Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.
A car driving at 45 km/h brakes with a constant deceleration of 7.5 m/s² until it stops. Find the braking distance s.
A ball is thrown straight up at 9 m/s from the edge of a roof 40 m above the ground. Ignore air resistance and use g = 9.81 m/s². How long after the throw does it hit the ground? Find t.
A 45 kg box is pushed up a ramp at 35° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². Find the pushing force F.
A block is given a push so that it starts sliding up a ramp at 35° with a speed of 4.5 m/s. The coefficient of kinetic friction is 0.15. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is launched from level ground at 34 m/s and must land 75 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 31 m/s and 50° 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 small ball on a string of length 1.6 m moves in a horizontal circle so that the string makes a constant angle of 15° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the time T for one revolution.
A curve of radius 110 m is banked at 25°, and the coefficient of static friction between tyres and road is 0.2. Use g = 9.81 m/s². Find the lowest speed vmin at which a car can take the curve without sliding down the bank.
A block slides from rest down a slope at 45° to the horizontal. The coefficient of kinetic friction is 0.1. Find its speed v after it has slid 2 m along the slope.
A 4 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 = 0.3) and then runs onto a spring (k = 500 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 4.5 kg glider moving at 6 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′1 of the first glider after the collision (positive in its original direction).
A 0.2 kg ball hits a wall at 17 m/s and bounces straight back at 17 m/s. The contact lasts 40 ms. Find the average force Favg of the wall on the ball.
A simple pendulum has a length of 1 m. Use g = 9.81 m/s². Find its period T for small swings.
A 1.6 kg mass on a spring with k = 390 N/m is pulled 4 cm from equilibrium and released. Find its maximum acceleration amax.
An object 2 cm tall stands 30 cm in front of a thin diverging lens with a focal length of 20 cm. Find the image height hi (negative for an inverted image).
Light of wavelength 600 nm falls at right angles on a grating with 400 lines per millimetre. Find the angle θ3 of the order-3 maximum.
Answer: s = 10.42 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 = 3.917 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.
Up is positive and the ground is h₀ below the starting point.
Only the positive root is physical.
Answer: F = 343.6 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: d = 1.482 m
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.
Both forces point down the slope; the mass cancels.
Answer: θ = 19.76 °
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 = 70.24°.
Answer: α = 63.82 °
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: T = 2.494 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.
Tension and gravity together give the inward force: tan θ = v²/(r·g).
Answer: vmin = 16.21 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.
Friction points up the slope.
Answer: v = 4.997 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 block drops a height d·sin θ while friction removes μmg·cos θ·d. The mass cancels.
Answer: x = 60.08 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′1 = 4.8 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: Favg = 170 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 = 2.006 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: amax = 9.75 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: hi = 0.8 cm
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: θ3 = 46.05 °
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, …
ExamProblems · examproblems.com/courses/physics-1-mechanics