Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 36 km/h sees an obstacle. Her reaction time is 0.9 s, after which the car brakes with a constant deceleration of 8 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 24 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. How long after launch does it hit the ground? Find t.
A 20 kg box is pushed up a ramp at 15° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.2. Use g = 9.81 m/s². Find the pushing force F.
A 1.5 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 acceleration a of the system.
A ball is launched at 45° 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 11 m/s and 40° above the horizontal from the top of a 25 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 block slides from rest down a slope at 25° to the horizontal. The coefficient of kinetic friction is 0.3. Find its speed v after it has slid 8 m along the slope.
A 1 kg block is released from rest at height 4.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 2.5 m long (μ = 0.3) and then runs onto a spring (k = 2500 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 1.5 kg glider moving at 4 m/s makes a perfectly elastic head-on collision with a 4 kg glider at rest on a frictionless track. Find the velocity v′1 of the first glider after the collision (positive in its original direction).
A 5 g bullet is fired horizontally into a 2.5 kg block hanging from light strings 1.8 m long, and stays in it. The strings swing out to a largest angle of 25° 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 4 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 3 m below the surface. Find the resultant hydrostatic force F on the gate.
Water drains from a large open tank through a sharp-edged orifice with a diameter of 50 mm. The water level is kept 2.5 m above the orifice and the discharge coefficient is 0.61. Find the flow rate Q. Use g = 9.81 m/s².
Water flows up through a pipe that narrows from 150 mm to 75 mm while rising 9 m. At section 1 the pressure is 370 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 10 L contains 100 g of CO₂ at 30 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of N₂ at 700 kPa and 30 °C, treating it as an ideal gas.
A reversible (Carnot) heat engine takes in 1900 kJ of heat from a reservoir at 800 °C and rejects 519 kJ per cycle. Find the temperature TL of the cold reservoir in °C.
A heat pump keeps a house at 21 °C while it is 0 °C outside, delivering 6 kW of heat to the house. Find the minimum power input Ẇmin.
An object stands 15 cm in front of a thin converging lens with a focal length of 20 cm. Find the image distance di. Use the sign convention 1/f = 1/do + 1/di (f < 0 for a diverging lens, di < 0 for a virtual image).
Light of wavelength 450 nm falls on two slits 0.15 mm apart. The screen is 3 m away. Find the distance y2 from the central bright fringe to bright fringe number 2.
Answer: d = 15.25 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: t = 12.7 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.
Solve −h₁ = v₁·t₂ − ½·g·t₂² for the positive root.
Answer: F = 88.68 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 = 6.466 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: t = 5.19 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: H = 27.55 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: v = 4.864 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 block drops a height d·sin θ while friction removes μmg·cos θ·d. The mass cancels.
Answer: x = 17.16 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′1 = −1.818 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.
The minus sign means the first glider bounces back.
Answer: v = 911.3 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 = 139.8 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 = 296.7 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: Q = 8.388 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 discharge coefficient accounts for the contraction of the jet and friction at the edge.
Answer: p2 = 279.8 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: P = 572.7 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 7.779 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: TL = 19.99 °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: Ẇmin = 0.4284 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: di = −60 cm
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: y2 = 18 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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