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 3.5 m/s² until it stops. Find the braking distance s.
Ball A is dropped from rest from a height of 25 m. At the same instant ball B is thrown straight up from the ground directly below at 14 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 45° to the horizontal. The coefficient of kinetic friction is 0.15. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A block is given a push so that it starts sliding up a ramp at 25° with a speed of 8 m/s. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is launched at 55° above horizontal from level ground. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A ball is launched at 13 m/s and 15° above the horizontal from the top of a 60 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 85 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 0.6 kg ball on a light string 1.1 m long is swung in a vertical circle. At the lowest point its speed is 8.5 m/s. Ignore air resistance and use g = 9.81 m/s². Find the tension Ttop in the string at the highest point.
A toy launcher has a vertical spring of stiffness 400 N/m. A 0.25 kg ball is placed on it and the spring is compressed by 8 cm, then released. Ignoring air resistance and the spring’s mass, find the height h the ball rises, measured from its position when the spring was compressed.
A 2 kg block is released from rest at height 5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3.5 m long (μ = 0.1) and then runs onto a spring (k = 1500 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 10 g bullet is fired horizontally into a 5 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 30 m/s and bounces straight back at 14 m/s. The contact lasts 5 ms. Find the average force Favg of the wall on the ball.
A uniform slender rod of mass 8.5 kg and length 1.7 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 person stands on a frictionless turntable and holds a 2 kg dumbbell in each hand, 0.7 m from the axis. The person and turntable alone have a moment of inertia of 5 kg·m² about the axis. They turn at 30 rpm, then pull the dumbbells in to 0.3 m from the axis. Treat the dumbbells as point masses. Find the new speed n2 in rpm.
A 5.5 kW motor running at 960 rpm drives a pinion with 36 teeth, which meshes with a gear with 45 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.
In a compound gear train, gear A (36 teeth) on the input shaft turning at 3000 rpm drives gear B (45 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D (30 teeth) on the output shaft. Find the output speed nout.
An astronaut on another planet finds that a simple pendulum 1.4 m long has a period of 3.85 s for small swings. Find the gravitational acceleration g on that planet.
A 2.2 kg mass on a spring with k = 490 N/m is pulled 19 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 25 mm is followed one cycle later by a peak of 18.3 mm. Find the damping ratio ζ.
A 5 kg mass hangs from spring k₃ = 30 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 = 80 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: s = 70.55 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 = 9.359 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 = 5.896 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: d = 4.697 m
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.
Both forces point down the slope; the mass cancels.
Answer: t = 6.346 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.
Starting and landing at the same height: t = 2vy/g.
Answer: v = 36.69 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 = 25.83 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: Ttop = 9.979 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.
Kinetic energy is traded for potential energy over a height 2r.
Tension and gravity both point down, towards the centre.
Answer: h = 0.5219 m
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: x = 34.88 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 = 800.1 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: Favg = 2640 N
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 = 8.656 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: n2 = 38.96 rpm
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 kinetic energy increases: the extra energy comes from the work the person does pulling the dumbbells in.
Answer: T2 = 66.34 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: nout = 1200 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.
total ratio = 2.5 : 1
Answer: g = 3.729 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 = 2.836 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 = 0.04959
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 = 10.81 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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