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 0.8 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 6 m/s from the edge of a roof 30 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 15 kg sled is pulled across level snow by a rope at 35° above the horizontal with a force of 150 N. The coefficient of kinetic friction is 0.4. Use g = 9.81 m/s². Find the acceleration a of the sled.
A 7.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 6 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 from level ground at 34 m/s and must land 95 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.
A ball is launched at 45° above horizontal from the top of a 10 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A 4 kg block slides 15 m down a slope at 45° to the horizontal, with a coefficient of kinetic friction of 0.25. Find the work Wf done against friction.
A 5 kg block is released from rest at height 4 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 1 m long (μ = 0.15) and then runs onto a spring (k = 1500 N/m) on a smooth surface. Find the maximum compression x of the spring.
An 8 g bullet is fired horizontally into a 3 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 8 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 1900 kg car runs into a parked 1800 kg car. The wrecks lock together and skid 19 m before stopping. The coefficient of kinetic friction between tyres and road is 0.6. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of water, with its top edge 1 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 2.5 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 50 mm. In the wide section the pressure is 160 kPa and the velocity is 2.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 50 mm and a gauge pressure of 80 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.
What volume V does 230 g of N₂ occupy at 550 kPa and 10 °C? Treat it as an ideal gas.
A fixed amount of N₂ goes from 350 kPa, 2 L, 30 °C to 100 kPa and 180 °C. Find the new volume V2.
A heat engine receives heat from a source at 450 °C and rejects heat to the surroundings at 10 °C. Its actual thermal efficiency is 37%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
A refrigerator keeps its cold space at 5 °C in a kitchen at 35 °C, removing heat from the cold space at 1.5 kW. Find the minimum power input Ẇmin.
Light travels in ethanol (n = 1.36) towards a flat boundary with water (n = 1.33). Find the critical angle θc beyond which it is totally reflected.
An optical fibre has a core with n₁ = 1.46 and a cladding with n₂ = 1.43. Find the largest angle θa to the fibre axis at which light from air can enter the end of the fibre and still be guided by total internal reflection.
Answer: d = 34.5 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 = 3.159 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 = 6.562 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 = 2.998 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: θ = 26.86 °
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.
The other solution is 90° − θ = 63.14°.
Answer: x = 77.77 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 = 104.1 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 = 50.18 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 = 471.1 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: v1 = 29.12 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 = 38.63 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 = 259.6 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 = 113.1 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: H = 8.369 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: V = 35.15 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 10.46 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: ηII = 60.81 %
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 = 0.1618 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 = 77.94 °
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: θa = 17.12 °
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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