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 8 m/s² until it stops. Find the braking distance s.
A stone is dropped from rest into a well. The splash is heard 2.7 s after the stone is released. The speed of sound is 343 m/s; ignore air resistance and use g = 9.81 m/s². Find the depth d of the well.
A 4 kg block slides down a slope at 30° to the horizontal. The coefficient of kinetic friction is 0.2. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 2 kg block on a ramp at 35° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 8 kg mass. The hanging mass moves down. The ramp is frictionless. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched at 30° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is launched at 50° above horizontal from the top of a 35 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A 1200 kg car drives up a 4° slope at a constant 72 km/h. Air resistance and rolling resistance together amount to 450 N. Find the power P the wheels must deliver, in kW.
A 3 kg block rests on a rough horizontal table (μ = 0.25). A light string runs from it over a smooth pulley at the edge of the table to a hanging 4 kg block. The system is released from rest. Find the speed v of the blocks after the hanging block has dropped 2.5 m.
An 11 kg projectile moving horizontally at 22 m/s explodes into two fragments that move along the same line. The 6 kg fragment continues forwards at 50 m/s. Find the velocity v′2 of the 5 kg fragment (positive in the original direction).
A 0.45 kg ball hits a wall at 19 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 2 m wide sits in the wall of a tank of water, with its top edge 0.5 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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
Water flows at 16 L/s through a horizontal pipe that narrows from 150 mm to 100 mm. The pressure in the narrow section is 200 kPa. Ignoring losses, find the pressure p1 in the wide section.
Water flows up through a pipe that narrows from 200 mm to 100 mm while rising 10 m. At section 1 the pressure is 270 kPa and the velocity is 2.25 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 5 L contains 460 g of He at 50 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of N₂ at 450 kPa and 80 °C, treating it as an ideal gas.
A heat engine receives heat from a source at 700 °C and rejects heat to the surroundings at 10 °C. Its actual thermal efficiency is 43%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
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 minimum power input Ẇmin.
An observer moves at 15 m/s away from a stationary source of 300 Hz. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.
An object is 30 cm in front of a concave mirror with a radius of curvature of 16 cm. Find the image distance di, positive for an image in front of the mirror (real) and negative for one behind it (virtual).
Answer: s = 25 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: d = 33.24 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.
Only the positive root is physical.
Answer: a = 3.206 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 = 6.723 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 = 11.47 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: t = 5.767 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.
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: P = 25.42 kW
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). Power is force times speed: P = F·v.
Answer: v = 4.772 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 = −11.6 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.
The minus sign means the second fragment flies backwards.
Answer: J = 13.05 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 = 58.86 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 = 111.2 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: p1 = 201.7 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.
Answer: p2 = 133.9 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: P = 61747 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 4.293 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηII = 60.65 %
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 = 0.5184 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: f′ = 286.9 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: di = 10.91 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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