Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 50 km/h brakes with a constant deceleration of 4.5 m/s² until it stops. Find the braking distance s.
A tram starts from rest and accelerates at 1 m/s² for 9 s, then runs at constant speed for 29 s, and finally brakes at 2 m/s² until it stops. Find the total distance s between the two stops.
An 11 kg block slides down a slope at 45° to the horizontal. The coefficient of kinetic friction is 0.15. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A block is given a push so that it starts sliding up a ramp at 20° with a speed of 7.5 m/s. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is launched at 25° above horizontal from level ground. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A ball is launched at 10 m/s and 25° above the horizontal from the top of a 60 m cliff. Ignore air resistance and use g = 9.81 m/s². Find the speed v with which it hits the ground.
A 1 kg ball hangs on a light string of length 1 m. It is pulled aside until the string makes 45° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A 10 kg block starts from rest at the bottom of a 15° slope and is pulled up the slope by a constant force of 45 N parallel to the slope. The coefficient of kinetic friction is 0.1. Find the speed v of the block after it has moved 7 m up the slope.
A 3 kg glider moving at 2 m/s makes a perfectly elastic head-on collision with a 2.5 kg glider at rest on a frictionless track. Find the velocity v′2 of the second glider after the collision.
A 2000 kg car runs into a parked 1500 kg car. The wrecks lock together and skid 4.5 m before stopping. The coefficient of kinetic friction between tyres and road is 0.8. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A vertical rectangular gate 3 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 1 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.
Water flows through a horizontal pipe that narrows from 100 mm to 75 mm. In the wide section the pressure is 100 kPa and the velocity is 1.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
Water flows up through a pipe that narrows from 150 mm to 100 mm while rising 10 m. At section 1 the pressure is 150 kPa and the velocity is 1 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 50 L contains 320 g of He at 75 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of O₂ at 950 kPa and 70 °C, treating it as an ideal gas.
A heat engine works between a hot reservoir at 550 °C and a cold reservoir at 25 °C. Find the highest possible thermal efficiency ηmax.
A heat pump delivers 10.5 kW of heat to a heating system at 21 °C, taking heat from a source at -5 °C. Its actual COP is 40% of the Carnot COP between these temperatures. Find the electrical power Ẇ it uses.
Light inside a transparent block is totally reflected at the boundary with ethanol (n = 1.36) for every angle of incidence above 30°. Find the refractive index n1 of the block.
Light of wavelength 450 nm falls on two slits 0.5 mm apart. The screen is 3 m away. Find the distance y1 from the central bright fringe to bright fringe number 1.
Answer: s = 21.43 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: s = 321.8 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 = 5.896 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: d = 4.969 m
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.
Both forces point down the slope; the mass cancels.
Answer: t = 2.24 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.
Starting and landing at the same height: t = 2vy/g.
Answer: v = 35.74 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.
Energy or v² = v₀² + 2as in the vertical direction.
Check with energy: v² = v₀² + 2gh.
Answer: T = 15.56 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) = 15.56 N.
Answer: v = 3.767 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 = 2.182 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: v1 = 14.71 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 = 88.29 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 = 24.52 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: p2 = 96.69 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 = 49.87 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 = 4628 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 10.66 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηmax = 63.78 %
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 = 2.32 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: n1 = 2.72
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: y1 = 2.7 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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