Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A driver moving at 108 km/h sees an obstacle. Her reaction time is 1.5 s, after which the car brakes with a constant deceleration of 7.5 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A ball is thrown straight up at 5 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 6 kg block slides down a slope at 35° 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 block is given a push so that it starts sliding up a ramp at 25° with a speed of 11 m/s. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is launched at 70° 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 24 m/s and 70° above the horizontal. It lands on the flat roof of a building 8 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 0.5 kg block slides 11 m down a slope at 45° to the horizontal, with a coefficient of kinetic friction of 0.3. Find the work Wf done against friction.
A 2 kg block starts from rest at the bottom of a 25° slope and is pulled up the slope by a constant force of 15 N parallel to the slope. The coefficient of kinetic friction is 0.2. Find the speed v of the block after it has moved 8 m up the slope.
A 5 kg glider moving at 11 m/s makes a perfectly elastic head-on collision with a 3.5 kg glider at rest on a frictionless track. Find the velocity v′1 of the first glider after the collision (positive in its original direction).
A 0.15 kg ball hits a wall at 24 m/s and bounces straight back at 8 m/s. The contact lasts 5 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 2.5 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 drains from a large open tank through a sharp-edged orifice with a diameter of 50 mm. The water level is kept 2 m above the orifice and the discharge coefficient is 0.62. Find the flow rate Q. Use g = 9.81 m/s².
A hose with an inner diameter of 50 mm siphons water from an open tank. The highest point C of the hose is 2.5 m above the water surface, and the outlet (open to the air) is 3 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the absolute pressure pC at point C. The atmospheric pressure is 101.3 kPa.
A tank of 50 L contains 240 g of CO₂ at 65 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of Ar goes from 150 kPa, 10 L, 10 °C to 700 kPa and 190 °C. Find the new volume V2.
A heat engine receives heat from a source at 400 °C and rejects heat to the surroundings at 40 °C. Its actual thermal efficiency is 37%. 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 -10 °C outside, delivering 6 kW of heat to the house. Find the highest possible coefficient of performance COPmax.
An object stands 20 cm in front of a thin converging lens with a focal length of 8 cm. Find the image distance di. Use the sign convention 1/f = 1/do + 1/di (f < 0 for a diverging lens, di < 0 for a virtual image).
A grating with 100 lines per millimetre shows the order-2 maximum of a spectral line at 6.9° from the straight-through direction. Find the wavelength λ of the line.
Answer: d = 105 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 = 2.824 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 = 4.02 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: d = 9.5 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 = 67.1 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 = 34.56 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 = 11.45 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 = 5.021 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′1 = 1.941 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: Favg = 960 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 = 29.43 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 = 445 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: Q = 7.626 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 discharge coefficient accounts for the contraction of the jet and friction at the edge.
Answer: pC = 47.34 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.
Surface and outlet are both open to the air, so the pressure terms cancel.
If this pressure falls to the vapour pressure of the water, the siphon stops working.
Answer: P = 306.6 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 3.505 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: ηII = 69.18 %
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 = 9.223
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: di = 13.33 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.
Answer: λ = 600.7 nm
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