Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A plane starts from rest and accelerates uniformly at 3 m/s² along a runway. Find its speed v after 1400 m.
A ball is thrown straight up at 6 m/s from the edge of a roof 20 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
A 15 kg block slides down a slope at 20° 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.
A 1.5 kg block on a ramp at 20° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 2.5 kg mass. The hanging mass moves down. The coefficient of kinetic friction between block and ramp is 0.15. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched from level ground at 70° above the horizontal and lands 80 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is thrown from ground level at 28 m/s and 65° above the horizontal. It lands on the flat roof of a building 23 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 car travelling at 30 km/h on a level road brakes hard and skids to a stop with its wheels locked. The coefficient of kinetic friction between the tyres and the road is 0.5. Find the stopping distance d.
A 1 kg block is released from rest at height 1.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3.5 m long (μ = 0.15) and then runs onto a spring (k = 1000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 0.5 kg glider moving at 10 m/s makes a perfectly elastic head-on collision with a 4 kg glider at rest on a frictionless track. Find the velocity v′2 of the second glider after the collision.
A 0.55 kg ball hits a wall at 25 m/s and bounces straight back at 5 m/s. The contact lasts 25 ms. Find the magnitude of the impulse J the wall gives the ball.
A vertical rectangular gate 2 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 3 m wide sits in the wall of a tank of seawater, with its top edge 3 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 75 mm. In the wide section the pressure is 250 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
Water flows up through a pipe that narrows from 200 mm to 75 mm while rising 7 m. At section 1 the pressure is 390 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 at section 2.
What volume V does 80 g of Ar occupy at 450 kPa and 30 °C? Treat it as an ideal gas.
A fixed amount of CH₄ goes from 400 kPa, 20 L, 25 °C to 650 kPa and 250 °C. Find the new volume V2.
A heat engine works between a hot reservoir at 425 °C and a cold reservoir at 40 °C. It takes in 2000 kJ of heat. Find the most work Wmax it could possibly produce.
A heat pump keeps a house at 22 °C while it is -5 °C outside, delivering 9 kW of heat to the house. Find the highest possible coefficient of performance COPmax.
A pipe open at both ends must have its harmonic number 2 at 858 Hz. The speed of sound is 343 m/s. Find the length L of the pipe.
A diffraction grating has 100 lines per millimetre. Light of wavelength 633 nm falls on it at right angles. What is the highest order mmax of bright maximum that can be seen?
Answer: v = 91.65 m/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: hmax = 21.83 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: a = 0.5897 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 = 13.64 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: v0 = 34.94 m/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.
Answer: x = 47.36 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: d = 7.079 m
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).
Doubling the speed makes the stopping distance four times longer.
Answer: x = 13.83 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 = 2.222 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 = 16.5 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 = 323.7 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 = 320.5 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: p2 = 249.4 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: p2 = 282.6 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 = 11.22 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 21.6 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Wmax = 1103 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: COPmax = 10.93
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 = 0.3998 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: mmax = 15
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, …
ExamProblems · examproblems.com/courses/physics-for-engineers