Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 36 km/h brakes with a constant deceleration of 3.5 m/s² until it stops. How long does braking take? Find t.
A tram starts from rest and accelerates at 2.5 m/s² for 10 s, then runs at constant speed for 22 s, and finally brakes at 2 m/s² until it stops. Find the total distance s between the two stops.
A 40 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 0.25. Use g = 9.81 m/s². Find the pushing force F.
A 9 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 4.5 kg mass. The coefficient of kinetic friction between block and table is 0.1. Use g = 9.81 m/s². Find the tension T in the string.
A ball is kicked from level ground at 38 m/s and 60° above the horizontal towards a wall 76 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 23 m/s and 20° 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 2.5 kg block slides 5 m down a slope at 35° to the horizontal, with a coefficient of kinetic friction of 0.3. Find the work Wf done against friction.
A 2 kg block is released from rest at height 2 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 2.5 m long (μ = 0.25) and then runs onto a spring (k = 1000 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.
An 800 kg car moving at 14 m/s runs into a stationary 1800 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?
A 0.15 kg ball hits a wall at 30 m/s and bounces straight back at 7 m/s. The contact lasts 10 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 3 m wide sits in the wall of a tank of glycerin, with its top edge 1 m below the surface. Find the depth ycp of the centre of pressure below the surface.
Water flows at 21 L/s through a horizontal pipe that narrows from 100 mm to 75 mm. The pressure in the narrow section is 240 kPa. Ignoring losses, find the pressure p1 in the wide section.
Water flows up through a pipe that narrows from 150 mm to 75 mm while rising 6 m. At section 1 the pressure is 270 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 50 L contains 320 g of He at 40 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of O₂ goes from 300 kPa, 2 L, 35 °C to 750 kPa and 140 °C. Find the new volume V2.
A reversible (Carnot) heat engine takes in 1000 kJ of heat from a reservoir at 675 °C and rejects 288 kJ per cycle. Find the temperature TL of the cold reservoir in °C.
A heat pump delivers 14.5 kW of heat to a heating system at 21 °C, taking heat from a source at 0 °C. Its actual COP is 30% of the Carnot COP between these temperatures. Find the electrical power Ẇ it uses.
Light inside a transparent block is totally reflected at the boundary with air (n = 1.00) for every angle of incidence above 38°. Find the refractive index n1 of the block.
Two slits 0.2 mm apart are lit with monochromatic light. On a screen 1.2 m away, bright fringe number 3 lies 9.9 mm from the central fringe. Find the wavelength λ.
Answer: t = 2.857 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: s = 831.3 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: F = 226.4 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: T = 32.37 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: y = 53.16 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 already on its way down at the wall.
Answer: H = 58.15 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: Wf = 30.13 J
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 = 5.194 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 ramp is smooth, so only the strip removes energy.
Answer: ΔK = 54277 J
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 lost energy goes into deformation, heat and sound.
Answer: J = 5.55 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 = 294.3 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 = 2.8 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: p1 = 247.7 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.
Answer: p2 = 194.3 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: P = 4163 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 1.073 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: TL = −0.0828 °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: Ẇ = 3.451 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: n1 = 1.624
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: λ = 550 nm
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, …
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