Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 80 km/h brakes with a constant deceleration of 5 m/s² until it stops. How long does braking take? 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 14 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 10 m/s and comes to rest after 20 m. Use g = 9.81 m/s². Find the coefficient of kinetic friction μ.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 110 N pulls the front 9 kg block; the 10 kg block is towed behind. The coefficient of kinetic friction is 0.25 for both blocks. Use g = 9.81 m/s². Find the tension T in the connecting string.
A ball is launched at 65° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A ball is launched at 35° 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 10 kg block slides 8 m down a slope at 40° to the horizontal, with a coefficient of kinetic friction of 0.2. Find the work Wf done against friction.
A 2.5 kg block is released from rest at height 3.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3 m long (μ = 0.25) and then runs onto a spring (k = 1500 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.
A 7 kg projectile moving horizontally at 31 m/s explodes into two fragments that move along the same line. The 2 kg fragment continues forwards at 42 m/s. Find the velocity v′2 of the 5 kg fragment (positive in the original direction).
A 0.3 kg ball hits a wall at 9 m/s and bounces straight back at 11 m/s. The contact lasts 20 ms. Find the magnitude of the impulse J the wall gives the ball.
A vertical rectangular gate 1 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 2 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 100 mm and a throat diameter of 60 mm. The pressure at the inlet is 29 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
Water flows up through a pipe that narrows from 75 mm to 25 mm while rising 8 m. At section 1 the pressure is 260 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 30 L contains 40 g of CO₂ at 35 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of CO₂ goes from 200 kPa, 2 L, 30 °C to 300 kPa and 200 °C. Find the new volume V2.
A heat engine works between a hot reservoir at 575 °C and a cold reservoir at 35 °C. It takes in 700 kJ of heat. Find the most work Wmax it could possibly produce.
A heat pump keeps a house at 20 °C while it is 7 °C outside, delivering 13 kW of heat to the house. Find the minimum power input Ẇmin.
An object stands 25 cm in front of a thin converging lens with a focal length of 15 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).
Two thin lenses share an axis, 60 cm apart: lens 1 with f₁ = 20 cm and lens 2 with f₂ = −10 cm. An object stands 40 cm in front of lens 1. Find the position di2 of the final image measured from lens 2 (positive beyond lens 2, real-is-positive convention).
Answer: t = 4.444 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: y = 9.369 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 = 0.2548
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: T = 57.89 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.
With the same μ on both blocks, friction drops out: T = F·m₁/(m₁ + m₂).
Answer: x = 61.22 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: t = 3.924 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: Wf = 120.2 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 = 7.345 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 = 26.6 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.
Answer: J = 6 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 = 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 = 80.93 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 = 23.08 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: p2 = 91.52 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 = 77.62 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 2.081 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Wmax = 445.7 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: Ẇmin = 0.5765 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: di = 37.5 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: di2 = −6.667 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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