Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A plane starts from rest and accelerates uniformly at 3 m/s² along a runway. Find its speed v after 1800 m.
A ball is thrown straight up at 12 m/s from the edge of a roof 30 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
A 20 kg sled is pulled across level snow by a rope at 35° above the horizontal with a force of 100 N. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². Find the acceleration a of the sled.
A 2 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 5 kg mass. The coefficient of kinetic friction between block and table is 0.1. Use g = 9.81 m/s². Find the tension T in the string.
A ball is kicked from level ground at 14 m/s and 25° above the horizontal towards a wall 5 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 11 m/s and 40° above the horizontal from the top of a 60 m cliff. Ignore air resistance and use g = 9.81 m/s². Find the greatest height H above the ground at the foot of the cliff.
In a fairground rotor, riders stand against the wall of a vertical cylinder of radius 2.2 m. When it spins fast enough the floor drops away and friction holds them up. The coefficient of static friction is 0.3. Use g = 9.81 m/s². Find the smallest rotation rate n in revolutions per minute.
A curve of radius 50 m is banked at 25°, and the coefficient of static friction between tyres and road is 0.2. Use g = 9.81 m/s². Find the highest speed vmax at which a car can take the curve without sliding up and out.
A 0.5 kg ball hangs on a light string of length 0.8 m. It is pulled aside until the string makes 20° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A 2.5 kg block is released from rest at height 3.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3.5 m long (μ = 0.15) and then runs onto a spring (k = 2000 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.
A 20 kg projectile moving horizontally at 5 m/s explodes into two fragments that move along the same line. The 12 kg fragment continues forwards at 46 m/s. Find the velocity v′2 of the 8 kg fragment (positive in the original direction).
A 0.1 kg ball hits a wall at 29 m/s and bounces straight back at 6 m/s. The contact lasts 10 ms. Find the average force Favg of the wall on the ball.
A uniform slender rod of mass 6.5 kg and length 1.9 m is pinned at one end and released from rest in a horizontal position. 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 solid sphere starts from rest and rolls without slipping down a slope inclined at 35° to the horizontal. Use g = 9.81 m/s². Its radius is 0.3 m. Find its angular velocity ω after it has rolled 2.5 m along the slope.
A worm with 4 start(s) turns at 1440 rpm and drives a worm wheel with 40 teeth. Find the speed nw of the wheel.
In a compound gear train, gear A (18 teeth) on the input shaft turns at 2880 rpm and drives gear B (50 teeth). Gear C (36 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 518.4 rpm. How many teeth ND must gear D have?
An astronaut on another planet finds that a simple pendulum 1.3 m long has a period of 5.63 s for small swings. Find the gravitational acceleration g on that planet.
A 0.8 kg mass on a spring with k = 360 N/m is pulled 4 cm from equilibrium and released. Find its maximum speed vmax.
A 5 kg mass on a spring with a stiffness of 15 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 spring k₃ = 5 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 5 kN/m and k₂ = 50 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: v = 103.9 m/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: hmax = 37.34 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.
Answer: a = 2.36 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.
Answer: T = 15.42 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 = 1.57 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 still rising at the wall.
Answer: H = 62.55 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.
Answer: n = 36.82 rpm
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: vmax = 18.99 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.
Friction points down the slope.
Answer: T = 5.497 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) = 5.497 N.
Answer: v = 7.64 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 = −56.5 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: Favg = 350 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: v = 7.478 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: ω = 14.94 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: nw = 144 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.
Answer: ND = 72
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 = 1.619 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: vmax = 0.8485 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: fd = 7.785 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: fn = 2.409 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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