Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car speeding at a constant 19 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 4.5 m/s². How long does it take the police car to catch up? Find t.
A ball is thrown straight up at 8 m/s from the edge of a roof 25 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
A 15 kg sled is pulled across level snow by a rope at 30° above the horizontal with a force of 90 N. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the acceleration a of the sled.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 65 N pulls the front 6 kg block; the 11 kg block is towed behind. The coefficient of kinetic friction is 0.15 for both blocks. Use g = 9.81 m/s². Find the acceleration a of the blocks.
A ball is launched at 45° 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 33 m/s and 30° above the horizontal from the top of a 60 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 ball hangs on a light string of length 0.8 m. It is pulled aside until the string makes 70° 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 1 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3.5 m long (μ = 0.1) and then runs onto a spring (k = 500 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 1600 kg car moving at 18 m/s runs into a stationary 900 kg car and the two lock together. How much kinetic energy ΔK is lost in the collision?
A 15 g bullet is fired horizontally into a 3.5 kg block hanging from light strings 1.3 m long, and stays in it. The strings swing out to a largest angle of 15° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 2.5 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 depth ycp of the centre of pressure below the surface.
Water flows through a horizontal pipe that narrows from 75 mm to 25 mm. In the wide section the pressure is 130 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 50 mm and a gauge pressure of 80 kPa. The nozzle exit diameter is 20 mm, and the height difference across the nozzle is negligible. Ignoring losses and air resistance, find the height H the jet rises above the nozzle. Use g = 9.81 m/s².
A tank of 10 L contains 210 g of He at 55 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of He at 900 kPa and 90 °C, treating it as an ideal gas.
A heat engine receives heat from a source at 650 °C and rejects heat to the surroundings at 35 °C. Its actual thermal efficiency is 37%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
Two Carnot engines work in series. Engine A receives heat from a reservoir at 600 °C and rejects heat to an intermediate reservoir at temperature Tm. Engine B takes all that heat and rejects heat to a reservoir at 20 °C. Both engines have the same thermal efficiency. Find Tm in °C.
One machine gives a sound level of 72 dB at a certain point. Find the level β at that point when 8 identical machines run at the same time (incoherent sources).
Light of wavelength 500 nm passes through a single slit 0.1 mm wide. Find the angle θ3 of dark fringe number 3 from the straight-through direction.
Answer: t = 8.444 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.
At that moment the police car moves at 2v = 38 m/s.
Answer: hmax = 28.26 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: a = 4.515 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: a = 2.352 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: x = 104.4 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 = 73.88 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: T = 22.72 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) = 22.72 N.
Answer: x = 11.29 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: ΔK = 93312 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: v = 218.5 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.
Answer: F = 95.65 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 = 3.595 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 = 40 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: H = 8.369 m
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.
Bernoulli along the jet: at the top the velocity is zero and the pressure is atmospheric again.
Answer: P = 14313 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 1.193 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηII = 55.54 %
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: Tm = 232.8 °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.
1 − Tm/TH = 1 − TL/Tm gives Tm² = TH·TL.
Answer: β = 81.03 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: θ3 = 0.8595 °
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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