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 m/s² until it stops. Find the braking distance s.
A ball is thrown straight up at 9 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.
An 11 kg block slides down a slope at 40° to the horizontal. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 2.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 1.5 kg mass. The coefficient of kinetic friction between block and table is 0.2. Use g = 9.81 m/s². Find the tension T in the string.
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 16 m/s and 35° above the horizontal from the top of a 35 m cliff. Ignore air resistance and use g = 9.81 m/s². At what angle α below the horizontal does it hit the ground?
A 1 kg block slides 16 m down a slope at 45° to the horizontal, with a coefficient of kinetic friction of 0.3. Find the work Wf done against friction.
A 4 kg block is released from rest at height 2 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 4.5 m long (μ = 0.15) and then runs onto a spring (k = 800 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.
A 5 g bullet is fired horizontally into a 1.5 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 14 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 10 g bullet is fired horizontally into a 5 kg block hanging from light strings 2 m long, and stays in it. The strings swing out to a largest angle of 10° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A vertical rectangular gate 1.5 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 1.5 m wide sits in the wall of a tank of glycerin, with its top edge 0.5 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 25 mm. The water level is kept 2.5 m above the orifice and the discharge coefficient is 0.6. Find the flow rate Q. Use g = 9.81 m/s².
A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 75 mm and a gauge pressure of 130 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 30 L contains 330 g of CO₂ at 100 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of CO₂ goes from 150 kPa, 20 L, 25 °C to 650 kPa and 130 °C. Find the new volume V2.
A heat engine works between a hot reservoir at 650 °C and a cold reservoir at 25 °C. It takes in 4300 kJ of heat. Find the most work Wmax it could possibly produce.
A refrigerator keeps its cold space at -20 °C in a kitchen at 25 °C, removing heat from the cold space at 4 kW. Find the minimum power input Ẇmin.
A pipe open at both ends is 0.65 m long. The speed of sound is 343 m/s. Find the frequency f2 of its harmonic number 2.
Two thin lenses share an axis, 25 cm apart: lens 1 with f₁ = 8 cm and lens 2 with f₂ = −10 cm. An object stands 40 cm in front of lens 1. Find the position di2 of the final image measured from lens 2 (positive beyond lens 2, real-is-positive convention).
Answer: s = 24.11 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 = 2.892 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 = 5.554 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: T = 11.04 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.
Answer: x = 19.98 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: α = 64.73 °
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.
Answer: Wf = 33.3 J
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 = 5.099 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 ramp is smooth, so only the strip removes energy.
Answer: v = 498.9 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: v = 386.8 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 = 242.8 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 = 55.62 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 = 2.063 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: H = 13.32 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 = 775.4 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 6.241 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Wmax = 2911 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 = 0.711 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: f2 = 527.7 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|.
Answer: di2 = −6 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.
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