Physics for Engineers: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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1Formulas and methods

Kinematics

Velocity: v = v₀ + a·t
Displacement: s = v₀·t + ½·a·t²
Without time: v² = v₀² + 2·a·s
Units: 1 m/s = 3.6 km/h

Method

List what you know: v₀, v, a, t and s, with units.
Pick the equation that contains your three knowns and the unknown.
Choose one positive direction and give opposing quantities a minus sign.
Solve and check that the answer is physically sensible.

Newton’s laws

Second law: ΣF = m·a
Weight: W = m·g
Kinetic friction: F_f = μ·N
On a slope: along: m·g·sin θ, normal: m·g·cos θ

Method

Draw a free-body diagram with every force on the body.
Choose a positive direction along the motion.
Split forces into components along and perpendicular to the motion (on a slope: m·g·sin θ along, m·g·cos θ into the slope).
Apply ΣF = m·a along the motion; friction is μ·N against the motion.

Projectile motion

Components: v_x = v₀cos θ, v_y = v₀sin θ
Position: x = v_xt, y = h + v_yt − ½gt²
Maximum height: H = h + v_y²/(2g)
Range (level ground): R = v₀²·sin 2θ/g

Method

Split the launch velocity into components: v_x = v₀·cos θ and v_y = v₀·sin θ.
The horizontal motion has constant speed; the vertical motion has acceleration −g.
Find the time of flight from the vertical motion.
Use that time in the horizontal motion to find the range.

Work & energy

Kinetic energy: E_k = ½mv²
Potential energy: E_p = mgh
Energy balance: E_k1 + E_p1 + W = E_k2 + E_p2 + losses
Work of friction: W = μ·N·d
Power: P = W/t

Momentum & collisions

Momentum: p = m·v
Conservation: Σm·v (before) = Σm·v (after)
Perfectly inelastic: v′ = (m₁v₁ + m₂v₂)/(m₁ + m₂)
Impulse: J = F·Δt = m·Δv

Method

Choose a positive direction and give each velocity its sign.
Write the total momentum before: Σm·v.
For a perfectly inelastic collision the bodies move together after; for an elastic one kinetic energy is also conserved.
Set momentum before equal to momentum after and solve.

Hydrostatic force

Pressure at depth: p = ρgh
Force on a vertical wall: F = ρg·h_c·A
Centre of pressure: y_p = y_c + I_c/(y_cA)
Rectangle: I_c = bH³/12

Continuity & Bernoulli

Continuity: A₁v₁ = A₂v₂
Bernoulli: p₁ + ½ρv₁² + ρgz₁ = p₂ + ½ρv₂² + ρgz₂
Flow rate: Q = A·v
Circle area: A = πd²/4

Ideal gas law

Ideal gas law: pV = nRT
Gas constant: R = 8.314 J/(mol·K)
Combined gas law: p₁V₁/T₁ = p₂V₂/T₂
Moles: n = m/M
Temperature: T[K] = T[°C] + 273.15

Heat engines & refrigerators

Efficiency: η = W/Q_H = 1 − Q_C/Q_H
Carnot efficiency: η_C = 1 − T_C/T_H (kelvin)
Refrigerator COP: COP_R = Q_C/W
Heat pump COP: COP_HP = Q_H/W = COP_R + 1
Energy balance: Q_H = W + Q_C

Waves, sound & optics

Wave speed: v = fλ, T = 1/f
String: v = √(T/μ), μ = m/L
Sound level: β = 10·log₁₀(I/I₀), I₀ = 10⁻¹² W/m²
Point source: I = P/(4πr²)
Standing waves: string or open pipe f_n = n·v/(2L); closed pipe f_n = n·v/(4L), n odd
Doppler effect: f′ = f·(v ± v_o)/(v ∓ v_s)
Refractive index: n = c/v, λ = λ₀/n
Snell’s law: n₁sin θ₁ = n₂sin θ₂
Critical angle: sin θ_c = n₂/n₁
Optical fibre: NA = sin θ_a = √(n₁² − n₂²)
Lenses and mirrors: 1/f = 1/d_o + 1/d_i, m = −d_i/d_o, mirror f = R/2
Double slit: d·sin θ = mλ, y_m = mλL/d
Grating: d = 1/N, d·sin θ = mλ, m_max ≤ d/λ
Single slit: a·sin θ = mλ (dark), w = 2λL/a

Method

Identify the kind of problem: wave speed, sound level, standing wave, Doppler, refraction, lens or mirror, or interference.
Write the relation first: v = fλ, β = 10·log(I/I₀), fₙ = n·v/(2L) or n·v/(4L), f′ = f(v ± v_o)/(v ∓ v_s), n₁sin θ₁ = n₂sin θ₂, 1/f = 1/d₀ + 1/dᵢ, d·sin θ = mλ.
Convert units: nm, mm and cm to m, and check whether angles are measured from the normal.
Check signs and plausibility: a virtual image has dᵢ < 0, an approaching source sounds higher, a closed pipe has only odd harmonics.

