Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 100 km/h brakes with a constant deceleration of 6 m/s² until it stops. How long does braking take? Find t.
A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 32 m/s² for 2 s, after which the engine stops and the rocket moves freely. Ignore air resistance and use g = 9.81 m/s². How long after launch does it hit the ground? Find t.
A 45 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 0.2. Use g = 9.81 m/s². Find the pushing force F.
A 3.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 2 kg mass. The coefficient of kinetic friction between block and table is 0.3. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is kicked from level ground at 20 m/s and 55° above the horizontal towards a wall 31 m away. Ignore air resistance and use g = 9.81 m/s². At what height y does the ball pass the wall?
A ball is launched at 20° above horizontal from the top of a 15 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A small ball on a string of length 1.5 m moves in a horizontal circle so that the string makes a constant angle of 40° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the time T for one revolution.
A 1.9 kg ball on a light string 1.2 m long is swung in a vertical circle. At the lowest point its speed is 9 m/s. Ignore air resistance and use g = 9.81 m/s². Find the tension Ttop in the string at the highest point.
A 1 kg ball hangs on a light string of length 1.5 m. It is pulled aside until the string makes 30° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A 1 kg block is released from rest at height 4.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 5 m long (μ = 0.2) and then runs onto a spring (k = 1000 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.
A 20 g bullet is fired horizontally into a 1.5 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 15 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 10 g bullet is fired horizontally into a 5 kg block hanging from light strings 1.7 m long, and stays in it. The strings swing out to a largest angle of 30° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A flywheel with a moment of inertia of 7 kg·m² spins at 1300 rpm. A brake applies a constant friction torque of 10 N·m. How long t does it take to stop?
A person stands on a frictionless turntable and holds a 2 kg dumbbell in each hand, 0.9 m from the axis. The person and turntable alone have a moment of inertia of 3.5 kg·m² about the axis. They turn at 40 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 1500 rpm drives a pinion with 16 teeth, which meshes with a gear with 24 teeth. The gear pair has an efficiency of 96%. Find the output torque T2 on the gear shaft.
In a compound gear train, gear A (16 teeth) on the input shaft turning at 1450 rpm drives gear B (60 teeth). Gear C (32 teeth) is fixed to the same shaft as B and drives gear D (50 teeth) on the output shaft. Find the output speed nout.
A 3 kg mass on a spring with k = 300 N/m oscillates on a frictionless surface with an amplitude of 20 cm. Find its speed v when it is 5 cm from equilibrium.
A 5 g bullet moving at 300 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 = 1000 N/m. Find the amplitude A of the resulting oscillation in cm.
A 15 kg mass is supported by a spring with a stiffness of 10 kN/m and a viscous damper with a damping coefficient of 230 N·s/m. Find the damping ratio ζ.
A 100 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 = 60 kN/m and k₂ = 60 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: t = 4.63 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: t = 15.98 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.
Solve −h₁ = v₁·t₂ − ½·g·t₂² for the positive root.
Answer: F = 351.4 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: a = 1.694 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: y = 8.453 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.
The ball is already on its way down at the wall.
Answer: x = 35.98 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: T = 2.15 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.
Tension and gravity together give the inward force: tan θ = v²/(r·g).
Answer: Ttop = 35.06 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: T = 12.44 N
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). On a circular path the net force towards the centre must equal mv²/r.
Check: T = mg(3 − 2cos θ) = 12.44 N.
Answer: v = 8.287 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 = 130.4 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: v = 1059 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: t = 95.29 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: n2 = 69.84 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 = 50.42 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 = 247.5 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 = 5.859 : 1
Answer: v = 1.936 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 = 2.37 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 = 0.2969
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 = 2.466 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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