Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 45 km/h sees an obstacle. Her reaction time is 0.6 s, after which the car brakes with a constant deceleration of 4.5 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A tram starts from rest and accelerates at 1 m/s² for 11 s, then runs at constant speed for 24 s, and finally brakes at 3 m/s² until it stops. Find the total distance s between the two stops.
A 25 kg box is pushed up a ramp at 20° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.4. Use g = 9.81 m/s². Find the pushing force F.
A 4 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 1.5 kg mass. The coefficient of kinetic friction between block and table is 0.2. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched at 50° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is launched at 20° above horizontal from the top of a 35 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A 2 kg ball hangs on a light string of length 2 m. It is pulled aside until the string makes 90° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A 0.5 kg block is released from rest at height 4.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 4 m long (μ = 0.15) and then runs onto a spring (k = 800 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 5 kg projectile moving horizontally at 8 m/s explodes into two fragments that move along the same line. The 4 kg fragment continues forwards at 55 m/s. Find the velocity v′2 of the 1 kg fragment (positive in the original direction).
A 1700 kg car runs into a parked 1800 kg car. The wrecks lock together and skid 8.5 m before stopping. The coefficient of kinetic friction between tyres and road is 0.6. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A vertical rectangular gate 2 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 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.
Water flows through a horizontal venturi meter with an inlet diameter of 250 mm and a throat diameter of 125 mm. The pressure at the inlet is 35 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
A hose with an inner diameter of 20 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 3 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the flow rate Q.
A tank of 5 L contains 100 g of CO₂ at 15 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of CH₄ at 850 kPa and 20 °C, treating it as an ideal gas.
A reversible (Carnot) heat engine takes in 1500 kJ of heat from a reservoir at 325 °C and rejects 748 kJ per cycle. Find the temperature TL of the cold reservoir in °C.
A Carnot heat engine receives 110 kW of heat from a source at 550 °C and rejects heat to the surroundings at 30 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at -5 °C and also rejects heat to the surroundings at 30 °C. Find the rate Q̇L at which heat is removed from the freezer.
One machine gives a sound level of 60 dB at a certain point. Find the level β at that point when 4 identical machines run at the same time (incoherent sources).
An optical fibre has a core with n₁ = 1.46 and a cladding with n₂ = 1.36. Find its numerical aperture NA.
Answer: d = 24.86 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 = 344.7 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: F = 176.1 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 = 1.249 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: H = 20.22 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.
At the top the vertical velocity is zero: vy² = 2gH.
Answer: t = 3.163 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: T = 58.86 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) = 58.86 N.
Answer: x = 21.87 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′2 = −180 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: v1 = 20.59 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 = 29.43 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 = 485.6 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: Q = 106 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 pipe is horizontal, so only pressure and velocity change.
Answer: Q = 2.41 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: P = 1089 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 5.594 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: TL = 25.13 °C
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 = 532.4 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: β = 66.02 dB
Waves: a periodic wave moves one wavelength in one period, so v = fλ with T = 1/f. On a stretched string v = √(T/μ), with T the tension and μ = m/L the mass per metre. Sound in air travels at about 343 m/s at 20 °C, light in vacuum at c = 3.00 × 10⁸ m/s. Loudness is measured on a logarithmic scale: β = 10·log₁₀(I/I₀) dB with I₀ = 10⁻¹² W/m², so 10 times the intensity adds 10 dB. A point source spreads its power P over a sphere: I = P/(4πr²).
Answer: NA = 0.531
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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