Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 54 km/h sees an obstacle. Her reaction time is 0.5 s, after which the car brakes with a constant deceleration of 5 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A ball is thrown straight up at 19 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 20 kg sled is pulled across level snow by a rope at 15° above the horizontal with a force of 110 N. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the sled.
A 2.5 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 0.3. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched at 20° 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 55° above horizontal from the top of a 10 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A block slides from rest down a slope at 25° to the horizontal. The coefficient of kinetic friction is 0.15. Find its speed v after it has slid 14 m along the slope.
A 2.5 kg block is released from rest at height 3 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 2 m long (μ = 0.3) and then runs onto a spring (k = 1000 N/m) on a smooth surface. Find the maximum compression x of the spring.
An 8 kg projectile moving horizontally at 12 m/s explodes into two fragments that move along the same line. The 4 kg fragment continues forwards at 53 m/s. Find the velocity v′2 of the 4 kg fragment (positive in the original direction).
A 1800 kg car runs into a parked 900 kg car. The wrecks lock together and skid 21.5 m before stopping. The coefficient of kinetic friction between tyres and road is 0.75. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A vertical rectangular gate 2 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 1 m wide sits in the wall of a tank of glycerin, with its top edge 1.5 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 100 mm to 75 mm. In the wide section the pressure is 230 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
A closed water tank has compressed air above the water at a gauge pressure of 40 kPa. Water leaves through an opening with a diameter of 25 mm, 2.5 m below the water surface, into the open air. Ignoring losses, find the jet velocity v. Use g = 9.81 m/s².
A tank of 20 L contains 20 g of CO₂ at 35 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of O₂ goes from 100 kPa, 20 L, 15 °C to 600 kPa and 90 °C. Find the new volume V2.
A heat engine works between a hot reservoir at 200 °C and a cold reservoir at 10 °C. It takes in 3100 kJ of heat. Find the most work Wmax it could possibly produce.
A heat pump keeps a house at 22 °C while it is 7 °C outside, delivering 13 kW of heat to the house. Find the minimum power input Ẇmin.
Light travels in acrylic (n = 1.49) towards a flat boundary with air (n = 1.00). Find the critical angle θc beyond which it is totally reflected.
A sensor at rest sends ultrasound of 25 kHz towards a car that moves away at 25 m/s and mixes the echo with the emitted signal. Find the beat frequency fb (speed of sound 343 m/s).
Answer: d = 30 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 = 4.546 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: a = 2.797 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 = 4.905 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 = 0.6973 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: t = 5.078 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: v = 8.874 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 = 34.31 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 = −29 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.
The minus sign means the second fragment flies backwards.
Answer: v1 = 26.68 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 = 36.79 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 = 2.333 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 = 229.4 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 = 11.36 m/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.
Answer: P = 58.21 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 4.201 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Wmax = 1245 kJ
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 = 0.6607 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: θc = 42.16 °
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: fb = 3397 Hz
Standing waves and the Doppler effect: a string fixed at both ends, or a pipe open at both ends, fits whole half-wavelengths: L = nλ/2, fn = n·v/(2L), n = 1, 2, 3, … A pipe closed at one end has a node at the closed end and an antinode at the open end: L = nλ/4 with odd n only, fn = n·v/(4L). When source or observer move along the line between them, f′ = f·(v ± vo)/(v ∓ vs): use the upper signs for motion towards each other. Two waves of nearly equal frequency beat at |f₁ − f₂|.
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