Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 60 km/h brakes with a constant deceleration of 4 m/s² until it stops. Find the braking distance s.
A tram starts from rest and accelerates at 1 m/s² for 6 s, then runs at constant speed for 15 s, and finally brakes at 3 m/s² until it stops. Find the total distance s between the two stops.
A 9 kg block slides down a slope at 25° to the horizontal. The coefficient of kinetic friction is 0.1. 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 25° with a speed of 9.5 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 32 m/s and must land 55 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 34 m/s and 35° above the horizontal from the top of a 10 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 2 kg ball hangs on a light string of length 1.5 m. It is pulled aside until the string makes 90° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A 1 kg block is released from rest at height 1 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 5 m long (μ = 0.1) and then runs onto a spring (k = 5000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 16 kg projectile moving horizontally at 32 m/s explodes into two fragments that move along the same line. The 12 kg fragment continues forwards at 82 m/s. Find the velocity v′2 of the 4 kg fragment (positive in the original direction).
A 1900 kg car runs into a parked 1600 kg car. The wrecks lock together and skid 8 m before stopping. The coefficient of kinetic friction between tyres and road is 0.8. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of water, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 2 m wide sits in the wall of a tank of glycerin, with its top edge 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
A pitot-static tube in a water pipe reads a difference of 10 kPa between the stagnation pressure and the static pressure. Find the water velocity v at the tip.
Water flows up through a pipe that narrows from 100 mm to 50 mm while rising 4 m. At section 1 the pressure is 280 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 20 L contains 330 g of CH₄ at 15 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of CO₂ at 700 kPa and 30 °C, treating it as an ideal gas.
A reversible (Carnot) heat engine takes in 2400 kJ of heat from a reservoir at 350 °C and rejects 1225 kJ per cycle. Find the temperature TL of the cold reservoir in °C.
A refrigerator keeps its cold space at -10 °C in a kitchen at 30 °C, removing heat from the cold space at 2 kW. Find the highest possible coefficient of performance COPmax.
An observer moves at 5 m/s towards a stationary source of 500 Hz. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.
Light of wavelength 589 nm passes through a single slit 0.12 mm wide. Find the width w of the central bright maximum on a screen 2.5 m away (between the first dark fringes on either side).
Answer: s = 34.72 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: s = 114 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 = 3.257 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 = 8.235 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 = 15.9 °
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 = 74.1°.
Answer: H = 29.38 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: T = 58.86 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.
Check: T = mg(3 − 2cos θ) = 58.86 N.
Answer: x = 4.429 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 = −118 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: v1 = 20.64 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: F = 66.22 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Answer: F = 83.43 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Answer: v = 4.472 m/s
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
The flow comes to rest at the tip, so its kinetic energy shows up as extra pressure.
Answer: p2 = 238.9 kPa
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
Work in Pa, then divide by 1000 to get kPa.
Answer: P = 2464 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 12.22 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: TL = 44.92 °C
Second-law limits: no cycle working between reservoirs at absolute temperatures TH and TL can beat the reversible (Carnot) cycle: ηmax = 1 − TL/TH, COPR,max = TL/(TH − TL) and COPHP,max = TH/(TH − TL). Energy is always conserved: QH = W + QL.
Answer: COPmax = 6.579
Second-law limits: no cycle working between reservoirs at absolute temperatures TH and TL can beat the reversible (Carnot) cycle: ηmax = 1 − TL/TH, COPR,max = TL/(TH − TL) and COPHP,max = TH/(TH − TL). Energy is always conserved: QH = W + QL.
Answer: f′ = 507.3 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: w = 24.54 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-for-engineers