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 31 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 2.5 m/s². How far from its starting point does the police car catch up? Find d.
A ball is thrown straight up at 5 m/s from the edge of a roof 35 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 40 kg box is pushed up a ramp at 35° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the pushing force F.
An 8.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 5.5 kg mass. The coefficient of kinetic friction between block and table is 0.3. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched from level ground at 28 m/s and must land 45 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 30° above horizontal from the top of a 5 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A small ball on a string of length 0.9 m moves in a horizontal circle so that the string makes a constant angle of 45° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the speed v of the ball.
A 1.1 kg ball on a string 0.8 m long is swung in a vertical circle. At the lowest point it moves at 4 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.
A 1 kg block slides 7 m down a slope at 35° to the horizontal, with a coefficient of kinetic friction of 0.2. Find the work Wf done against friction.
A 20 kg block starts from rest at the bottom of a 20° slope and is pulled up the slope by a constant force of 175 N parallel to the slope. The coefficient of kinetic friction is 0.3. Find the speed v of the block after it has moved 2 m up the slope.
A 1.5 kg glider moving at 8 m/s makes a perfectly elastic head-on collision with a 2 kg glider at rest on a frictionless track. Find the velocity v′1 of the first glider after the collision (positive in its original direction).
A 1800 kg car runs into a parked 1600 kg car. The wrecks lock together and skid 4.5 m before stopping. The coefficient of kinetic friction between tyres and road is 0.8. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A 5 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 15.5 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 angular acceleration α of the pulley.
A solid sphere starts from rest and rolls without slipping down a slope inclined at 20° to the horizontal. Use g = 9.81 m/s². Find its speed v after it has rolled 5.5 m along the slope.
A 1.1 kW motor running at 1450 rpm drives a pinion with 12 teeth, which meshes with a gear with 30 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.
A 2.2 kW motor at 1450 rpm drives gear A (12 teeth), which meshes with gear B (72 teeth). Gear C (20 teeth) on the same shaft as B drives gear D (36 teeth) on the output shaft. Each gear mesh has an efficiency of 96%. Find the output torque Tout.
A 1.4 kg mass on a spring with k = 310 N/m is pulled 6 cm in the positive direction and released from rest at t = 0. Find its displacement x from equilibrium at t = 0.8 s (with sign).
A 0.5 kg mass is attached to the lower end of an unstretched vertical spring with k = 440 N/m and released from rest. Use g = 9.81 m/s². Find the greatest speed vmax of the mass.
A 20 kg mass on a spring with a stiffness of 25 kN/m has a viscous damper giving a damping ratio of 0.7. Find the damped natural frequency fd of its free vibration in Hz.
A 40 kg mass hangs from spring k₃ = 10 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 15 kN/m and k₂ = 100 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.
Answer: d = 768.8 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 = 62 m/s.
Answer: t = 3.229 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: F = 257.2 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.
Constant speed: the forces balance.
Answer: T = 42.59 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: θ = 17.13 °
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 = 72.87°.
Answer: x = 77 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.
Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.
Answer: v = 2.499 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.
Tension and gravity together give the inward force: tan θ = v²/(r·g).
Answer: T = 32.79 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 (up) is positive.
Answer: Wf = 11.25 J
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 = 3.243 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′1 = −1.143 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.
The minus sign means the first glider bounces back.
Answer: v1 = 15.87 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: α = 25.65 rad/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.
Answer: v = 5.134 m/s
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.
Rolling without slipping: ω = v/r, and friction does no work at the contact point.
Answer: T2 = 17.57 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 = 144.2 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: x = 4.733 cm
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: vmax = 0.3307 m/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.
The speed is largest at equilibrium.
Answer: fd = 4.018 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 = 2.414 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.
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
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