Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 54 km/h brakes with a constant deceleration of 7.5 m/s² until it stops. How long does braking take? Find t.
Ball A is dropped from rest from a height of 45 m. At the same instant ball B is thrown straight up from the ground directly below at 18 m/s. Ignore air resistance and use g = 9.81 m/s². At what height y above the ground do they meet?
A 15 kg box is pushed up a ramp at 15° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.15. Use g = 9.81 m/s². Find the pushing force F.
A block is given a push so that it starts sliding up a ramp at 15° with a speed of 9.5 m/s. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is launched at 50° above horizontal from level ground. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A ball is thrown from ground level at 35 m/s and 70° above the horizontal. It lands on the flat roof of a building 18 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 car takes a flat curve of radius 80 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 1.4 kg ball on a string 1.4 m long is swung in a vertical circle. At the lowest point it moves at 6.5 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.
A block slides from rest down a slope at 35° to the horizontal. The coefficient of kinetic friction is 0.4. Find its speed v after it has slid 20 m along the slope.
A 2 kg cart is released from rest at a height of 16 m on a smooth track that leads into a vertical circular loop of radius 5 m. Find the normal force N the track exerts on the cart at the top of the loop.
A 20 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 23 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 0.15 kg ball hits a wall at 29 m/s and bounces straight back at 5 m/s. The contact lasts 5 ms. Find the average force Favg of the wall on the ball.
A 12.5 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 16 kg and radius 0.45 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the tension T in the cord.
A uniform slender rod of mass 3.5 kg and length 1.9 m hangs at rest from a frictionless pivot at its top end. A 20 g bullet moving horizontally at 140 m/s hits the rod 1.76 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.
An 11 kW motor running at 1440 rpm drives a pinion with 15 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 (15 teeth) on the input shaft turns at 1500 rpm and drives gear B (48 teeth). Gear C (24 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 375 rpm. How many teeth ND must gear D have?
A simple pendulum has a length of 0.2 m. Use g = 9.81 m/s². Find its period T for small swings.
A 3.2 kg mass on a spring with k = 140 N/m is pulled 11 cm from equilibrium and released. Find the total energy E of the oscillation.
A 200 kg mass on a spring with a stiffness of 100 kN/m has a viscous damper giving a damping ratio of 0.45. Find the damped natural frequency fd of its free vibration in Hz.
A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 80 kN/m and k₂ = 12 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
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.
Answer: y = 14.34 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: F = 59.41 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: d = 9.194 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 = 1.562 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: x = 73.07 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: vmax = 25.06 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 = 55.98 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 (up) is positive.
Answer: v = 9.823 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: N = 27.47 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.
The cart just makes it round when h = 2.5R; then N = 0 at the top.
Answer: v = 533.2 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: Favg = 1020 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: T = 47.85 N
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: φ = 23.96 °
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: T2 = 112 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: T = 0.8971 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: E = 0.847 J
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: fd = 3.178 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.
Answer: keq = 92 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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