Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 72 km/h brakes with a constant deceleration of 7.5 m/s² until it stops. How long does braking take? Find t.
A ball is thrown straight up at 20 m/s from the edge of a roof 25 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 40 kg box is pushed up a ramp at 20° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the pushing force F.
A block is given a push so that it starts sliding up a ramp at 30° with a speed of 6 m/s. The coefficient of kinetic friction is 0.15. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is launched at 30° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A ball is thrown from ground level at 36 m/s and 65° above the horizontal. It lands on the flat roof of a building 26 m high, on its way down. Ignore air resistance and use g = 9.81 m/s². Find the horizontal distance x from the launch point to where it lands.
A 4 kg block slides 9 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 rests on a rough horizontal table (μ = 0.15). A light string runs from it over a smooth pulley at the edge of the table to a hanging 6 kg block. The system is released from rest. Find the speed v of the blocks after the hanging block has dropped 2.5 m.
A 0.5 kg glider moving at 9 m/s makes a perfectly elastic head-on collision with a 5 kg glider at rest on a frictionless track. Find the velocity v′2 of the second glider after the collision.
A 0.3 kg ball hits a wall at 25 m/s and bounces straight back at 10 m/s. The contact lasts 50 ms. Find the magnitude of the impulse J the wall gives the ball.
A vertical rectangular gate 3 m wide sits in the wall of a tank of water, with its top edge 3 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 1.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
Water flows through a horizontal venturi meter with an inlet diameter of 200 mm and a throat diameter of 150 mm. The pressure at the inlet is 38 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
Water flows up through a pipe that narrows from 75 mm to 50 mm while rising 8 m. At section 1 the pressure is 180 kPa and the velocity is 2.25 m/s. Ignoring losses, find the pressure p2 at section 2.
What volume V does 190 g of He occupy at 500 kPa and 75 °C? Treat it as an ideal gas.
A fixed amount of He goes from 100 kPa, 10 L, 35 °C to 100 kPa and 250 °C. Find the new volume V2.
A heat engine receives heat from a source at 850 °C and rejects heat to the surroundings at 40 °C. Its actual thermal efficiency is 36%. 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 28 °C, removing heat from the cold space at 4.5 kW. Find the highest possible coefficient of performance COPmax.
A string fixed at both ends must have its harmonic number 5 at 500 Hz; waves travel on it at 100 m/s. Find its length L.
Light of wavelength 450 nm falls on two slits 0.15 mm apart. The screen is 1.5 m away. Find the distance y1 from the central bright fringe to bright fringe number 1.
Answer: t = 2.667 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.
Answer: t = 5.081 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: F = 171.1 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: d = 2.913 m
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.
Both forces point down the slope; the mass cancels.
Answer: x = 22.6 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: x = 87.12 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.
Larger root: the ball is descending.
Answer: Wf = 62.43 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 = 4.838 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).
Answer: v′2 = 1.636 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.
Answer: J = 10.5 N·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 = 165.5 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 = 2.042 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: Q = 186.3 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 pipe is horizontal, so only pressure and velocity change.
Answer: p2 = 91.24 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 = 274.8 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 16.98 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: ηII = 49.92 %
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: COPmax = 12.09
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: L = 0.5 m
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: y1 = 4.5 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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