Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car speeding at a constant 15 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 5.5 m/s². How far from its starting point does the police car catch up? Find d.
A tram starts from rest and accelerates at 2.5 m/s² for 5 s, then runs at constant speed for 24 s, and finally brakes at 5 m/s² until it stops. Find the total distance s between the two stops.
A 2 kg block slides down a slope at 30° to the horizontal. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 7.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 3.5 kg mass. The table is frictionless. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched at 30° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A ball is launched at 13 m/s and 40° above the horizontal from the top of a 30 m cliff. Ignore air resistance and use g = 9.81 m/s². At what angle α below the horizontal does it hit the ground?
A car drives over the top of a hump in the road, which is a circular arc of radius 70 m. Use g = 9.81 m/s². Above what speed vmax do the wheels leave the road at the top?
A small cart is released from rest on a frictionless track and runs into a vertical loop of radius 2 m. From what minimum height hmin above the bottom of the loop must it start to stay on the track at the top of the loop?
A 0.2 kg ball hangs on a light string of length 1.5 m. It is pulled aside until the string makes 60° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A 4 kg block is released from rest at height 2.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 4.5 m long (μ = 0.25) and then runs onto a spring (k = 4000 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.
A 2.5 kg glider moving at 3 m/s makes a perfectly elastic head-on collision with a 4 kg glider at rest on a frictionless track. Find the velocity v′1 of the first glider after the collision (positive in its original direction).
A 0.25 kg ball hits a wall at 18 m/s and bounces straight back at 15 m/s. The contact lasts 40 ms. Find the magnitude of the impulse J the wall gives the ball.
A uniform slender rod of mass 11 kg and length 1.8 m is pinned at one end and held horizontal. It is released from rest. Use g = 9.81 m/s². Find the acceleration atip of the free end just after release.
A hollow sphere starts from rest and rolls without slipping down a slope inclined at 20° to the horizontal. Use g = 9.81 m/s². Its radius is 0.3 m. Find its angular velocity ω after it has rolled 7.5 m along the slope.
A 0.75 kW motor running at 1500 rpm drives a pinion with 32 teeth, which meshes with a gear with 45 teeth. The gear pair has an efficiency of 96%. Find the torque T1 on the motor shaft.
An open belt connects a 120 mm driving pulley at 960 rpm to a 300 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.35 and the tight-side tension may not exceed 1000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
How long must a simple pendulum be to have a period of 0.9 s? Use g = 9.81 m/s². Find L.
A body oscillates in simple harmonic motion with period 3.2 s and amplitude 17 cm. What is the shortest time t it needs to move from its maximum displacement of 17 cm to a displacement of 8.5 cm?
In a free-vibration test of a damped mass–spring system, a displacement peak of 15 mm is followed 4 cycles later by a peak of 7.1 mm. Find the damping ratio ζ.
A 20 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 20 kN/m and the lower one k₂ = 20 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
Answer: d = 81.82 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 that moment the police car moves at 2v = 30 m/s.
Answer: s = 346.9 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: a = 2.356 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.
Down the slope is positive.
Answer: T = 23.41 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.
Answer: x = 22.6 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.
Starting and landing at the same height: t = 2vy/g.
Answer: α = 68.79 °
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.
Energy or v² = v₀² + 2as in the vertical direction.
Answer: vmax = 26.2 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: hmin = 5 m
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 = 3.924 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) = 3.924 N.
Answer: v = 5.194 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′1 = −0.6923 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.
The minus sign means the first glider bounces back.
Answer: J = 8.25 N·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: atip = 14.71 m/s²
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.
atip = 1.5·g for any length: the free end falls faster than a dropped stone.
Answer: ω = 18.32 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 = 4.775 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: P = 3.752 kW
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Answer: L = 0.2013 m
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: t = 0.5333 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: ζ = 0.02975
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 = 10 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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