.5 m/s² until it stops. How long does braking take? Find t.
Ball A is dropped from rest from a height of 30 m. At the same instant ball B is thrown straight up from the ground directly below at 29 m/s. Ignore air resistance and use g = 9.81 m/s². When do the balls meet? Find t.
A 15 kg block slides down a slope at 25° to the horizontal. The coefficient of kinetic friction is 0.2. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A block is given a push so that it starts sliding up a ramp at 15° with a speed of 11 m/s. The coefficient of kinetic friction is 0.2. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is launched from level ground at 26 m/s and must land 55 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 19 m/s and 40° above the horizontal from the top of a 50 m cliff. Ignore air resistance and use g = 9.81 m/s². Find the speed v with which it hits the ground.
A 1700 kg car drives over the top of a hump in the road at 9 m/s. The top of the hump is a circular arc of radius 75 m. Use g = 9.81 m/s². Find the normal force N of the road on the car at the top.
A 2.5 kg cart is released from rest at a height of 10.5 m on a frictionless track and runs through a vertical loop of radius 2.5 m. Use g = 9.81 m/s². Find the normal force Ntop of the track on the cart at the top of the loop.
A block slides from rest down a slope at 45° 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 spring of stiffness 2000 N/m at the bottom of a 30° slope is compressed by 8 cm against a 3 kg block and released. The coefficient of kinetic friction between block and slope is 0.1. How far d up the slope does the block travel from the release point before it stops? (Ignore the short distance over which the spring pushes, i.e. take friction and gravity over the whole distance d.)
A 3 kg glider moving at 9 m/s makes a perfectly elastic head-on collision with a 4 kg glider at rest on a frictionless track. Find the velocity v′2 of the second glider after the collision.
A 20 g bullet is fired horizontally into a 3.5 kg block hanging from light strings 1.3 m long, and stays in it. The strings swing out to a largest angle of 20° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 10.5 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 12 kg and radius 0.15 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the tension T in the cord.
A uniform solid disc (mass 39 kg, radius 0.3 m) spins freely at 500 rpm about a vertical frictionless shaft. A thin ring of mass 7 kg and the same radius is dropped onto it from rest, centred on the shaft, and the two quickly spin together. Find their common speed n2 in rpm.
A 1.1 kW motor running at 1500 rpm drives a pinion with 30 teeth, which meshes with a gear with 72 teeth. The gear pair has an efficiency of 95%. Find the output torque T2 on the gear shaft.
In a compound gear train, gear A (36 teeth) on the input shaft turning at 960 rpm drives gear B (60 teeth). Gear C (25 teeth) is fixed to the same shaft as B and drives gear D (40 teeth) on the output shaft. Find the output speed nout.
An astronaut on another planet finds that a simple pendulum 0.5 m long has a period of 0.89 s for small swings. Find the gravitational acceleration g on that planet.
A 1.8 kg mass on a spring with k = 470 N/m is pulled 11 cm from equilibrium and released. Find the total energy E of the oscillation.
A 250 kg mass is supported by a spring with a stiffness of 40 kN/m and a viscous damper with a damping coefficient of 950 N·s/m. Find the damping ratio ζ.
A 50 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 80 kN/m and k₂ = 60 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
Answer: t = 1.852 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 = 1.034 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, the ground is y = 0.
Answer: a = 2.368 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: d = 13.64 m
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.
Both forces point down the slope; the mass cancels.
Answer: θ = 26.48 °
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 = 63.52°.
Answer: v = 36.63 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: N = 14841 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 (down) is positive.
The car feels lighter: the road pushes up less than its weight.
Answer: Ntop = 83.38 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: v = 14.57 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: d = 0.3707 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).
Answer: v′2 = 7.714 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: v = 218.3 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 = 37.46 N
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: n2 = 367.9 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.
Answer: T2 = 15.97 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: nout = 360 rpm
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.
total ratio = 2.667 : 1
Answer: g = 24.92 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 = 2.844 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: ζ = 0.1502
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 = 140 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
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