2Practice problems

1.KinematicsMedium

A plane needs a speed of 250 km/h to take off and the usable runway is 1900 m long. Starting from rest, what constant acceleration a does it need?

v₀ = 0v = 250 km/hs = 1900 m

a = m/s²

2.KinematicsHard

A ball is thrown straight up at 5 m/s from the edge of a roof 5 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.

v₀ = 5 m/sh₀ = 5 mg = 9.81 m/s²

t = s

3.Newton’s lawsMedium

A 5 kg block slides down a slope at 20° to the horizontal. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.

m = 5 kgθ = 20°μ = 0.3g = 9.81 m/s²

a = m/s²

4.Newton’s lawsHard

A 7 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 3 kg mass. The table is frictionless. Use g = 9.81 m/s². Find the acceleration a of the system.

m₁ = 7 kgm₂ = 3 kgμ = 0g = 9.81 m/s²

a = m/s²

5.Projectile motionMedium

A ball is launched at 50° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.

v₀ = 24 m/sθ = 50°g = 9.81 m/s²

H = m

6.Projectile motionHard

A ball is thrown from ground level at 32 m/s and 40° above the horizontal. It lands on the flat roof of a building 7 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.

v₀ = 32 m/sθ = 40°h_roof = 7 mg = 9.81 m/s²

x = m

7.Work & energyMedium

A 5 kg ball hangs on a light string of length 1 m. It is pulled aside until the string makes 30° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.

m = 5 kgL = 1 mθ = 30°g = 9.81 m/s²

T = N

8.Work & energyHard

A 5 kg block starts from rest at the bottom of a 25° slope and is pulled up the slope by a constant force of 70 N parallel to the slope. The coefficient of kinetic friction is 0.3. Find the speed v of the block after it has moved 8 m up the slope.

m = 5 kgF = 70 Nθ = 25°μ = 0.3d = 8 mg = 9.81 m/s²

v = m/s

9.Momentum & collisionsMedium

A 5 kg glider moving at 3 m/s makes a perfectly elastic head-on collision with a 2.5 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).

m₁ = 5 kgm₂ = 2.5 kgv₁ = 3 m/sv₂ = 0

v′₁ = m/s

10.Momentum & collisionsHard

A 0.1 kg ball hits a wall at 26 m/s and bounces straight back at 20 m/s. The contact lasts 25 ms. Find the average force F_avg of the wall on the ball.

m = 0.1 kgv_in = 26 m/sv_out = 20 m/sΔt = 25 ms

F_avg = N

11.Hydrostatic forceMedium

A vertical rectangular gate 3 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.

d = 4 mH = 1.5 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

12.Hydrostatic forceHard

A vertical rectangular gate 3 m wide sits in the wall of a tank of oil, with its top edge 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 1.5 mH = 3 mb = 3 mρ = 880 kg/m³g = 9.81 m/s²

F = kN

13.Continuity & BernoulliMedium

Water drains from a large open tank through a sharp-edged orifice with a diameter of 20 mm. The water level is kept 5.5 m above the orifice and the discharge coefficient is 0.64. Find the flow rate Q. Use g = 9.81 m/s².

d = 20 mmh = 5.5 mC_d = 0.64g = 9.81 m/s²

Q = L/s

14.Continuity & BernoulliHard

A horizontal venturi meter (inlet 80 mm, throat 40 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 300 mm. The discharge coefficient is 0.97. Find the flow rate Q. Use ρ = 1000 kg/m³ for water, 13 600 kg/m³ for mercury and g = 9.81 m/s².

d₁ = 80 mmd₂ = 40 mmΔh = 300 mmC_d = 0.97ρ_Hg = 13 600 kg/m³g = 9.81 m/s²

Q = L/s

15.Ideal gas lawMedium

What volume V does 460 g of N₂ occupy at 800 kPa and 20 °C? Treat it as an ideal gas.

m = 460 gM = 28.01 g/molP = 800 kPaT = 20 °CR = 8.314 J/(mol·K)

V = L

16.Ideal gas lawHard

Find the density ρ of CO₂ at 650 kPa and 90 °C, treating it as an ideal gas.

M = 44.01 g/molP = 650 kPaT = 90 °CR = 8.314 J/(mol·K)

ρ = kg/m³

17.Heat engines & refrigeratorsMedium

A heat engine works between a hot reservoir at 800 °C and a cold reservoir at 30 °C. Find the highest possible thermal efficiency η_max.

T_H = 800 °CT_L = 30 °C

η_max = %

18.Heat engines & refrigeratorsHard

A refrigerator keeps its cold space at -20 °C in a kitchen at 28 °C, removing heat from the cold space at 1 kW. Find the minimum power input Ẇ_min.

