Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 50 km/h brakes with a constant deceleration of 4.5 m/s² until it stops. Find the braking distance s.
Ball A is dropped from rest from a height of 60 m. At the same instant ball B is thrown straight up from the ground directly below at 20 m/s. Ignore air resistance and use g = 9.81 m/s². At what height y above the ground do they meet?
A 4 kg block slides down a slope at 40° to the horizontal. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 90 N pulls the front 9 kg block; the 12 kg block is towed behind. The coefficient of kinetic friction is 0.3 for both blocks. Use g = 9.81 m/s². Find the acceleration a of the blocks.
A ball is launched at 40° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is launched at 19 m/s and 40° above the horizontal from the top of a 5 m cliff. Ignore air resistance and use g = 9.81 m/s². Find the speed v with which it hits the ground.
A car takes a flat curve of radius 95 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 1.2 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 2.5 kg block slides 13 m down a slope at 25° to the horizontal, with a coefficient of kinetic friction of 0.1. Find the work Wf done against friction.
A 10 kg block is released from rest at height 3.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 5 m long (μ = 0.25) and then runs onto a spring (k = 500 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.
A 12 g bullet is fired horizontally into a 4 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 11 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 1100 kg car runs into a parked 800 kg car. The wrecks lock together and skid 6 m before stopping. The coefficient of kinetic friction between tyres and road is 0.6. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A uniform slender rod of mass 1.5 kg and length 1 m is pinned at one end and released from rest in a horizontal position. Ignore friction and use g = 9.81 m/s². Find the angular velocity ω of the rod as it swings through the vertical position.
A solid sphere of mass 10.5 kg rolls without slipping at 7 m/s. What percentage p of its total kinetic energy is rotational?
A 1.1 kW motor running at 1450 rpm drives a pinion with 16 teeth, which meshes with a gear with 30 teeth. Ignore losses. Find the output torque T2 on the gear shaft.
A 2.2 kW motor at 1440 rpm drives gear A (30 teeth), which meshes with gear B (36 teeth). Gear C (16 teeth) on the same shaft as B drives gear D (32 teeth) on the output shaft. Each gear mesh has an efficiency of 95%. Find the output torque Tout.
How long must a simple pendulum be to have a period of 1.3 s? Use g = 9.81 m/s². Find L.
A 2.2 kg mass on a spring with k = 140 N/m is pulled 10 cm from equilibrium and released. Find its maximum speed vmax.
In a free-vibration test of a damped mass–spring system, a displacement peak of 15 mm is followed 5 cycles later by a peak of 1.2 mm. Find the damping ratio ζ.
A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 20 kN/m and k₂ = 40 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
Answer: s = 21.43 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: y = 15.86 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.
Up is positive, the ground is y = 0.
Answer: a = 5.554 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 = 1.343 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: H = 8.424 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.
At the top the vertical velocity is zero: vy² = 2gH.
Answer: v = 21.43 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.
Energy or v² = v₀² + 2as in the vertical direction.
Check with energy: v² = v₀² + 2gh.
Answer: vmax = 21.59 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 = 3.431 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: Wf = 28.9 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 = 6.644 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 ramp is smooth, so only the strip removes energy.
Answer: v = 491.2 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: v1 = 14.52 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: ω = 5.425 rad/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 = 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: T2 = 13.58 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 = 31.6 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: L = 0.4199 m
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 = 0.7977 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: ζ = 0.08014
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 = 3.898 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
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