Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.
A car speeding at a constant 19 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 3.5 m/s². How far from its starting point does the police car catch up? Find d.
A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 26 m/s² for 2 s, after which the engine stops and the rocket moves freely. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax the rocket reaches.
Masses of 10 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 3.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 1 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 launched from level ground at 25° above the horizontal and lands 95 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 25° above horizontal from the top of a 10 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A car drives over the top of a hump in the road, which is a circular arc of radius 30 m. Use g = 9.81 m/s². Above what speed vmax do the wheels leave the road at the top?
A ball on a string 1.5 m long is swung in a vertical circle. Use g = 9.81 m/s². What is the smallest speed vmin at the top for which the string just stays tight?
A 2 kg block slides 10 m down a slope at 25° to the horizontal, with a coefficient of kinetic friction of 0.15. Find the work Wf done against friction.
A 1 kg block is released from rest at height 3.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 4 m long (μ = 0.15) and then runs onto a spring (k = 2000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 6 kg projectile moving horizontally at 7 m/s explodes into two fragments that move along the same line. The 2 kg fragment continues forwards at 23 m/s. Find the velocity v′2 of the 4 kg fragment (positive in the original direction).
A 0.1 kg ball hits a wall at 22 m/s and bounces straight back at 20 m/s. The contact lasts 50 ms. Find the average force Favg of the wall on the ball.
A simple pendulum has a length of 1.9 m. Use g = 9.81 m/s². Find its period T for small swings.
An 8 g bullet moving at 200 m/s hits and stays in a 2.5 kg block resting on a frictionless surface. The block is attached to a horizontal spring with k = 2000 N/m. Find the amplitude A of the resulting oscillation in cm.
Light travels in diamond (n = 2.42) towards a flat boundary with acrylic (n = 1.49). Find the critical angle θc beyond which it is totally reflected.
Light of wavelength 589 nm passes through a single slit 0.15 mm wide. Find the width w of the central bright maximum on a screen 2 m away (between the first dark fringes on either side).
Answer: d = 206.3 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 = 38 m/s.
Answer: hmax = 189.8 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 the top the velocity is zero.
Answer: T = 32.7 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.
The net driving force is the difference of the weights; it moves the total mass.
Answer: T = 8.393 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: v0 = 34.88 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 = 78.04 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 = 17.16 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: vmin = 3.836 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: Wf = 26.67 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: x = 16.87 cm
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 = −1 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 = 84 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: T = 2.765 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 = 2.259 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: θc = 38 °
Geometric optics: at a boundary, n₁·sin θ₁ = n₂·sin θ₂ (Snell's law), with angles measured from the normal and n = c/v. Going into a slower medium the ray bends towards the normal. From a denser medium (n₁ > n₂) beyond the critical angle θc = arcsin(n₂/n₁) all light is reflected: total internal reflection. Thin lenses and spherical mirrors obey 1/f = 1/do + 1/di with the magnification m = −di/do. Sign convention: f > 0 for a converging lens and a concave mirror (f = R/2), f < 0 for a diverging lens and a convex mirror; di > 0 for a real image, di < 0 for a virtual one, and m < 0 for an inverted image.
Answer: w = 15.71 mm
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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