Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 80 km/h sees an obstacle. Her reaction time is 0.9 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 9 m/s from the edge of a roof 15 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.
A 9 kg block slides down a slope at 25° to the horizontal. The coefficient of kinetic friction is 0.15. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 4 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 8.5 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 at 20° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is launched at 26 m/s and 35° above the horizontal from the top of a 35 m cliff. Ignore air resistance and use g = 9.81 m/s². At what angle α below the horizontal does it hit the ground?
A 1 kg block slides 11 m down a slope at 45° to the horizontal, with a coefficient of kinetic friction of 0.2. Find the work Wf done against friction.
A 2 kg block is released from rest at height 1 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 5 m long (μ = 0.1) and then runs onto a spring (k = 2500 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.
A 14 kg projectile moving horizontally at 17 m/s explodes into two fragments that move along the same line. The 3 kg fragment continues forwards at 70 m/s. Find the velocity v′2 of the 11 kg fragment (positive in the original direction).
A 0.1 kg ball hits a wall at 25 m/s and bounces straight back at 9 m/s. The contact lasts 10 ms. Find the average force Favg of the wall on the ball.
A vertical rectangular gate 1 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 seawater, with its top edge 3 m below the surface. Find the resultant hydrostatic force F on the gate.
Water flows at 38 L/s through a horizontal pipe that narrows from 200 mm to 100 mm. The pressure in the narrow section is 120 kPa. Ignoring losses, find the pressure p1 in the wide section.
Water flows up through a pipe that narrows from 100 mm to 50 mm while rising 1 m. At section 1 the pressure is 180 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 50 L contains 340 g of N₂ at 60 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of O₂ at 250 kPa and 20 °C, treating it as an ideal gas.
A reversible (Carnot) heat engine takes in 2300 kJ of heat from a reservoir at 800 °C and rejects 628 kJ per cycle. Find the temperature TL of the cold reservoir in °C.
Two Carnot engines work in series. Engine A receives heat from a reservoir at 850 °C and rejects heat to an intermediate reservoir at temperature Tm. Engine B takes all that heat and rejects heat to a reservoir at 40 °C. Both engines have the same thermal efficiency. Find Tm in °C.
A source of 440 Hz moving at 10 m/s and an observer moving at 5 m/s approach each other. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.
An object is 6 cm in front of a concave mirror with a radius of curvature of 50 cm. Find the magnification m (negative for an inverted image).
Answer: d = 69.38 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: hmax = 19.13 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 = 2.812 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 = 45.27 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: H = 3.434 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: α = 54.76 °
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.
Energy or v² = v₀² + 2as in the vertical direction.
Answer: Wf = 15.26 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 = 3.132 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).
The ramp is smooth, so only the strip removes energy.
Answer: v′2 = 2.545 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 = 340 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 = 79.71 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 = 106.8 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 = 131 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 = 166 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 = 672.4 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 3.282 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: TL = 19.87 °C
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: Tm = 319.9 °C
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.
1 − Tm/TH = 1 − TL/Tm gives Tm² = TH·TL.
Answer: f′ = 459.8 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: m = 1.316
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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