Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car speeding at a constant 30 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 5 m/s². How long does it take the police car to catch up? Find t.
Ball A is dropped from rest from a height of 15 m. At the same instant ball B is thrown straight up from the ground directly below at 15 m/s. Ignore air resistance and use g = 9.81 m/s². At what height y above the ground do they meet?
A box slides across a level floor with an initial speed of 6 m/s and comes to rest after 4.6 m. Use g = 9.81 m/s². Find the coefficient of kinetic friction μ.
A 6 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 5 kg mass. The coefficient of kinetic friction between block and table is 0.25. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is kicked from level ground at 16 m/s and 60° above the horizontal towards a wall 5 m away. Ignore air resistance and use g = 9.81 m/s². At what height y does the ball pass the wall?
A ball is launched at 34 m/s and 45° above the horizontal from the top of a 55 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 toy launcher has a vertical spring of stiffness 1500 N/m. A 0.25 kg ball is placed on it and the spring is compressed by 5 cm, then released. Ignoring air resistance and the spring’s mass, find the height h the ball rises, measured from its position when the spring was compressed.
A 5 kg block starts from rest at the bottom of a 15° slope and is pulled up the slope by a constant force of 40 N parallel to the slope. The coefficient of kinetic friction is 0.2. Find the speed v of the block after it has moved 3 m up the slope.
A 14 kg projectile moving horizontally at 12 m/s explodes into two fragments that move along the same line. The 1 kg fragment continues forwards at 51 m/s. Find the velocity v′2 of the 13 kg fragment (positive in the original direction).
A 0.2 kg ball hits a wall at 11 m/s and bounces straight back at 13 m/s. The contact lasts 25 ms. Find the average force Favg of the wall on the ball.
A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 1 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 water, with its top edge 2 m below the surface. Find the resultant hydrostatic force F on the gate.
Water flows through a horizontal venturi meter with an inlet diameter of 200 mm and a throat diameter of 100 mm. The pressure at the inlet is 23 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
A closed water tank has compressed air above the water at a gauge pressure of 150 kPa. Water leaves through an opening with a diameter of 25 mm, 3 m below the water surface, into the open air. Ignoring losses, find the jet velocity v. Use g = 9.81 m/s².
A tank of 30 L contains 490 g of N₂ at 95 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of O₂ goes from 150 kPa, 5 L, 35 °C to 500 kPa and 240 °C. Find the new volume V2.
A reversible refrigerator keeps a cold room at 2 °C by removing 6 kW of heat, and rejects heat to the surroundings at 35 °C. Find the rate Q̇H at which it rejects heat.
Two Carnot engines work in series. Engine A receives heat from a reservoir at 700 °C and rejects heat to an intermediate reservoir at temperature Tm. Engine B takes all that heat and rejects heat to a reservoir at 35 °C. Both engines produce the same work. Find Tm in °C.
An object stands 20 cm in front of a thin converging lens with a focal length of 8 cm. Find the magnification m (negative for an inverted image).
An object is 20 cm in front of a concave mirror with a radius of curvature of 30 cm. Find the magnification m (negative for an inverted image).
Answer: t = 12 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.
At that moment the police car moves at 2v = 60 m/s.
Answer: y = 10.09 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.
Up is positive, the ground is y = 0.
Answer: μ = 0.3989
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: a = 3.121 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: y = 6.744 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.
The ball is still rising at the wall.
Answer: α = 59.43 °
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: h = 0.7645 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).
Answer: v = 4.625 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 = 9 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 = 192 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 = 40.47 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 = 55.02 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: v = 18.94 m/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.
Answer: P = 1785 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 2.498 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Q̇H = 6.72 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.
Power input: Ẇ = 0.7196 kW.
Answer: Tm = 367.5 °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.
W = Q·(1 − Tlow/Thigh) and Q/T is the same at every reservoir, so W is proportional to the temperature drop.
Answer: m = −0.6667
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: m = −3
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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