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 3 m/s² until it stops. Find the braking distance s.
A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 30 m/s² for 3 s, after which the engine stops and the rocket moves freely. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax the rocket reaches.
Masses of 5.5 kg and 4.5 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 acceleration a of the masses.
A 5 kg block on a ramp at 30° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 4 kg mass. The hanging mass moves down. The coefficient of kinetic friction between block and ramp is 0.15. Use g = 9.81 m/s². Find the tension T in the string.
A ball is kicked from level ground at 34 m/s and 45° above the horizontal towards a wall 94 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 45° above horizontal from the top of a 30 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A car takes a flat curve of radius 75 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 1.8 kg ball on a string 0.6 m long is swung in a vertical circle. At the lowest point it moves at 5 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.
A car travelling at 90 km/h on a level road brakes hard and skids to a stop with its wheels locked. The coefficient of kinetic friction between the tyres and the road is 0.8. Find the stopping distance d.
A 2 kg block is released from rest at height 3.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 1 m long (μ = 0.15) and then runs onto a spring (k = 2000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 5 kg glider moving at 11 m/s makes a perfectly elastic head-on collision with a 1.5 kg glider at rest on a frictionless track. Find the velocity v′2 of the second glider after the collision.
A 0.45 kg ball hits a wall at 28 m/s and bounces straight back at 14 m/s. The contact lasts 25 ms. Find the average force Favg of the wall on the ball.
A uniform slender rod of mass 16 kg and length 2.4 m is pinned at one end and held horizontal. It is released from rest. Use g = 9.81 m/s². Find the vertical force R that the pin exerts on the rod just after release.
A solid cylinder of mass 8 kg rolls without slipping at 7 m/s. What percentage p of its total kinetic energy is rotational?
A 3 kW motor running at 2900 rpm drives a pinion with 15 teeth, which meshes with a gear with 50 teeth. The gear pair has an efficiency of 95%. Find the torque T1 on the motor shaft.
In a compound gear train, gear A (16 teeth) on the input shaft turning at 1500 rpm drives gear B (80 teeth). Gear C (45 teeth) is fixed to the same shaft as B and drives gear D (48 teeth) on the output shaft. Find the output speed nout.
A 0.7 kg mass on a spring with k = 310 N/m oscillates on a frictionless surface with an amplitude of 15 cm. Find its speed v when it is 12 cm from equilibrium.
A 15 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 = 1300 N/m. Find the amplitude A of the resulting oscillation in cm.
In a free-vibration test of a damped mass–spring system, a displacement peak of 25 mm is followed 2 cycles later by a peak of 9.1 mm. Find the damping ratio ζ.
A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 10 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 = 128.6 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: hmax = 547.8 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.
At the top the velocity is zero.
Answer: a = 0.981 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.
The net driving force is the difference of the weights; it moves the total mass.
Answer: T = 35.53 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: y = 19.02 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 = 122 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: vmax = 24.26 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 = 92.66 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: d = 39.82 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).
Doubling the speed makes the stopping distance four times longer.
Answer: x = 25.64 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′2 = 16.92 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.
Solving momentum and energy conservation together gives these two results.
Answer: Favg = 756 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: R = 39.24 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.
At the moment of release ω = 0, so there is no centripetal acceleration and the centre accelerates straight down.
R = m·g/4, independent of the length.
Answer: p = 33.33 %
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: T1 = 9.879 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 = 281.3 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 = 5.333 : 1
Answer: v = 1.894 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 = 6.229 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: ζ = 0.08016
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 = 11.25 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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