Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 36 km/h sees an obstacle. Her reaction time is 1.1 s, after which the car brakes with a constant deceleration of 8 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A tram starts from rest and accelerates at 1.5 m/s² for 11 s, then runs at constant speed for 18 s, and finally brakes at 4.5 m/s² until it stops. Find the total distance s between the two stops.
A 13 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 6 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 3 kg mass. The coefficient of kinetic friction between block and table is 0.25. Use g = 9.81 m/s². Find the tension T in the string.
A ball is kicked from level ground at 40 m/s and 65° above the horizontal towards a wall 87 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 40° above horizontal from the top of a 15 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A toy launcher has a vertical spring of stiffness 1000 N/m. A 0.2 kg ball is placed on it and the spring is compressed by 8 cm, then released. Ignoring air resistance and the spring’s mass, find the height h the ball rises, measured from its position when the spring was compressed.
A 10 kg cart is released from rest at a height of 4.8 m on a smooth track that leads into a vertical circular loop of radius 1.5 m. Find the speed v of the cart at the top of the loop.
A 1400 kg car moving at 16 m/s runs into a stationary 1800 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?
A 0.35 kg ball hits a wall at 9 m/s and bounces straight back at 13 m/s. The contact lasts 5 ms. Find the magnitude of the impulse J the wall gives the ball.
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of water, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 1 m wide sits in the wall of a tank of glycerin, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
Water flows through a horizontal pipe that narrows from 100 mm to 75 mm. In the wide section the pressure is 260 kPa and the velocity is 1.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
A hose with an inner diameter of 25 mm siphons water from an open tank. The highest point C of the hose is 3 m above the water surface, and the outlet (open to the air) is 3.5 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the flow rate Q.
What volume V does 240 g of O₂ occupy at 550 kPa and 20 °C? Treat it as an ideal gas.
Find the density ρ of Ar at 900 kPa and 60 °C, treating it as an ideal gas.
A heat engine works between a hot reservoir at 625 °C and a cold reservoir at 25 °C. Find the highest possible thermal efficiency ηmax.
A heat engine takes in 3700 kJ of heat from a reservoir at 700 °C, produces 1998 kJ of work and rejects the rest to a reservoir at 40 °C, per cycle. Find the total entropy generated Sgen per cycle.
An object stands 40 cm in front of a thin converging lens with a focal length of 12 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 450 nm falls at right angles on a grating with 600 lines per millimetre. Find the angle θ1 of the order-1 maximum.
Answer: d = 17.25 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 = 418 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 = 3.216 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: T = 24.52 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: y = 56.66 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: x = 125 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: h = 1.631 m
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: v = 5.943 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). 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: ΔK = 100800 J
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 lost energy goes into deformation, heat and sound.
Answer: J = 7.7 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: F = 71.74 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 = 148.3 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: p2 = 256.7 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: Q = 4.068 L/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.
Surface and outlet are both open to the air, so the pressure terms cancel.
Answer: V = 33.24 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 12.98 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηmax = 66.8 %
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: Sgen = 1.633 kJ/K
Entropy generation: over a full cycle the working fluid has no net entropy change. The hot reservoir loses QH/TH and the cold reservoir gains QL/TL. Their sum is the entropy generated, which is zero for a reversible cycle and positive for every real one.
Sgen > 0, so the engine is possible but irreversible.
Answer: di = 17.14 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 = 15.66 °
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