Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 90 km/h brakes with a constant deceleration of 5.5 m/s² until it stops. How long does braking take? Find t.
A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 38 m/s² for 2 s, after which the engine stops and the rocket moves freely. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax the rocket reaches.
A 19 kg block slides down a slope at 25° to the horizontal. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 4 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 4 kg mass. The coefficient of kinetic friction between block and table is 0.25. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched at 30° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A ball is launched at 31 m/s and 35° above the horizontal from the top of a 45 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.
An 800 kg car drives up a 6° slope at a constant 54 km/h. Air resistance and rolling resistance together amount to 200 N. Find the power P the wheels must deliver, in kW.
A 5 kg block is released from rest at height 2.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3.5 m long (μ = 0.15) and then runs onto a spring (k = 800 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.
A 1700 kg car moving at 22 m/s runs into a stationary 1600 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?
A 0.1 kg ball hits a wall at 8 m/s and bounces straight back at 12 m/s. The contact lasts 5 ms. Find the magnitude of the impulse J the wall gives the ball.
A vertical rectangular gate 1 m wide sits in the wall of a tank of water, with its top edge 0.5 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 2 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 100 mm to 75 mm. In the wide section the pressure is 130 kPa and the velocity is 2.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
Water flows up through a pipe that narrows from 75 mm to 25 mm while rising 5 m. At section 1 the pressure is 140 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 5 L contains 460 g of CO₂ at 30 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of N₂ at 950 kPa and 10 °C, treating it as an ideal gas.
A reversible refrigerator keeps a cold room at 0 °C by removing 0.5 kW of heat, and rejects heat to the surroundings at 28 °C. Find the rate Q̇H at which it rejects heat.
A heat pump keeps a house at 21 °C while it is 7 °C outside, delivering 13 kW of heat to the house. Find the minimum power input Ẇmin.
A small source sends sound equally in all directions. The level is 60 dB at 5 m from it. Find the level β2 at 15 m, ignoring absorption and reflections.
Light of wavelength 589 nm falls on two slits 0.1 mm apart. The screen is 1.2 m away. Find the distance y2 from the central bright fringe to bright fringe number 2.
Answer: t = 4.545 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: hmax = 370.4 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.
At the top the velocity is zero.
Answer: a = 1.923 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: T = 24.52 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: x = 12.71 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.
Starting and landing at the same height: t = 2vy/g.
Answer: H = 61.11 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: P = 15.31 kW
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). Power is force times speed: P = F·v.
Answer: v = 6.225 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 = 199467 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 = 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 = 58.86 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 = 3.714 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 = 121.8 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 = 80.95 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 = 5269 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 11.3 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: Q̇H = 0.5513 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.
Power input: Ẇ = 0.05125 kW.
Answer: Ẇmin = 0.6187 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.
The least work is needed by a reversible machine, which has the highest COP.
Answer: β2 = 50.46 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).
Every doubling of the distance lowers the level by about 6 dB.
Answer: y2 = 14.14 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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