Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 120 km/h sees an obstacle. Her reaction time is 1.5 s, after which the car brakes with a constant deceleration of 6 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A ball is thrown straight up at 17 m/s from the edge of a roof 35 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
A 30 kg sled is pulled across level snow by a rope at 20° above the horizontal with a force of 90 N. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the acceleration a of the sled.
A block is given a push so that it starts sliding up a ramp at 30° with a speed of 8.5 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 14 m/s and 30° above the horizontal towards a wall 5 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 18 m/s and 50° 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 5 kg ball hangs on a light string of length 3 m. It is pulled aside until the string makes 60° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A spring of stiffness 5000 N/m at the bottom of a 40° slope is compressed by 5 cm against a 1 kg block and released. The coefficient of kinetic friction between block and slope is 0.25. 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 20 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 20 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 0.45 kg ball hits a wall at 11 m/s and bounces straight back at 15 m/s. The contact lasts 20 ms. Find the average force Favg of the wall on the ball.
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of water, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of oil, with its top edge 1 m below the surface. Find the depth ycp of the centre of pressure below the surface.
Water drains from a large open tank through a sharp-edged orifice with a diameter of 25 mm. The water level is kept 2 m above the orifice and the discharge coefficient is 0.61. Find the flow rate Q. Use g = 9.81 m/s².
A closed water tank has compressed air above the water at a gauge pressure of 80 kPa. Water leaves through an opening with a diameter of 40 mm, 2.5 m below the water surface, into the open air. Ignoring losses, find the jet velocity v. Use g = 9.81 m/s².
A tank of 30 L contains 210 g of O₂ at 95 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of N₂ goes from 300 kPa, 2 L, 10 °C to 350 kPa and 210 °C. Find the new volume V2.
A heat engine works between a hot reservoir at 725 °C and a cold reservoir at 10 °C. It takes in 4100 kJ of heat. Find the most work Wmax it could possibly produce.
A refrigerator keeps its cold space at -25 °C in a kitchen at 30 °C, removing heat from the cold space at 0.5 kW. Find the minimum power input Ẇmin.
A pipe closed at one end is 0.6 m long. The speed of sound is 343 m/s. Find the frequency f3 of its harmonic number 3.
An optical fibre has a core with n₁ = 1.46 and a cladding with n₂ = 1.44. Find its numerical aperture NA.
Answer: d = 142.6 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: hmax = 49.73 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 = 1.941 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: d = 5.139 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 = 2.053 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 still rising at the wall.
Answer: H = 64.69 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 = 98.1 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) = 98.1 N.
Answer: d = 0.7636 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 = 200.1 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: Favg = 585 N
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 = 38.63 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 = 2.481 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: Q = 1.876 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.
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Answer: v = 14.46 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.
Answer: P = 669.5 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 2.925 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Wmax = 2937 kJ
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: Ẇmin = 0.1108 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.
The least work is needed by a reversible machine, which has the highest COP.
Answer: f3 = 428.8 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: NA = 0.2408
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.
ExamProblems · examproblems.com/courses/physics-for-engineers