Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A plane needs a speed of 300 km/h to take off and the usable runway is 2400 m long. Starting from rest, what constant acceleration a does it need?
A ball is thrown straight up at 16 m/s from the edge of a roof 40 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 box slides across a level floor with an initial speed of 7.5 m/s and comes to rest after 9.6 m. Use g = 9.81 m/s². Find the coefficient of kinetic friction μ.
A 4 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 3.5 kg mass. The coefficient of kinetic friction between block and table is 0.25. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched at 50° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is launched at 55° above horizontal from the top of a 60 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A car takes a flat curve of radius 35 m. The coefficient of static friction between tyres and road is 0.8. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A 4.5 kg cart is released from rest at a height of 8.25 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 Nbottom of the track on the cart at the bottom of the loop.
A car travelling at 120 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 2.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 2 m long (μ = 0.15) and then runs onto a spring (k = 800 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 2.5 kg glider moving at 11 m/s makes a perfectly elastic head-on collision with a 1.5 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 4 kg block hanging from light strings 1.9 m long, and stays in it. The strings swing out to a largest angle of 10° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A uniform slender rod of mass 8 kg and length 1.4 m is pinned at one end and released from rest in a horizontal position. Ignore friction and use g = 9.81 m/s². Find the speed v of the free end as the rod swings through the vertical position.
A thin hoop of mass 18.5 kg rolls without slipping at 8 m/s. What percentage p of its total kinetic energy is rotational?
A 1.1 kW motor at 1450 rpm drives a worm with 4 start(s). The worm wheel has 80 teeth and the drive has an efficiency of 80%. Find the output torque Tw on the wheel shaft.
An open belt connects a 200 mm pulley and a 400 mm pulley whose centres are 600 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
A simple pendulum has a length of 2 m. Use g = 9.81 m/s². Find its period T for small swings.
A 2.6 kg mass on a spring with k = 280 N/m is pulled 9 cm from equilibrium and released. Find its maximum acceleration amax.
In a free-vibration test of a damped mass–spring system, a displacement peak of 12 mm is followed 2 cycles later by a peak of 8.2 mm. Find the damping ratio ζ.
A 10 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 40 kN/m and the lower one k₂ = 50 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
Answer: a = 1.447 m/s²
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 = 4.92 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: μ = 0.2986
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: T = 22.89 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: H = 43.19 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: t = 5.194 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: vmax = 16.57 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: Nbottom = 335.5 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: d = 113.3 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 = 23.23 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 = 13.75 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 = 151.3 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: v = 6.419 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: p = 50 %
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 share depends only on the shape, not on the mass or the speed.
Answer: Tw = 115.9 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: θ = 160.8 °
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: T = 2.837 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: amax = 9.692 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: ζ = 0.03029
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: keq = 22.22 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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