Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car speeding at a constant 33 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 ball is thrown straight up at 5 m/s from the edge of a roof 15 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
A 20 kg sled is pulled across level snow by a rope at 40° above the horizontal with a force of 100 N. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the sled.
A 2.5 kg block on a ramp at 25° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 5.5 kg mass. The hanging mass moves down. The coefficient of kinetic friction between block and ramp is 0.2. 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 time of flight t until the ball hits the ground.
A ball is launched at 40° above horizontal from the top of a 15 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A 1200 kg car drives up a 3° slope at a constant 54 km/h. Air resistance and rolling resistance together amount to 800 N. Find the power P the wheels must deliver, in kW.
A spring of stiffness 2000 N/m at the bottom of a 25° slope is compressed by 8 cm against a 0.5 kg block and released. The coefficient of kinetic friction between block and slope is 0.3. 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 12 kg projectile moving horizontally at 23 m/s explodes into two fragments that move along the same line. The 6 kg fragment continues forwards at 68 m/s. Find the velocity v′2 of the 6 kg fragment (positive in the original direction).
A 0.35 kg ball hits a wall at 12 m/s and bounces straight back at 17 m/s. The contact lasts 40 ms. Find the magnitude of the impulse J the wall gives the ball.
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 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 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
Water flows through a horizontal pipe that narrows from 75 mm to 50 mm. In the wide section the pressure is 290 kPa and the velocity is 2.25 m/s. Ignoring losses, find the pressure p2 in the narrow section.
Water flows up through a pipe that narrows from 150 mm to 100 mm while rising 4 m. At section 1 the pressure is 370 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 at section 2.
What volume V does 370 g of CO₂ occupy at 150 kPa and 85 °C? Treat it as an ideal gas.
A fixed amount of CH₄ goes from 150 kPa, 20 L, 35 °C to 200 kPa and 80 °C. Find the new volume V2.
A reversible refrigerator keeps a cold room at -5 °C by removing 3.5 kW of heat, and rejects heat to the surroundings at 20 °C. Find the rate Q̇H at which it rejects heat.
A refrigerator keeps its cold space at -10 °C in a kitchen at 28 °C, removing heat from the cold space at 2 kW. Find the highest possible coefficient of performance COPmax.
Light travels in acrylic (n = 1.49) towards a flat boundary with glycerol (n = 1.47). Find the critical angle θc beyond which it is totally reflected.
An optical fibre has a core with n₁ = 1.52 and a cladding with n₂ = 1.44. Find its numerical aperture NA.
Answer: d = 363 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 = 66 m/s.
Answer: hmax = 16.27 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 = 1.851 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 = 27.04 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: t = 2.371 s
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: t = 2.797 s
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: P = 21.24 kW
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). Power is force times speed: P = F·v.
Answer: d = 1.879 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 = −22 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: J = 10.15 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 = 209.7 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 = 38.85 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 = 279.7 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: p2 = 329.6 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: V = 166.9 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 17.19 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Q̇H = 3.826 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.
Power input: Ẇ = 0.3263 kW.
Answer: COPmax = 6.925
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: θc = 80.6 °
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: NA = 0.4866
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.
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