T_H = 28 °CT_L = -20 °CQ̇_L = 1 kW

Ẇ_min = kW

19.Waves, sound & opticsMedium

An object stands 20 cm in front of a thin converging lens with a focal length of 5 cm. Find the image distance d_i. Use the sign convention 1/f = 1/d_o + 1/d_i (f < 0 for a diverging lens, d_i < 0 for a virtual image).

d_o = 20 cmf = 5 cm

d_i = cm

20.Waves, sound & opticsHard

Light of wavelength 532 nm falls at right angles on a grating with 100 lines per millimetre. Find the angle θ₂ of the order-2 maximum.

N = 100 mm⁻¹λ = 532 nmm = 2

θ₂ = °

3Solutions

1.Kinematics

Answer: a = 1.269 m/s²

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.

Units

v = 250/3.6 = 69.44 m/s

Acceleration

a = v²/(2s) = 69.44²/(2·1900) = 1.269 m/s²

Answer

a = 1.269 m/s²

2.Kinematics

Answer: t = 1.641 s

Equation of motion

Up is positive and the ground is h₀ below the starting point.

−5 = 5·t − ½·9.81·t²

4.905·t² − 5·t − 5 = 0

Solve the quadratic

Only the positive root is physical.

t = (5 + √(5² + 2·9.81·5))/9.81 = 1.641 s

Answer

t = 1.641 s

3.Newton’s laws

Answer: a = 0.5897 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.

Forces along the slope

Down the slope is positive.

m·a = m·g·sin θ − μ·m·g·cos θ

Acceleration

a = 9.81·(sin 20° − 0.3·cos 20°) = 9.81·(0.342 − 0.2819) = 0.5897 m/s²

Answer

a = 0.5897 m/s²

4.Newton’s laws

Answer: a = 2.943 m/s²

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (29.43 − 0)/10 = 2.943 m/s²

Answer

a = 2.943 m/s²

5.Projectile motion

Answer: H = 17.23 m

Projectile motion: with no air resistance, x = v_xt and y = h + v_yt − ½gt². The two directions are independent except that they share the same time.

Split the launch velocity

Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.

v_x = v₀·cos θ = 24·cos 50° = 15.43 m/s

v_y = v₀·sin θ = 24·sin 50° = 18.39 m/s

Highest point

At the top the vertical velocity is zero: v_y² = 2gH.

H = v_y²/(2g) = 18.39²/(2·9.81) = 17.23 m

Answer

H = 17.23 m

6.Projectile motion

Answer: x = 93.64 m

Projectile motion: with no air resistance, x = v_xt and y = h + v_yt − ½gt². The two directions are independent except that they share the same time.

Split the launch velocity

v_x = 24.51 m/s, v_y = 20.57 m/s

Time to land on the roof

Larger root: the ball is descending.

7 = 20.57·t − 4.905·t²

t = (20.57 + √(20.57² − 2·9.81·7))/9.81 = 3.82 s

Horizontal distance

x = v_x·t = 24.51·3.82 = 93.64 m

Answer

x = 93.64 m

7.Work & energy

Answer: T = 62.19 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.

Speed at the bottom

v² = 2gL(1 − cos θ) = 2·9.81·1·(1 − cos 30°) = 2.629 m²/s²

Forces at the bottom

m·g = 5·9.81 = 49.05 N

T = mg + mv²/L = 49.05 + 5·2.629/1 = 62.19 N

Answer

T = 62.19 N

Check: T = mg(3 − 2cos θ) = 62.19 N.

8.Work & energy

Answer: v = 10.72 m/s

Forces along the slope

m·g = 5·9.81 = 49.05 N

mg·sin θ = 20.73 N

μ·mg·cos θ = 13.34 N

Work–energy

½mv² = (70 − 34.07)·8 = 287.5 J

v = √(2·287.5/5) = 10.72 m/s

Answer

v = 10.72 m/s

9.Momentum & collisions

Answer: v′₁ = 1 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.

Elastic collision with a target at rest

Solving momentum and energy conservation together gives these two results.

v′₁ = (5 − 2.5)/(5 + 2.5)·3 = 1 m/s

v′₂ = 2·5/(5 + 2.5)·3 = 4 m/s

Answer

v′₁ = 1 m/s

10.Momentum & collisions

Answer: F_avg = 184 N

Change in momentum

J = m·Δv = 0.1·(20 + 26) = 4.6 N·s

Average force

F_avg = J/Δt = 4.6/0.025 = 184 N

Answer

F_avg = 184 N

11.Hydrostatic force

Answer: F = 209.7 kN

Hydrostatic force on a plane surface: F = ρ·g·y_c·A. It acts at the centre of pressure, y_cp = y_c + I_c/(y_cA), which always lies below the centroid.

