Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 80 km/h brakes with a constant deceleration of 7 m/s² until it stops. How long does braking take? Find t.
A ball is thrown straight up at 17 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 19 kg block slides down a slope at 45° 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 5 m/s. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is launched at 70° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A ball is launched at 20 m/s and 35° above the horizontal from the top of a 35 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 block slides from rest down a slope at 45° to the horizontal. The coefficient of kinetic friction is 0.25. Find its speed v after it has slid 17 m along the slope.
A spring of stiffness 1000 N/m at the bottom of a 15° slope is compressed by 15 cm against a 5 kg block and released. The coefficient of kinetic friction between block and slope is 0.15. How far d up the slope does the block travel from the release point before it stops? (Ignore the short distance over which the spring pushes, i.e. take friction and gravity over the whole distance d.)
A 15 kg projectile moving horizontally at 24 m/s explodes into two fragments that move along the same line. The 5 kg fragment continues forwards at 56 m/s. Find the velocity v′2 of the 10 kg fragment (positive in the original direction).
A 12 g bullet is fired horizontally into a 4 kg block hanging from light strings 1.3 m long, and stays in it. The strings swing out to a largest angle of 10° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 3 m wide sits in the wall of a tank of water, with its top edge 4 m below the surface. Find the depth ycp of the centre of pressure below the surface.
A pitot-static tube in a water pipe reads a difference of 22 kPa between the stagnation pressure and the static pressure. Find the water velocity v at the tip.
A hose with an inner diameter of 25 mm siphons water from an open tank. The highest point C of the hose is 0.5 m above the water surface, and the outlet (open to the air) is 4 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the absolute pressure pC at point C. The atmospheric pressure is 101.3 kPa.
A tank of 20 L contains 220 g of He at 35 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of Ar at 450 kPa and 160 °C, treating it as an ideal gas.
A heat engine works between a hot reservoir at 625 °C and a cold reservoir at 40 °C. Find the highest possible thermal efficiency ηmax.
A Carnot heat engine receives 190 kW of heat from a source at 1000 °C and rejects heat to the surroundings at 27 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at -20 °C and also rejects heat to the surroundings at 27 °C. Find the rate Q̇L at which heat is removed from the freezer.
An object stands 4 cm in front of a thin converging lens with a focal length of 8 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 500 nm falls at right angles on a grating with 800 lines per millimetre. Find the angle θ2 of the order-2 maximum.
Answer: t = 3.175 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.
Answer: t = 4.753 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 = 4.856 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 = 1.677 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: x = 67.1 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.
Starting and landing at the same height: t = 2vy/g.
Answer: α = 60.2 °
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: v = 13.3 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: d = 0.5681 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′2 = 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.
Answer: v = 208.1 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 = 36.79 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: ycp = 4.519 m
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.
Pressure grows with depth, so the resultant acts below the centroid.
Answer: v = 6.633 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: pC = 57.15 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.
Surface and outlet are both open to the air, so the pressure terms cancel.
If this pressure falls to the vapour pressure of the water, the siphon stops working.
Answer: P = 7040 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 4.992 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηmax = 65.13 %
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: Q̇L = 782.1 kW
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: di = −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: θ2 = 53.13 °
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