Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 90 km/h brakes with a constant deceleration of 3 m/s² until it stops. How long does braking take? Find t.
A ball is thrown straight up at 10 m/s from the edge of a roof 30 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.
Masses of 8 kg and 7 kg hang on either side of a light, frictionless pulley, connected by a light string. The system is released from rest. Use g = 9.81 m/s². Find the tension T in the string.
A 5 kg block on a ramp at 30° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 6.5 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 at 25° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is thrown from ground level at 26 m/s and 50° above the horizontal. It lands on the flat roof of a building 13 m high, on its way down. Ignore air resistance and use g = 9.81 m/s². Find the horizontal distance x from the launch point to where it lands.
In a fairground rotor, riders stand against the wall of a vertical cylinder of radius 2.1 m. When it spins fast enough the floor drops away and friction holds them up. The coefficient of static friction is 0.6. Use g = 9.81 m/s². Find the smallest angular speed ω in rad/s.
A 0.9 kg ball on a light string 1.5 m long is swung in a vertical circle. At the lowest point its speed is 10.5 m/s. Ignore air resistance and use g = 9.81 m/s². Find the tension Ttop in the string at the highest point.
A car travelling at 80 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 0.7. Find the stopping distance d.
A 2 kg block starts from rest at the bottom of a 35° slope and is pulled up the slope by a constant force of 25 N parallel to the slope. The coefficient of kinetic friction is 0.3. Find the speed v of the block after it has moved 9 m up the slope.
An 800 kg car moving at 13 m/s runs into a stationary 1000 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?
A 0.5 kg ball hits a wall at 17 m/s and bounces straight back at 11 m/s. The contact lasts 10 ms. Find the average force Favg of the wall on the ball.
A uniform slender rod of mass 12 kg and length 1 m is pinned at one end and held horizontal. It is released from rest. Use g = 9.81 m/s². Find the vertical force R that the pin exerts on the rod just after release.
A hollow sphere of mass 19.5 kg rolls without slipping at 9 m/s. Find its total kinetic energy Ek.
A 125 mm pulley at 960 rpm drives a 250 mm pulley by a belt with 2 % slip. Find the speed n2 of the driven pulley.
In a compound gear train, gear A (45 teeth) on the input shaft turning at 2900 rpm drives gear B (60 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D (16 teeth) on the output shaft. Find the output speed nout.
A simple pendulum has a length of 0.4 m. Use g = 9.81 m/s². Find its period T for small swings.
A mass on a spring with k = 130 N/m oscillates on a frictionless surface with an amplitude of 6 cm. Find its kinetic energy K when it is 5 cm from equilibrium.
A 200 kg mass is supported by a spring with a stiffness of 20 kN/m and a viscous damper with a damping coefficient of 800 N·s/m. Find the damping ratio ζ.
A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 80 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
Answer: t = 8.333 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 = 3.694 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: T = 73.25 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.
The net driving force is the difference of the weights; it moves the total mass.
Answer: a = 2.673 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: H = 13.15 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.
At the top the vertical velocity is zero: vy² = 2gH.
Answer: x = 54.21 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.
Larger root: the ball is descending.
Answer: ω = 2.79 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: Ttop = 22 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.
Kinetic energy is traded for potential energy over a height 2r.
Tension and gravity both point down, towards the centre.
Answer: d = 35.96 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: v = 8.962 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).
Answer: ΔK = 37556 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 = 1400 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: R = 29.43 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.
At the moment of release ω = 0, so there is no centripetal acceleration and the centre accelerates straight down.
R = m·g/4, independent of the length.
Answer: Ek = 1316 J
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: n2 = 470.4 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: nout = 2039 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 = 1.422 : 1
Answer: T = 1.269 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: K = 0.0715 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 = 0.2
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 = 120 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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