Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A plane starts from rest and accelerates uniformly at 3.5 m/s² along a runway. Find its speed v after 1900 m.
A ball is thrown straight up at 16 m/s from the edge of a roof 40 m above the ground. Ignore air resistance and use g = 9.81 m/s². How long after the throw does it hit the ground? Find t.
A 20 kg block slides down a slope at 35° to the horizontal. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 1 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 4.5 kg mass. The table is frictionless. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched at 45° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is launched at 22 m/s and 60° above the horizontal from the top of a 30 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 1 kg block slides 20 m down a slope at 40° to the horizontal, with a coefficient of kinetic friction of 0.2. Find the work Wf done against friction.
A 10 kg block starts from rest at the bottom of a 30° slope and is pulled up the slope by a constant force of 120 N parallel to the slope. The coefficient of kinetic friction is 0.3. Find the speed v of the block after it has moved 5 m up the slope.
A 1.5 kg glider moving at 2 m/s makes a perfectly elastic head-on collision with a 4.5 kg glider at rest on a frictionless track. Find the velocity v′2 of the second glider after the collision.
A 0.5 kg ball hits a wall at 22 m/s and bounces straight back at 10 m/s. The contact lasts 50 ms. Find the average force Favg of the wall on the ball.
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of water, with its top edge 2 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 2.5 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 41 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
Water flows up through a pipe that narrows from 75 mm to 50 mm while rising 1 m. At section 1 the pressure is 220 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure p2 at section 2.
What volume V does 230 g of He occupy at 100 kPa and 30 °C? Treat it as an ideal gas.
Find the density ρ of CH₄ at 550 kPa and 180 °C, treating it as an ideal gas.
A reversible (Carnot) heat engine takes in 3500 kJ of heat from a reservoir at 375 °C and rejects 1529 kJ per cycle. Find the temperature TL of the cold reservoir in °C.
A refrigerator keeps its cold space at -10 °C in a kitchen at 30 °C, removing heat from the cold space at 4.5 kW. Find the highest possible coefficient of performance COPmax.
A source of 800 Hz moves at 34 m/s away from a stationary observer. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.
An optical fibre has a core with n₁ = 1.55 and a cladding with n₂ = 1.54. Find the largest angle θa to the fibre axis at which light from air can enter the end of the fibre and still be guided by total internal reflection.
Answer: v = 115.3 m/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: t = 4.92 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.
Up is positive and the ground is h₀ below the starting point.
Only the positive root is physical.
Answer: a = 3.216 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.
Down the slope is positive.
Answer: a = 8.026 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 = 29.46 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: H = 48.5 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 = 30.06 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 = 6.743 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).
Answer: v′2 = 1 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.
Answer: Favg = 320 N
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 = 60.7 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 = 347.6 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 = 114.8 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: p2 = 205.6 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 = 1448 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 2.342 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: TL = 9.999 °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: COPmax = 6.579
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: f′ = 727.9 Hz
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: θa = 10.12 °
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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