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 3 m/s² until it stops. Find the braking distance s.
Ball A is dropped from rest from a height of 20 m. At the same instant ball B is thrown straight up from the ground directly below at 22 m/s. Ignore air resistance and use g = 9.81 m/s². At what height y above the ground do they meet?
A 25 kg box is pushed up a ramp at 25° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the pushing force F.
A 6.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 3.5 kg mass. The coefficient of kinetic friction between block and table is 0.3. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched from level ground at 50° above the horizontal and lands 145 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 30° above horizontal from the top of a 30 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A car travelling at 110 km/h on a level road brakes hard and skids to a stop with its wheels locked. The coefficient of kinetic friction between the tyres and the road is 0.6. Find the stopping distance d.
A 2.5 kg block is released from rest at height 1.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 1.5 m long (μ = 0.2) and then runs onto a spring (k = 5000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 1.5 kg glider moving at 3 m/s makes a perfectly elastic head-on collision with a 5 kg glider at rest on a frictionless track. Find the velocity v′1 of the first glider after the collision (positive in its original direction).
A 0.6 kg ball hits a wall at 28 m/s and bounces straight back at 19 m/s. The contact lasts 25 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 2.5 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 oil, with its top edge 3 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 320 kPa and the velocity is 2.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
Water flows up through a pipe that narrows from 100 mm to 50 mm while rising 8 m. At section 1 the pressure is 150 kPa and the velocity is 2 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 20 L contains 120 g of Ar at 70 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of CH₄ goes from 150 kPa, 5 L, 20 °C to 700 kPa and 110 °C. Find the new volume V2.
A heat engine receives heat from a source at 600 °C and rejects heat to the surroundings at 25 °C. Its actual thermal efficiency is 26%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
A heat engine takes in 800 kJ of heat from a reservoir at 550 °C, produces 256 kJ of work and rejects the rest to a reservoir at 25 °C, per cycle. Find the total entropy generated Sgen per cycle.
A pipe closed at one end must have its harmonic number 3 at 396 Hz. The speed of sound is 343 m/s. Find the length L of the pipe.
Two thin lenses share an axis, 10 cm apart: lens 1 with f₁ = 15 cm and lens 2 with f₂ = 12 cm. An object stands 40 cm in front of lens 1. Find the position di2 of the final image measured from lens 2 (positive beyond lens 2, real-is-positive convention).
Answer: s = 46.3 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: y = 15.95 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.
Up is positive, the ground is y = 0.
Answer: F = 170.3 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.
Constant speed: the forces balance.
Answer: a = 1.521 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: v0 = 38.01 m/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.
Answer: t = 4.593 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: d = 79.31 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).
Doubling the speed makes the stopping distance four times longer.
Answer: x = 10.85 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′1 = −1.615 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.
Solving momentum and energy conservation together gives these two results.
The minus sign means the first glider bounces back.
Answer: J = 28.2 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 = 88.29 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 = 183.4 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 = 307.3 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 = 41.52 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 = 428.5 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 1.4 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: ηII = 39.48 %
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 = 0.8527 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: L = 0.6496 m
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: di2 = 6.462 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.
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