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 16 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.
A tram starts from rest and accelerates at 2 m/s² for 12 s, then runs at constant speed for 25 s, and finally brakes at 3.5 m/s² until it stops. Find the total distance s between the two stops.
A 15 kg sled is pulled across level snow by a rope at 40° above the horizontal with a force of 130 N. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². Find the acceleration a of the sled.
A 1.5 kg block on a ramp at 40° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 5 kg mass. The hanging mass moves down. The coefficient of kinetic friction between block and ramp is 0.1. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched at 20° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is launched at 55° 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 takes a flat curve of radius 135 m. The coefficient of static friction between tyres and road is 0.6. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A 1.8 kg ball on a light string 1.1 m long is swung in a vertical circle. At the lowest point its speed is 8 m/s. Ignore air resistance and use g = 9.81 m/s². Find the tension Ttop in the string at the highest point.
A 1 kg block slides 5 m down a slope at 45° to the horizontal, with a coefficient of kinetic friction of 0.3. Find the work Wf done against friction.
A 0.5 kg cart is released from rest at a height of 16 m on a smooth track that leads into a vertical circular loop of radius 5 m. Find the normal force N the track exerts on the cart at the top of the loop.
A 12 g bullet is fired horizontally into a 2 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 3 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
At a crossing a 1700 kg car travelling east at 8 m/s collides with a 1300 kg car travelling north at 13 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 1.3 m long has a period of 1.44 s for small swings. Find the gravitational acceleration g on that planet.
A 2.5 kg mass on a spring with k = 310 N/m is pulled 20 cm from equilibrium and released. Find its maximum speed vmax.
A source of 400 Hz moves at 25 m/s towards a stationary observer. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.
Light of wavelength 600 nm falls on two slits 0.25 mm apart. The screen is 1.5 m away. Find the distance y2 from the central bright fringe to bright fringe number 2.
Answer: d = 85.33 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 = 32 m/s.
Answer: s = 826.3 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.
Answer: a = 5.579 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 = 19.46 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: H = 3.434 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.
At the top the vertical velocity is zero: vy² = 2gH.
Answer: x = 104.7 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 = 28.19 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: Ttop = 16.44 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.
Kinetic energy is traded for potential energy over a height 2r.
Tension and gravity both point down, towards the centre.
Answer: Wf = 10.41 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: N = 6.867 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.
The cart just makes it round when h = 2.5R; then N = 0 at the top.
Answer: v = 128.6 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: θ = 51.18 °
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 = 24.75 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: vmax = 2.227 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: f′ = 431.4 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: y2 = 7.2 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, …
ExamProblems · examproblems.com/courses/physics-1-mechanics