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 15 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 5 m/s². How long does it take the police car to catch up? Find t.
A ball is thrown straight up at 18 m/s from the edge of a roof 15 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 15 kg box is pushed up a ramp at 30° 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 7 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 5.5 kg mass. The coefficient of kinetic friction between block and table is 0.2. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched from level ground at 60° above the horizontal and lands 30 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 40° above horizontal from the top of a 30 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A car travelling at 30 km/h on a level road brakes hard and skids to a stop with its wheels locked. The coefficient of kinetic friction between the tyres and the road is 0.6. Find the stopping distance d.
A 4 kg block is released from rest at height 4.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3 m long (μ = 0.25) and then runs onto a spring (k = 800 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 0.5 kg glider moving at 5 m/s makes a perfectly elastic head-on collision with a 3 kg glider at rest on a frictionless track. Find the velocity v′2 of the second glider after the collision.
At a crossing a 1700 kg car travelling east at 20 m/s collides with a 2000 kg car travelling north at 18 m/s. The cars lock together. Find the direction θ of the wreck just after the collision, measured north of east.
A vertical rectangular gate 1.5 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 glycerin, with its top edge 1 m below the surface. Find the depth ycp of the centre of pressure below the surface.
Water flows through a horizontal venturi meter with an inlet diameter of 100 mm and a throat diameter of 60 mm. The pressure at the inlet is 53 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 25 mm while rising 5 m. At section 1 the pressure is 390 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 at section 2.
What volume V does 50 g of He occupy at 700 kPa and 25 °C? Treat it as an ideal gas.
Find the density ρ of Ar at 450 kPa and 90 °C, treating it as an ideal gas.
A heat engine works between a hot reservoir at 700 °C and a cold reservoir at 5 °C. Find the highest possible thermal efficiency ηmax.
A heat pump delivers 7 kW of heat to a heating system at 35 °C, taking heat from a source at -5 °C. Its actual COP is 30% of the Carnot COP between these temperatures. Find the electrical power Ẇ it uses.
A small source sends sound equally in all directions. The level is 100 dB at 5 m from it. Find the level β2 at 20 m, ignoring absorption and reflections.
An optical fibre has a core with n₁ = 1.48 and a cladding with n₂ = 1.38. Find its numerical aperture NA.
Answer: t = 6 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 = 30 m/s.
Answer: t = 4.37 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: F = 105.4 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: a = 3.218 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: v0 = 18.43 m/s
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 = 4.304 s
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: d = 5.899 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).
Doubling the speed makes the stopping distance four times longer.
Answer: x = 60.65 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: v′2 = 1.429 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: θ = 46.64 °
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 = 82.77 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 = 1.556 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: Q = 31.2 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 = 278.4 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 = 44.23 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 5.954 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηmax = 71.42 %
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 = 3.029 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: β2 = 87.96 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: NA = 0.5348
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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