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 4 m/s² until it stops. How long does braking take? Find t.
A ball is thrown straight up at 5 m/s from the edge of a roof 15 m above the ground. Ignore air resistance and use g = 9.81 m/s². How long after the throw does it hit the ground? Find t.
A box slides across a level floor with an initial speed of 6.5 m/s and comes to rest after 8.6 m. Use g = 9.81 m/s². Find the coefficient of kinetic friction μ.
A block is given a push so that it starts sliding up a ramp at 15° with a speed of 5.5 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 from level ground at 40° above the horizontal and lands 15 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 15° above horizontal from the top of a 10 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A car takes a flat curve of radius 30 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 1.9 kg ball on a light string 1 m long is swung in a vertical circle. At the lowest point its speed is 9.5 m/s. Ignore air resistance and use g = 9.81 m/s². Find the tension Ttop in the string at the highest point.
A 2.5 kg block slides 6 m down a slope at 40° to the horizontal, with a coefficient of kinetic friction of 0.4. Find the work Wf done against friction.
A 0.5 kg block is released from rest at height 1.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3 m long (μ = 0.15) 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 6 kg projectile moving horizontally at 28 m/s explodes into two fragments that move along the same line. The 5 kg fragment continues forwards at 68 m/s. Find the velocity v′2 of the 1 kg fragment (positive in the original direction).
A 1900 kg car runs into a parked 1700 kg car. The wrecks lock together and skid 13.5 m before stopping. The coefficient of kinetic friction between tyres and road is 0.7. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A uniform slender rod of mass 4.5 kg and length 2.5 m is pinned at one end and released from rest in a horizontal position. A small ball of mass 4 kg is fixed to its free end. Ignore friction and use g = 9.81 m/s². Find the speed v of the free end as the rod swings through the vertical position.
A thin hoop rolls without slipping down a slope inclined at 10° to the horizontal. Use g = 9.81 m/s². Find the smallest coefficient of static friction μs that prevents it from slipping.
A 0.75 kW motor running at 1450 rpm drives a pinion with 18 teeth, which meshes with a gear with 32 teeth. The gear pair has an efficiency of 98%. Find the output torque T2 on the gear shaft.
A 1.5 kW motor at 2900 rpm drives gear A (30 teeth), which meshes with gear B (60 teeth). Gear C (36 teeth) on the same shaft as B drives gear D (45 teeth) on the output shaft. Each gear mesh has an efficiency of 97%. Find the output torque Tout.
An astronaut on another planet finds that a simple pendulum 1.5 m long has a period of 2.3 s for small swings. Find the gravitational acceleration g on that planet.
A 15 g bullet moving at 450 m/s hits and stays in a 3 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.
In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed one cycle later by a peak of 3.9 mm. Find the damping ratio ζ.
A 25 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 80 kN/m and the lower one k₂ = 80 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
Answer: t = 6.944 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 = 2.331 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 and the ground is h₀ below the starting point.
Only the positive root is physical.
Answer: μ = 0.2504
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: d = 3.082 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: v0 = 12.22 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.
Answer: t = 2.543 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.
Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.
Answer: vmax = 15.34 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 = 78.28 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: Wf = 45.09 J
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 = 4.539 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′2 = −172 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: v1 = 25.8 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 = 7.466 m/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: μs = 0.08816
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.
Answer: T2 = 8.605 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: Tout = 11.62 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: g = 11.19 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: A = 12.29 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 = 0.1482
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 = 40 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
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