Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 72 km/h sees an obstacle. Her reaction time is 0.7 s, after which the car brakes with a constant deceleration of 7.5 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 20 m/s² for 2.5 s, after which the engine stops and the rocket moves freely. Ignore air resistance and use g = 9.81 m/s². How long after launch does it hit the ground? Find t.
A 40 kg box is pushed up a ramp at 20° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.4. Use g = 9.81 m/s². Find the pushing force F.
A 10 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 1 kg mass. The table is frictionless. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is kicked from level ground at 38 m/s and 35° above the horizontal towards a wall 28 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 35 m/s and 20° above the horizontal from the top of a 55 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 takes a flat curve of radius 110 m. The coefficient of static friction between tyres and road is 0.3. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A 0.7 kg ball on a light string 1 m long is swung in a vertical circle. At the lowest point its speed is 8 m/s. Ignore air resistance and use g = 9.81 m/s². Find the tension Ttop in the string at the highest point.
A 1 kg ball hangs on a light string of length 2 m. It is pulled aside until the string makes 30° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A 1 kg cart is released from rest at a height of 13.5 m on a smooth track that leads into a vertical circular loop of radius 3 m. Find the normal force N the track exerts on the cart at the top of the loop.
A 1.5 kg glider moving at 6 m/s makes a perfectly elastic head-on collision with a 4.5 kg glider at rest on a frictionless track. Find the velocity v′2 of the second glider after the collision.
At a crossing a 1900 kg car travelling east at 9 m/s collides with a 1700 kg car travelling north at 13 m/s. The cars lock together. Find the speed v′ of the wreck just after the collision.
A flywheel with a moment of inertia of 19.5 kg·m² spins at 1200 rpm. A brake applies a constant friction torque of 55 N·m. How long t does it take to stop?
A uniform slender rod of mass 10 kg and length 1.6 m hangs at rest from a frictionless pivot at its top end. A 30 g bullet moving horizontally at 340 m/s hits the rod 0.89 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.
A belt runs on a 200 mm driving pulley at 1450 rpm. The tension is 1500 N on the tight side and 450 N on the slack side. Find the power P transmitted, in kW.
In a compound gear train, gear A (20 teeth) on the input shaft turning at 1440 rpm drives gear B (50 teeth). Gear C (40 teeth) is fixed to the same shaft as B and drives gear D (60 teeth) on the output shaft. Find the output speed nout.
A simple pendulum has a length of 1.1 m. Use g = 9.81 m/s². Find its period T for small swings.
A 1 kg mass on a spring with k = 150 N/m is pulled 19 cm from equilibrium and released. Find the total energy E of the oscillation.
A 20 kg mass is supported by a spring with a stiffness of 80 kN/m and a viscous damper with a damping coefficient of 130 N·s/m. Find the damping ratio ζ.
A 20 kg mass hangs from spring k₃ = 20 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 = 30 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: d = 40.67 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: t = 13.82 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.
Solve −h₁ = v₁·t₂ − ½·g·t₂² for the positive root.
Answer: F = 281.7 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: a = 0.8918 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: y = 15.64 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: α = 46.75 °
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 = 17.99 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.
Answer: Ttop = 10.46 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: T = 12.44 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) = 12.44 N.
Answer: N = 39.24 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′2 = 3 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: v′ = 7.762 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: t = 44.55 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: φ = 20.14 °
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: P = 15.94 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: nout = 384 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 = 3.75 : 1
Answer: T = 2.104 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 = 2.708 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: ζ = 0.05139
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.
ζ < 1, so the system is underdamped and oscillates as it decays.
Answer: fn = 4.015 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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