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 19 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 5 m/s². How long does it take the police car to catch up? Find t.
A tram starts from rest and accelerates at 2 m/s² for 6 s, then runs at constant speed for 30 s, and finally brakes at 3.5 m/s² until it stops. Find the total travel time t.
A box slides across a level floor with an initial speed of 11 m/s and comes to rest after 15 m. Use g = 9.81 m/s². Find the coefficient of kinetic friction μ.
A 5.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 1.5 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 at 55° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A ball is launched at 20° above horizontal from the top of a 30 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A 1100 kg car drives over the top of a hump in the road at 13 m/s. The top of the hump is a circular arc of radius 25 m. Use g = 9.81 m/s². Find the normal force N of the road on the car at the top.
A curve of radius 210 m is banked at 25°, and the coefficient of static friction between tyres and road is 0.3. Use g = 9.81 m/s². Find the lowest speed vmin at which a car can take the curve without sliding down the bank.
An 800 kg car drives up a 5° slope at a constant 72 km/h. Air resistance and rolling resistance together amount to 650 N. Find the power P the wheels must deliver, in kW.
A 10 kg block starts from rest at the bottom of a 35° slope and is pulled up the slope by a constant force of 130 N parallel to the slope. The coefficient of kinetic friction is 0.1. Find the speed v of the block after it has moved 6 m up the slope.
A 600 kg car moving at 20 m/s runs into a stationary 1100 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?
A 0.45 kg ball hits a wall at 22 m/s and bounces straight back at 16 m/s. The contact lasts 40 ms. Find the magnitude of the impulse J the wall gives the ball.
A 5 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 17.5 kg and radius 0.4 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the speed v of the block after it has fallen 4.5 m.
A solid cylinder of mass 16.5 kg rolls without slipping at 9 m/s. What percentage p of its total kinetic energy is rotational?
A 1.1 kW motor running at 960 rpm drives a pinion with 12 teeth, which meshes with a gear with 48 teeth. The gear pair has an efficiency of 98%. Find the torque T1 on the motor shaft.
An open belt connects a 160 mm driving pulley at 720 rpm to a 500 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 1500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
An astronaut on another planet finds that a simple pendulum 0.6 m long has a period of 0.98 s for small swings. Find the gravitational acceleration g on that planet.
An 8 g bullet moving at 400 m/s hits and stays in a 3 kg block resting on a frictionless surface. The block is attached to a horizontal spring with k = 900 N/m. Find the amplitude A of the resulting oscillation in cm.
A 200 kg mass is supported by a spring with a stiffness of 50 kN/m and a viscous damper with a damping coefficient of 1260 N·s/m. Find the damping ratio ζ.
A 5 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 30 kN/m and the lower one k₂ = 5 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
Answer: t = 7.6 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.
At that moment the police car moves at 2v = 38 m/s.
Answer: t = 39.43 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: μ = 0.4111
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: a = 1.331 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: x = 138.3 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.
Starting and landing at the same height: t = 2vy/g.
Answer: x = 95.64 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: N = 3355 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 (down) is positive.
The car feels lighter: the road pushes up less than its weight.
Answer: vmin = 17.34 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.
Friction points up the slope.
Answer: P = 26.68 kW
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). Power is force times speed: P = F·v.
Answer: v = 8.879 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 = 77647 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: J = 17.1 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: v = 5.666 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 = 33.33 %
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: T1 = 10.94 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: P = 4.619 kW
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: g = 24.66 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: A = 6.15 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: ζ = 0.1992
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: fn = 4.66 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.
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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