Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car speeding at a constant 27 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 4.5 m/s². How far from its starting point does the police car catch up? Find d.
A stone is dropped from rest into a well. The splash is heard 2.1 s after the stone is released. The speed of sound is 343 m/s; ignore air resistance and use g = 9.81 m/s². Find the depth d of the well.
A 20 kg sled is pulled across level snow by a rope at 25° above the horizontal with a force of 110 N. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². Find the acceleration a of the sled.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 65 N pulls the front 5 kg block; the 7 kg block is towed behind. The coefficient of kinetic friction is 0.25 for both blocks. Use g = 9.81 m/s². Find the tension T in the connecting string.
A ball is launched from level ground at 75° above the horizontal and lands 40 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is thrown from ground level at 28 m/s and 60° above the horizontal. It lands on the flat roof of a building 2 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.
A car takes a flat curve of radius 100 m. The coefficient of static friction between tyres and road is 0.5. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A ball on a string 1.5 m long is swung in a vertical circle. Use g = 9.81 m/s². What is the smallest speed vmin at the top for which the string just stays tight?
A toy launcher has a vertical spring of stiffness 500 N/m. A 0.2 kg ball is placed on it and the spring is compressed by 4 cm, then released. Ignoring air resistance and the spring’s mass, find the height h the ball rises, measured from its position when the spring was compressed.
A 2 kg block rests on a rough horizontal table (μ = 0.25). A light string runs from it over a smooth pulley at the edge of the table to a hanging 5 kg block. The system is released from rest. Find the speed v of the blocks after the hanging block has dropped 2 m.
A 1 kg glider moving at 10 m/s makes a perfectly elastic head-on collision with a 2.5 kg glider at rest on a frictionless track. Find the velocity v′2 of the second glider after the collision.
A 0.45 kg ball hits a wall at 12 m/s and bounces straight back at 9 m/s. The contact lasts 40 ms. Find the magnitude of the impulse J the wall gives the ball.
A motor delivers a torque of 30 N·m at 1000 rpm. Find its power output P in kW.
A uniform slender rod of mass 7.5 kg and length 1.3 m hangs at rest from a frictionless pivot at its top end. A 15 g bullet moving horizontally at 500 m/s hits the rod 1.27 m below the pivot and stays embedded. Use g = 9.81 m/s². Find the largest angle φ from the vertical through which the rod swings.
A 2.2 kW motor running at 1440 rpm drives a pinion with 16 teeth, which meshes with a gear with 36 teeth. The gear pair has an efficiency of 98%. Find the output torque T2 on the gear shaft.
A 15 kW motor at 1440 rpm drives gear A (15 teeth), which meshes with gear B (48 teeth). Gear C (12 teeth) on the same shaft as B drives gear D (15 teeth) on the output shaft. Each gear mesh has an efficiency of 95%. Find the output torque Tout.
A simple pendulum has a length of 2.1 m. Use g = 9.81 m/s². Find its period T for small swings.
A 2.5 kg mass on a spring with k = 420 N/m is pulled 8 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 80 kN/m and a viscous damper with a damping coefficient of 4470 N·s/m. Find the damping ratio ζ.
A 10 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 = 80 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
Answer: d = 324 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.
At that moment the police car moves at 2v = 54 m/s.
Answer: d = 20.42 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.
Only the positive root is physical.
Answer: a = 3.113 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: T = 37.92 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.
With the same μ on both blocks, friction drops out: T = F·m₁/(m₁ + m₂).
Answer: v0 = 28.01 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: x = 68.04 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: vmax = 22.15 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: vmin = 3.836 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: h = 0.2039 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 = 5.023 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: v′2 = 5.714 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: J = 9.45 N·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: P = 3.142 kW
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: φ = 38.95 °
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 pivot force passes through the pivot, so it has no moment there during the impact.
Answer: T2 = 32.17 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: Tout = 359.1 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: T = 2.907 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: E = 1.344 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 = 0.4998
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 = 10.43 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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