Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 80 km/h brakes with a constant deceleration of 4.5 m/s² until it stops. Find the braking distance s.
A tram starts from rest and accelerates at 2 m/s² for 6 s, then runs at constant speed for 29 s, and finally brakes at 3 m/s² until it stops. Find the total distance s between the two stops.
A 15 kg box is pushed up a ramp at 40° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.4. Use g = 9.81 m/s². Find the pushing force F.
A 3.5 kg block on a ramp at 30° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 5 kg mass. The hanging mass moves down. The ramp is frictionless. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched from level ground at 70° above the horizontal and lands 120 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 15° 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.
In a fairground rotor, riders stand against the wall of a vertical cylinder of radius 2.3 m. When it spins fast enough the floor drops away and friction holds them up. The coefficient of static friction is 0.4. Use g = 9.81 m/s². Find the smallest angular speed ω in rad/s.
A 5 kg cart is released from rest at a height of 9.2 m on a frictionless track and runs through a vertical loop of radius 2 m. Use g = 9.81 m/s². Find the normal force Ntop of the track on the cart at the top of the loop.
A block slides from rest down a slope at 25° to the horizontal. The coefficient of kinetic friction is 0.35. Find its speed v after it has slid 11 m along the slope.
A 0.5 kg block is released from rest at height 3.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 4.5 m long (μ = 0.25) and then runs onto a spring (k = 5000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 1.5 kg glider moving at 12 m/s makes a perfectly elastic head-on collision with a 3.5 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 0.55 kg ball hits a wall at 24 m/s and bounces straight back at 4 m/s. The contact lasts 20 ms. Find the magnitude of the impulse J the wall gives the ball.
A uniform slender rod of mass 3.5 kg and length 0.5 m is pinned at one end and held horizontal. It is released from rest. Use g = 9.81 m/s². Find its angular acceleration α just after release.
A uniform slender rod of mass 4.5 kg and length 1.8 m hangs at rest from a frictionless pivot at its top end. A 25 g bullet moving horizontally at 180 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 22 kW motor running at 960 rpm drives a pinion with 16 teeth and a module of 2 mm. Find the tangential force Ft on the pinion teeth.
An open belt connects a 100 mm driving pulley at 2880 rpm to a 200 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 2000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
A 0.7 kg mass on a spring with k = 370 N/m oscillates on a frictionless surface with an amplitude of 7 cm. Find its speed v when it is 5 cm from equilibrium.
A 10 g bullet moving at 500 m/s hits and stays in a 4 kg block resting on a frictionless surface. The block is attached to a horizontal spring with k = 100 N/m. Find the amplitude A of the resulting oscillation in cm.
In a free-vibration test of a damped mass–spring system, a displacement peak of 12 mm is followed 5 cycles later by a peak of 1.8 mm. Find the damping ratio ζ.
A 40 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 60 kN/m and k₂ = 20 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
Answer: s = 54.87 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: s = 408 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: F = 139.7 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 = 30.3 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: v0 = 42.79 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: t = 3.8 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: ω = 3.265 rad/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: Ntop = 206 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: v = 4.77 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).
The block drops a height d·sin θ while friction removes μmg·cos θ·d. The mass cancels.
Answer: x = 6.826 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′1 = −4.8 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: J = 15.4 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: α = 29.43 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.
At the moment of release ω = 0, so there is no centripetal acceleration and the centre accelerates straight down.
Answer: φ = 23.54 °
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: Ft = 13677 N
Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.
Answer: P = 17.8 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: v = 1.126 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.
Answer: A = 24.97 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 = 0.06028
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 = 80 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
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