Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.
A car driving at 90 km/h brakes with a constant deceleration of 4 m/s² until it stops. Find the braking distance s.
A tram starts from rest and accelerates at 2.5 m/s² for 7 s, then runs at constant speed for 26 s, and finally brakes at 4 m/s² until it stops. Find the total distance s between the two stops.
Masses of 9.5 kg and 2 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 tension T in the string.
A 1.5 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 0.25. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched from level ground at 26 m/s and must land 45 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.
A ball is launched at 20 m/s and 45° above the horizontal from the top of a 45 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.
A car takes a flat curve of radius 30 m. The coefficient of static friction between tyres and road is 0.6. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A small cart is released from rest on a frictionless track and runs into a vertical loop of radius 7.5 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 2 kg block slides 15 m down a slope at 30° to the horizontal, with a coefficient of kinetic friction of 0.3. Find the work Wf done against friction.
A 10 kg block starts from rest at the bottom of a 35° slope and is pulled up the slope by a constant force of 110 N parallel to the slope. The coefficient of kinetic friction is 0.15. Find the speed v of the block after it has moved 10 m up the slope.
A 3 kg glider moving at 8 m/s makes a perfectly elastic head-on collision with a 0.5 kg glider at rest on a frictionless track. Find the velocity v′2 of the second glider after the collision.
At a crossing a 1300 kg car travelling east at 11 m/s collides with an 800 kg car travelling north at 11 m/s. The cars lock together. Find the direction θ of the wreck just after the collision, measured north of east.
An astronaut on another planet finds that a simple pendulum 1.2 m long has a period of 2.06 s for small swings. Find the gravitational acceleration g on that planet.
A 2.2 kg mass on a spring with k = 140 N/m is pulled 18 cm from equilibrium and released. Find the total energy E of the oscillation.
An observer moves at 5 m/s away from a stationary source of 600 Hz. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.
Light of wavelength 550 nm passes through a single slit 0.12 mm wide. Find the angle θ1 of dark fringe number 1 from the straight-through direction.
Answer: s = 78.13 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: s = 554.5 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 = 32.42 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.
The net driving force is the difference of the weights; it moves the total mass.
Answer: T = 14.15 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: θ = 20.39 °
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 other solution is 90° − θ = 69.61°.
Answer: H = 55.19 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: vmax = 13.29 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 = 18.75 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: Wf = 76.46 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 = 9.13 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).
Answer: v′2 = 13.71 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: θ = 31.61 °
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: g = 11.16 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: E = 2.268 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: f′ = 591.3 Hz
Standing waves and the Doppler effect: a string fixed at both ends, or a pipe open at both ends, fits whole half-wavelengths: L = nλ/2, fn = n·v/(2L), n = 1, 2, 3, … A pipe closed at one end has a node at the closed end and an antinode at the open end: L = nλ/4 with odd n only, fn = n·v/(4L). When source or observer move along the line between them, f′ = f·(v ± vo)/(v ∓ vs): use the upper signs for motion towards each other. Two waves of nearly equal frequency beat at |f₁ − f₂|.
Answer: θ1 = 0.2626 °
Interference and diffraction: light from two slits a distance d apart reaches a point on the screen with a path difference d·sin θ. Bright fringes appear where d·sin θ = mλ (m = 0, ±1, ±2, …). For a screen at distance L ≫ d, the m-th bright fringe is at y = mλL/d and the fringe spacing is Δy = λL/d. A grating with N lines per millimetre has d = 1/N and sharp maxima at d·sin θ = mλ; since sin θ ≤ 1, the highest order is the largest whole number m ≤ d/λ. A single slit of width a has dark fringes at a·sin θ = mλ, m = ±1, ±2, …
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