Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 60 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 30 m/s. Ignore air resistance and use g = 9.81 m/s². When do the balls meet? Find t.
Masses of 7.5 kg and 7 kg hang on either side of a light, frictionless pulley, connected by a light string. The system is released from rest. Use g = 9.81 m/s². Find the tension T in the string.
A 2 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 3 kg mass. The coefficient of kinetic friction between block and table is 0.2. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched at 70° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is thrown from ground level at 38 m/s and 70° above the horizontal. It lands on the flat roof of a building 17 m high, on its way down. Ignore air resistance and use g = 9.81 m/s². Find the horizontal distance x from the launch point to where it lands.
A curve of radius 120 m is banked at 8°. At what speed v can a car take it with no friction needed at all? Use g = 9.81 m/s².
A ball on a string 1 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 0.5 kg ball hangs on a light string of length 0.5 m. It is pulled aside until the string makes 50° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
An 8 kg block rests on a rough horizontal table (μ = 0.15). A light string runs from it over a smooth pulley at the edge of the table to a hanging 2 kg block. The system is released from rest. Find the speed v of the blocks after the hanging block has dropped 3 m.
A 16 kg projectile moving horizontally at 35 m/s explodes into two fragments that move along the same line. The 13 kg fragment continues forwards at 91 m/s. Find the velocity v′2 of the 3 kg fragment (positive in the original direction).
An 8 g bullet is fired horizontally into a 1.5 kg block hanging from light strings 1.5 m long, and stays in it. The strings swing out to a largest angle of 20° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A shaft transmits 3.5 kW at 2200 rpm. Find the torque T in the shaft.
A hollow sphere starts from rest and rolls without slipping down a slope inclined at 10° to the horizontal. Use g = 9.81 m/s². Its radius is 0.3 m. Find its angular velocity ω after it has rolled 7 m along the slope.
A 15 kW motor running at 2900 rpm drives a pinion with 20 teeth, which meshes with a gear with 60 teeth. The gear pair has an efficiency of 96%. Find the torque T1 on the motor shaft.
In a compound gear train, gear A (12 teeth) on the input shaft turns at 3000 rpm and drives gear B (25 teeth). Gear C (20 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 960 rpm. How many teeth ND must gear D have?
A 2.4 kg mass on a spring with k = 130 N/m is pulled 10 cm in the positive direction and released from rest at t = 0. Find its displacement x from equilibrium at t = 0.8 s (with sign).
A 2.8 kg mass on a spring with k = 440 N/m is pulled 3 cm from equilibrium and released. Find its maximum acceleration amax.
A 15 kg mass is supported by a spring with a stiffness of 15 kN/m and a viscous damper with a damping coefficient of 280 N·s/m. Find the damping ratio ζ.
A 25 kg mass hangs from spring k₃ = 60 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 12 kN/m and k₂ = 30 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 = 30.86 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.
Answer: t = 2 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.
Up is positive, the ground is y = 0.
Answer: T = 71.04 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.
The net driving force is the difference of the weights; it moves the total mass.
Answer: T = 14.13 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: H = 21.78 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: x = 87.96 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.
Larger root: the ball is descending.
Answer: v = 12.86 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 = 3.132 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: T = 8.409 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) = 8.409 N.
Answer: v = 2.17 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: v′2 = −207.7 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.
The minus sign means the second fragment flies backwards.
Answer: v = 251.1 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: T = 15.19 N·m
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: ω = 12.61 rad/s
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.
Rolling without slipping: ω = v/r, and friction does no work at the contact point.
Answer: T1 = 49.39 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: ND = 30
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: x = 9.229 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: amax = 4.714 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: ζ = 0.2951
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 = 5.003 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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