Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.
A plane starts from rest and accelerates uniformly at 2.25 m/s² along a runway. Find its speed v after 2000 m.
A tram starts from rest and accelerates at 2.5 m/s² for 5 s, then runs at constant speed for 21 s, and finally brakes at 2.5 m/s² until it stops. Find the total distance s between the two stops.
A 20 kg sled is pulled across level snow by a rope at 35° above the horizontal with a force of 120 N. The coefficient of kinetic friction is 0.4. Use g = 9.81 m/s². Find the acceleration a of the sled.
An 8 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 0.2. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched from level ground at 65° above the horizontal and lands 10 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 20° above horizontal from the top of a 40 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A car takes a flat curve of radius 90 m. The coefficient of static friction between tyres and road is 0.5. 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 8.5 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 1 kg ball hangs on a light string of length 3 m. It is pulled aside until the string makes 90° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A 10 kg block is released from rest at height 5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 1.5 m long (μ = 0.3) 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 1300 kg car moving at 20 m/s runs into a stationary 900 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?
A 5 g bullet is fired horizontally into a 2 kg block hanging from light strings 1.5 m long, and stays in it. The strings swing out to a largest angle of 25° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 0.9 kg mass on a spring with k = 160 N/m oscillates on a frictionless surface with an amplitude of 18 cm. Find its speed v when it is 2 cm from equilibrium.
A 10 g bullet moving at 100 m/s hits and stays in a 3.5 kg block resting on a frictionless surface. The block is attached to a horizontal spring with k = 600 N/m. Find the amplitude A of the resulting oscillation in cm.
A small source sends sound equally in all directions. The level is 90 dB at 2 m from it. Find the level β2 at 20 m, ignoring absorption and reflections.
Light of wavelength 600 nm passes through a single slit 0.12 mm wide. Find the angle θ3 of dark fringe number 3 from the straight-through direction.
Answer: v = 94.87 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: s = 325 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 = 2.367 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: a = 2.566 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: v0 = 11.32 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: x = 100.2 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.
Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.
Answer: vmax = 21.01 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 = 21.25 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 = 29.43 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) = 29.43 N.
Answer: v = 9.448 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: ΔK = 106364 J
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 lost energy goes into deformation, heat and sound.
Answer: v = 665.9 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 = 2.385 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: A = 2.179 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: β2 = 70 dB
Waves: a periodic wave moves one wavelength in one period, so v = fλ with T = 1/f. On a stretched string v = √(T/μ), with T the tension and μ = m/L the mass per metre. Sound in air travels at about 343 m/s at 20 °C, light in vacuum at c = 3.00 × 10⁸ m/s. Loudness is measured on a logarithmic scale: β = 10·log₁₀(I/I₀) dB with I₀ = 10⁻¹² W/m², so 10 times the intensity adds 10 dB. A point source spreads its power P over a sphere: I = P/(4πr²).
Every doubling of the distance lowers the level by about 6 dB.
Answer: θ3 = 0.8595 °
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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