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 ball is thrown straight up at 8 m/s from the edge of a roof 30 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 4 kg block slides down a slope at 35° to the horizontal. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A block is given a push so that it starts sliding up a ramp at 30° with a speed of 9 m/s. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is kicked from level ground at 28 m/s and 35° above the horizontal towards a wall 45 m away. Ignore air resistance and use g = 9.81 m/s². At what height y does the ball pass the wall?
A ball is launched at 11 m/s and 45° above the horizontal from the top of a 55 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 140 m. The coefficient of static friction between tyres and road is 0.3. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A curve of radius 100 m is banked at 20°, and the coefficient of static friction between tyres and road is 0.3. 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 2.5 kg block slides 8 m down a slope at 35° to the horizontal, with a coefficient of kinetic friction of 0.35. Find the work Wf done against friction.
A 2.5 kg block is released from rest at height 2.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 1.5 m long (μ = 0.1) and then runs onto a spring (k = 2500 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 4.5 kg glider moving at 3 m/s makes a perfectly elastic head-on collision with a 2.5 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 0.1 kg ball hits a wall at 18 m/s and bounces straight back at 8 m/s. The contact lasts 5 ms. Find the magnitude of the impulse J the wall gives the ball.
How long must a simple pendulum be to have a period of 1.9 s? Use g = 9.81 m/s². Find L.
A 1.6 kg mass on a spring with k = 130 N/m is pulled 3 cm from equilibrium and released. Find its maximum acceleration amax.
An object stands 15 cm in front of a thin converging lens with a focal length of 10 cm. Find the image distance di. Use the sign convention 1/f = 1/do + 1/di (f < 0 for a diverging lens, di < 0 for a virtual image).
Light of wavelength 600 nm passes through a single slit 0.05 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: t = 3.42 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: a = 3.216 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.
Down the slope is positive.
Answer: d = 5.762 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: y = 12.63 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.
The ball is already on its way down at the wall.
Answer: H = 58.08 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 = 20.3 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 = 7.522 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: Wf = 56.25 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 = 21.47 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 = 0.8571 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: J = 2.6 N·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: L = 0.897 m
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 = 2.437 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: di = 30 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: θ1 = 0.6876 °
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