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 32 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 6 m/s². How far from its starting point does the police car catch up? Find d.
Ball A is dropped from rest from a height of 50 m. At the same instant ball B is thrown straight up from the ground directly below at 27 m/s. Ignore air resistance and use g = 9.81 m/s². At what height y above the ground do they meet?
A 15 kg sled is pulled across level snow by a rope at 30° above the horizontal with a force of 150 N. The coefficient of kinetic friction is 0.4. Use g = 9.81 m/s². Find the acceleration a of the sled.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 115 N pulls the front 6 kg block; the 9 kg block is towed behind. The coefficient of kinetic friction is 0.2 for both blocks. Use g = 9.81 m/s². Find the acceleration a of the blocks.
A ball is launched at 45° above horizontal from level ground. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A ball is launched at 27 m/s and 70° above the horizontal from the top of a 25 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 150 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 0.3 kg ball on a string 1 m long is swung in a vertical circle. At the lowest point it moves at 5.5 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.
A 1500 kg car drives up a 5° slope at a constant 90 km/h. Air resistance and rolling resistance together amount to 550 N. Find the power P the wheels must deliver, in kW.
A 10 kg block is released from rest at height 2 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 2.5 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 1.5 kg glider moving at 7 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 1800 kg car travelling east at 8 m/s collides with a 1000 kg car travelling north at 22 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 1.8 m long has a period of 4.37 s for small swings. Find the gravitational acceleration g on that planet.
A 2.2 kg mass is attached to the lower end of an unstretched vertical spring with k = 320 N/m and released from rest. Use g = 9.81 m/s². Find the greatest speed vmax of the mass.
A pipe open at both ends is 0.6 m long. The speed of sound is 343 m/s. Find the frequency f3 of its harmonic number 3.
An optical fibre has a core with n₁ = 1.52 and a cladding with n₂ = 1.44. Find its numerical aperture NA.
Answer: d = 341.3 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 = 64 m/s.
Answer: y = 33.18 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.
Up is positive, the ground is y = 0.
Answer: a = 6.736 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 = 5.705 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 = 3.172 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.
Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.
Starting and landing at the same height: t = 2vy/g.
Answer: H = 57.81 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 = 27.12 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: T = 12.02 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.
Towards the centre (up) is positive.
Answer: P = 45.81 kW
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). Power is force times speed: P = F·v.
Answer: v = 5.646 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 = 10.5 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.
Solving momentum and energy conservation together gives these two results.
Answer: θ = 56.79 °
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 = 3.721 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 = 0.8134 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.
The speed is largest at equilibrium.
Answer: f3 = 857.5 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: NA = 0.4866
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.
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