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 23 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 3 m/s². How far from its starting point does the police car catch up? Find d.
A tram starts from rest and accelerates at 2.5 m/s² for 8 s, then runs at constant speed for 7 s, and finally brakes at 4.5 m/s² until it stops. Find the total travel time t.
A 19 kg block slides down a frictionless slope at 20° to the horizontal. 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 15° with a speed of 9 m/s. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?
A ball is launched at 40° above horizontal from level ground. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A ball is launched at 20° above horizontal from the top of a 25 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A toy launcher has a vertical spring of stiffness 2000 N/m. A 0.15 kg ball is placed on it and the spring is compressed by 3 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 2 kg block is released from rest at height 2.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 5 m long (μ = 0.25) and then runs onto a spring (k = 500 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 4.5 kg glider moving at 7 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 0.6 kg ball hits a wall at 25 m/s and bounces straight back at 17 m/s. The contact lasts 5 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 3 m wide sits in the wall of a tank of seawater, with its top edge 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
Water flows at 36 L/s through a horizontal pipe that narrows from 150 mm to 75 mm. The pressure in the narrow section is 190 kPa. Ignoring losses, find the pressure p1 in the wide section.
A closed water tank has compressed air above the water at a gauge pressure of 190 kPa. Water leaves through an opening with a diameter of 20 mm, 3.5 m below the water surface, into the open air. Ignoring losses, find the flow rate Q. Use g = 9.81 m/s².
A tank of 20 L contains 130 g of N₂ at 40 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of CO₂ goes from 200 kPa, 20 L, 15 °C to 400 kPa and 200 °C. Find the new volume V2.
A reversible refrigerator keeps a cold room at 4 °C by removing 0.5 kW of heat, and rejects heat to the surroundings at 30 °C. Find the rate Q̇H at which it rejects heat.
A refrigerator keeps its cold space at -18 °C in a kitchen at 30 °C, removing heat from the cold space at 1.5 kW. Find the highest possible coefficient of performance COPmax.
An observer moves at 20 m/s away from a stationary source of 400 Hz. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.
A sensor at rest sends ultrasound of 20 kHz towards a car that approaches at 15 m/s and mixes the echo with the emitted signal. Find the beat frequency fb (speed of sound 343 m/s).
Answer: d = 352.7 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 = 46 m/s.
Answer: t = 19.44 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: a = 3.355 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 = 7.525 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: t = 2.621 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.
Starting and landing at the same height: t = 2vy/g.
Answer: t = 3.244 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: h = 0.6116 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: x = 31.32 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 = 0.875 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 = 5040 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 = 181 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: p1 = 221.1 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.
Answer: Q = 6.654 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.
Answer: P = 604.2 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 16.42 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Q̇H = 0.5469 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 = 0.04691 kW.
Answer: COPmax = 5.316
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: f′ = 376.7 Hz
Standing waves and the Doppler effect: a string fixed at both ends, or a pipe open at both ends, fits whole half-wavelengths: L = nλ/2, fn = n·v/(2L), n = 1, 2, 3, … A pipe closed at one end has a node at the closed end and an antinode at the open end: L = nλ/4 with odd n only, fn = n·v/(4L). When source or observer move along the line between them, f′ = f·(v ± vo)/(v ∓ vs): use the upper signs for motion towards each other. Two waves of nearly equal frequency beat at |f₁ − f₂|.
Answer: fb = 1829 Hz
Standing waves and the Doppler effect: a string fixed at both ends, or a pipe open at both ends, fits whole half-wavelengths: L = nλ/2, fn = n·v/(2L), n = 1, 2, 3, … A pipe closed at one end has a node at the closed end and an antinode at the open end: L = nλ/4 with odd n only, fn = n·v/(4L). When source or observer move along the line between them, f′ = f·(v ± vo)/(v ∓ vs): use the upper signs for motion towards each other. Two waves of nearly equal frequency beat at |f₁ − f₂|.
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