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 20 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 5 m/s². How far from its starting point does the police car catch up? Find d.
A tram starts from rest and accelerates at 1 m/s² for 6 s, then runs at constant speed for 17 s, and finally brakes at 4 m/s² until it stops. Find the total travel time t.
Masses of 5.5 kg and 4 kg hang on either side of a light, frictionless pulley, connected by a light string. The system is released from rest. Use g = 9.81 m/s². Find the tension T in the string.
A 4 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 4 kg mass. The coefficient of kinetic friction between block and table is 0.1. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched from level ground at 28 m/s and must land 55 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.
A ball is thrown from ground level at 31 m/s and 65° above the horizontal. It lands on the flat roof of a building 26 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 5 kg ball hangs on a light string of length 1 m. It is pulled aside until the string makes 70° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A 5 kg block is released from rest at height 3 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3 m long (μ = 0.15) and then runs onto a spring (k = 500 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.
A 9 kg projectile moving horizontally at 27 m/s explodes into two fragments that move along the same line. The 7 kg fragment continues forwards at 86 m/s. Find the velocity v′2 of the 2 kg fragment (positive in the original direction).
At a crossing a 1100 kg car travelling east at 19 m/s collides with a 1500 kg car travelling north at 24 m/s. The cars lock together. Find the speed v′ of the wreck just after the collision.
A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 1.5 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 seawater, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
Water flows through a horizontal pipe that narrows from 150 mm to 75 mm. In the wide section the pressure is 240 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 in the narrow section.
Water flows up through a pipe that narrows from 150 mm to 50 mm while rising 2 m. At section 1 the pressure is 300 kPa and the velocity is 1 m/s. Ignoring losses, find the pressure p2 at section 2.
A tank of 5 L contains 150 g of N₂ at 85 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of O₂ at 550 kPa and 180 °C, treating it as an ideal gas.
A heat engine works between a hot reservoir at 650 °C and a cold reservoir at 20 °C. Find the highest possible thermal efficiency ηmax.
Two Carnot engines work in series. Engine A receives heat from a reservoir at 600 °C and rejects heat to an intermediate reservoir at temperature Tm. Engine B takes all that heat and rejects heat to a reservoir at 15 °C. Both engines have the same thermal efficiency. Find Tm in °C.
An object stands 8 cm in front of a thin converging lens with a focal length of 12 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, 30 cm apart: lens 1 with f₁ = 20 cm and lens 2 with f₂ = 15 cm. An object stands 18 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: d = 160 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.
At that moment the police car moves at 2v = 40 m/s.
Answer: t = 24.5 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: T = 45.44 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.
The net driving force is the difference of the weights; it moves the total mass.
Answer: a = 4.415 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: θ = 21.74 °
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 other solution is 90° − θ = 68.26°.
Answer: x = 59.84 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: T = 113.6 N
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). On a circular path the net force towards the centre must equal mv²/r.
Check: T = mg(3 − 2cos θ) = 113.6 N.
Answer: v = 7.073 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 = −179.5 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: v′ = 16.01 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: F = 176.6 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 = 362 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: p2 = 228.3 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: p2 = 240.4 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 = 3189 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 4.672 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηmax = 68.24 %
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 = 228.4 °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: di = −24 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 = 16.15 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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