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 4.5 m/s² until it stops. How long does braking take? Find t.
Ball A is dropped from rest from a height of 40 m. At the same instant ball B is thrown straight up from the ground directly below at 24 m/s. Ignore air resistance and use g = 9.81 m/s². At what height y above the ground do they meet?
A 20 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 0.25. Use g = 9.81 m/s². Find the pushing force F.
A 2 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 6 kg mass. The coefficient of kinetic friction between block and table is 0.3. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched from level ground at 40 m/s and must land 95 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.
A ball is launched at 20° above horizontal from the top of a 55 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A 0.2 kg ball hangs on a light string of length 0.5 m. It is pulled aside until the string makes 30° 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 15° slope is compressed by 5 cm against a 3 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.)
An 11 kg projectile moving horizontally at 35 m/s explodes into two fragments that move along the same line. The 9 kg fragment continues forwards at 56 m/s. Find the velocity v′2 of the 2 kg fragment (positive in the original direction).
At a crossing a 1700 kg car travelling east at 15 m/s collides with a 1100 kg car travelling north at 17 m/s. The cars lock together. Find the speed v′ of the wreck just after 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 oil, with its top edge 2.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
Water flows through a horizontal pipe that narrows from 200 mm to 100 mm. In the wide section the pressure is 140 kPa and the velocity is 0.5 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 4 m. At section 1 the pressure is 360 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 20 L contains 400 g of N₂ at 100 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of Ar at 450 kPa and 80 °C, treating it as an ideal gas.
A heat engine works between a hot reservoir at 575 °C and a cold reservoir at 35 °C. It takes in 2800 kJ of heat. Find the most work Wmax it could possibly produce.
A heat pump keeps a house at 22 °C while it is 5 °C outside, delivering 7 kW of heat to the house. Find the highest possible coefficient of performance COPmax.
An object 2 cm tall stands 4 cm in front of a thin converging lens with a focal length of 25 cm. Find the image height hi (negative for an inverted image).
Light of wavelength 480 nm falls on two slits 0.4 mm apart. The screen is 3 m away. Find the spacing Δy between neighbouring bright fringes.
Answer: t = 2.222 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: y = 26.38 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 = 127.4 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 = 19.13 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: θ = 17.81 °
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 other solution is 90° − θ = 72.19°.
Answer: x = 68.24 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.
Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.
Answer: T = 2.488 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) = 2.488 N.
Answer: d = 0.4245 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 = −59.5 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: v′ = 11.29 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 = 78.48 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 = 3.308 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: p2 = 138.1 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 = 303.9 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 = 2215 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 6.123 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: Wmax = 1783 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: COPmax = 17.36
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: hi = 2.381 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.
Answer: Δy = 3.6 mm
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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