Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 100 km/h sees an obstacle. Her reaction time is 0.5 s, after which the car brakes with a constant deceleration of 4.5 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A stone is dropped from rest into a well. The splash is heard 3.8 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 16 kg block slides down a frictionless slope at 35° to the horizontal. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 7 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 5 kg mass. The coefficient of kinetic friction between block and table is 0.1. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched from level ground at 40 m/s and must land 130 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 31 m/s and 60° above the horizontal from the top of a 10 m cliff. Ignore air resistance and use g = 9.81 m/s². At what angle α below the horizontal does it hit the ground?
A car takes a flat curve of radius 150 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 kg cart is released from rest at a height of 1.5 m on a frictionless track and runs through a vertical loop of radius 0.5 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 2 kg block slides 17 m down a slope at 15° to the horizontal, with a coefficient of kinetic friction of 0.1. Find the work Wf done against friction.
A 3 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 4 kg block. The system is released from rest. Find the speed v of the blocks after the hanging block has dropped 2 m.
A 10 g bullet is fired horizontally into a 2 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 13 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 0.3 kg ball hits a wall at 23 m/s and bounces straight back at 15 m/s. The contact lasts 5 ms. Find the magnitude of the impulse J the wall gives the ball.
A motor brings a flywheel with a moment of inertia of 10 kg·m² from rest up to 600 rpm in 8 s. Ignoring losses, find the average power Pavg it delivers, in kW.
A solid sphere of mass 19 kg rolls without slipping at 2.5 m/s. What percentage p of its total kinetic energy is rotational?
A worm with 3 start(s) turns at 960 rpm and drives a worm wheel with 48 teeth. Find the speed nw of the wheel.
In a compound gear train, gear A (15 teeth) on the input shaft turns at 2880 rpm and drives gear B (16 teeth). Gear C (18 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 675 rpm. How many teeth ND must gear D have?
An astronaut on another planet finds that a simple pendulum 1.4 m long has a period of 2.22 s for small swings. Find the gravitational acceleration g on that planet.
A 1.2 kg mass is attached to the lower end of an unstretched vertical spring with k = 300 N/m and released from rest. Use g = 9.81 m/s². Find the greatest speed vmax of the mass.
A 50 kg mass on a spring with a stiffness of 20 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 25 kg mass hangs from spring k₃ = 15 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 20 kN/m and k₂ = 5 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 = 99.62 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.
Answer: d = 64.04 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 = 5.627 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: a = 3.515 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: θ = 26.43 °
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 = 63.57°.
Answer: α = 62.89 °
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.
Answer: vmax = 34.31 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 = 274.7 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: Wf = 32.22 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 = 4.268 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 = 321 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: J = 11.4 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: Pavg = 2.467 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: p = 28.57 %
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: nw = 60 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.
Answer: ND = 72
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: g = 11.21 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: vmax = 0.6204 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 = 2.273 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 = 3.082 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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