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 32 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 3 m/s². How long does it take the police car to catch up? Find t.
A ball is thrown straight up at 10 m/s from the edge of a roof 10 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 14 kg block slides down a slope at 25° 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 3.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 4.5 kg mass. The coefficient of kinetic friction between block and table is 0.25. Use g = 9.81 m/s². Find the tension T in the string.
A ball is launched at 40° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.
A ball is launched at 24 m/s and 70° above the horizontal from the top of a 10 m cliff. Ignore air resistance and use g = 9.81 m/s². Find the greatest height H above the ground at the foot of the cliff.
A car travelling at 50 km/h on a level road brakes hard and skids to a stop with its wheels locked. The coefficient of kinetic friction between the tyres and the road is 0.4. Find the stopping distance d.
A 0.5 kg block is released from rest at height 4.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 2.5 m long (μ = 0.1) and then runs onto a spring (k = 1000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 2.5 kg glider moving at 4 m/s makes a perfectly elastic head-on collision with a 1 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 0.6 kg ball hits a wall at 16 m/s and bounces straight back at 18 m/s. The contact lasts 20 ms. Find the average force Favg of the wall on the ball.
A vertical rectangular gate 1.5 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 1 m wide sits in the wall of a tank of water, with its top edge 4 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 1.5 m above the orifice and the discharge coefficient is 0.64. Find the flow rate Q. Use g = 9.81 m/s².
Water flows up through a pipe that narrows from 100 mm to 75 mm while rising 5 m. At section 1 the pressure is 130 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 at section 2.
What volume V does 160 g of O₂ occupy at 300 kPa and 10 °C? Treat it as an ideal gas.
A fixed amount of Ar goes from 350 kPa, 20 L, 35 °C to 350 kPa and 150 °C. Find the new volume V2.
A heat engine works between a hot reservoir at 550 °C and a cold reservoir at 25 °C. Find the highest possible thermal efficiency ηmax.
A Carnot heat engine receives 40 kW of heat from a source at 1000 °C and rejects heat to the surroundings at 27 °C. All of its work drives a Carnot refrigerator that removes heat from a freezer at -20 °C and also rejects heat to the surroundings at 27 °C. Find the rate Q̇L at which heat is removed from the freezer.
A small loudspeaker radiates 0.2 W of sound power equally in all directions. Find the sound intensity level β at 8 m.
An object is 8 cm in front of a convex mirror with a radius of curvature of 20 cm. Find the image distance di, positive for an image in front of the mirror (real) and negative for one behind it (virtual).
Answer: t = 21.33 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 = 64 m/s.
Answer: t = 2.774 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 = 2.368 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 = 24.14 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 = 4.128 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: H = 35.92 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.
Answer: d = 24.58 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).
Doubling the speed makes the stopping distance four times longer.
Answer: x = 20.42 cm
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′1 = 1.714 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 = 1020 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 = 44.15 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 = 161.9 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 = 6.817 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: p2 = 80.68 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: V = 39.24 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 27.46 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: ηmax = 63.78 %
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: Q̇L = 164.7 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.
Answer: β = 83.96 dB
Waves: a periodic wave moves one wavelength in one period, so v = fλ with T = 1/f. On a stretched string v = √(T/μ), with T the tension and μ = m/L the mass per metre. Sound in air travels at about 343 m/s at 20 °C, light in vacuum at c = 3.00 × 10⁸ m/s. Loudness is measured on a logarithmic scale: β = 10·log₁₀(I/I₀) dB with I₀ = 10⁻¹² W/m², so 10 times the intensity adds 10 dB. A point source spreads its power P over a sphere: I = P/(4πr²).
Answer: di = −4.444 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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