Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 90 km/h brakes with a constant deceleration of 7 m/s² until it stops. Find the braking distance s.
A ball is thrown straight up at 20 m/s from the edge of a roof 20 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 7 kg block slides down a slope at 35° to the horizontal. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 60 N pulls the front 4 kg block; the 7 kg block is towed behind. The coefficient of kinetic friction is 0.3 for both blocks. Use g = 9.81 m/s². Find the acceleration a of the blocks.
A ball is launched at 35° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is launched at 60° above horizontal from the top of a 20 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A 5 kg block slides 20 m down a slope at 40° to the horizontal, with a coefficient of kinetic friction of 0.2. Find the work Wf done against friction.
A 4 kg block is released from rest at height 4 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 5 m long (μ = 0.15) and then runs onto a spring (k = 4000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 20 kg projectile moving horizontally at 40 m/s explodes into two fragments that move along the same line. The 1 kg fragment continues forwards at 82 m/s. Find the velocity v′2 of the 19 kg fragment (positive in the original direction).
A 0.5 kg ball hits a wall at 25 m/s and bounces straight back at 11 m/s. The contact lasts 10 ms. Find the average force Favg of the wall on the ball.
A vertical rectangular gate 2 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.
A vertical rectangular gate 1 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 flows through a horizontal pipe that narrows from 150 mm to 75 mm. In the wide section the pressure is 130 kPa and the velocity is 2.25 m/s. Ignoring losses, find the pressure p2 in the narrow section.
Water flows up through a pipe that narrows from 75 mm to 40 mm while rising 6 m. At section 1 the pressure is 260 kPa and the velocity is 2.75 m/s. Ignoring losses, find the pressure p2 at section 2.
What volume V does 50 g of N₂ occupy at 450 kPa and 25 °C? Treat it as an ideal gas.
Find the density ρ of O₂ at 700 kPa and 80 °C, treating it as an ideal gas.
A heat engine works between a hot reservoir at 675 °C and a cold reservoir at 10 °C. Find the highest possible thermal efficiency ηmax.
A refrigerator keeps its cold space at 5 °C in a kitchen at 35 °C, removing heat from the cold space at 4 kW. Find the minimum power input Ẇmin.
Light travels in diamond (n = 2.42) towards a flat boundary with flint glass (n = 1.62). Find the critical angle θc beyond which it is totally reflected.
A sensor at rest sends ultrasound of 30 kHz towards a car that approaches at 30 m/s and mixes the echo with the emitted signal. Find the beat frequency fb (speed of sound 343 m/s).
Answer: s = 44.64 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 = 4.908 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 = 3.216 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: a = 2.512 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: H = 15.09 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.
At the top the vertical velocity is zero: vy² = 2gH.
Answer: x = 64.98 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.
Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.
Answer: Wf = 150.3 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: x = 25.25 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 = 37.79 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: Favg = 1800 N
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 = 80.93 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 = 23.18 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 = 92.03 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 = 158.2 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 = 9.833 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 7.629 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηmax = 70.14 %
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 = 0.4314 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 = 42.02 °
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 = 5751 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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