Centroid depth and area

y_c = d + H/2 = 4 + 1.5/2 = 4.75 m

A = H·b = 1.5·3 = 4.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4.75·4.5 = 209.7 kN

Answer

F = 209.7 kN

12.Hydrostatic force

Answer: F = 233.1 kN

Hydrostatic force on a plane surface: F = ρ·g·y_c·A. It acts at the centre of pressure, y_cp = y_c + I_c/(y_cA), which always lies below the centroid.

Centroid depth and area

y_c = d + H/2 = 1.5 + 3/2 = 3 m

A = H·b = 3·3 = 9 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·3·9 = 233.1 kN

Answer

F = 233.1 kN

13.Continuity & Bernoulli

Answer: Q = 2.089 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.

Ideal jet velocity

√(2gh) = √(2·9.81·5.5) = 10.39 m/s

Actual flow rate

The discharge coefficient accounts for the contraction of the jet and friction at the edge.

Q = C_d·A·√(2gh) = 0.64·0.0003142·10.39 = 0.002089 m³/s = 2.089 L/s

Answer

Q = 2.089 L/s

14.Continuity & Bernoulli

Answer: Q = 10.84 L/s

Pressure difference from the manometer

Δp = (13600 − 1000)·9.81·0.3 = 37082 Pa

Throat velocity

Continuity and Bernoulli together give the ideal throat velocity.

β = 40/80 = 0.5

v₂ = √(2·37082/(1000·(1 − 0.0625))) = 8.894 m/s

Actual flow rate

Q = C_d·A₂·v₂ = 0.97·0.001257·8.894 = 0.01084 m³/s = 10.84 L/s

Answer

Q = 10.84 L/s

15.Ideal gas law

Answer: V = 50.03 L

Ideal gas law: PV = nRT, with n = m/M.

Moles

n = 460/28.01 = 16.42 mol

T = 20 + 273.15 = 293.1 K

Volume

V = nRT/P = 16.42·8.314·293.1/800 = 50.03 L

Answer

V = 50.03 L

16.Ideal gas law

Answer: ρ = 9.475 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)

T = 90 + 273.15 = 363.1 K

Density

ρ = 650·44.01/(8.314·363.1) = 9.475 kg/m³

Answer

ρ = 9.475 kg/m³

17.Heat engines & refrigerators

Answer: η_max = 71.75 %

Second-law limits: no cycle working between reservoirs at absolute temperatures T_H and T_L can beat the reversible (Carnot) cycle: η_max = 1 − T_L/T_H, COP_R,max = T_L/(T_H − T_L) and COP_HP,max = T_H/(T_H − T_L). Energy is always conserved: Q_H = W + Q_L.

Absolute temperatures

T_H = 800 + 273.15 = 1073 K

T_L = 30 + 273.15 = 303.1 K

Carnot efficiency

η_max = 1 − T_L/T_H = 1 − 303.1/1073 = 71.75 %

Answer

η_max = 71.75 %

18.Heat engines & refrigerators

Answer: Ẇ_min = 0.1896 kW

Absolute temperatures

T_H = 28 + 273.15 = 301.1 K

T_L = -20 + 273.15 = 253.1 K

Best possible COP

COP_R,max = T_L/(T_H − T_L) = 253.1/(301.1 − 253.1) = 5.274

Minimum power

The least work is needed by a reversible machine, which has the highest COP.

Ẇ_min = Q̇_L/COP = 1/5.274 = 0.1896 kW

Answer

Ẇ_min = 0.1896 kW

19.Waves, sound & optics

Answer: d_i = 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 θ_c = arcsin(n₂/n₁) all light is reflected: total internal reflection. Thin lenses and spherical mirrors obey 1/f = 1/d_o + 1/d_i with the magnification m = −d_i/d_o. 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; d_i > 0 for a real image, d_i < 0 for a virtual one, and m < 0 for an inverted image.

Thin-lens equation

1/d_i = 1/f − 1/d_o = 1/5 − 1/20

d_i = 6.667 cm

Ray diagram

real, on the far side of the lens

Answer

d_i = 6.667 cm

20.Waves, sound & optics

Answer: θ₂ = 6.108 °

Interference and diffraction: light from two slits a distance d apart reaches a point on the screen with a path difference d·sin θ. Bright fringes appear where d·sin θ = mλ (m = 0, ±1, ±2, …). For a screen at distance L ≫ d, the m-th bright fringe is at y = mλL/d and the fringe spacing is Δy = λL/d. A grating with N lines per millimetre has d = 1/N and sharp maxima at d·sin θ = mλ; since sin θ ≤ 1, the highest order is the largest whole number m ≤ d/λ. A single slit of width a has dark fringes at a·sin θ = mλ, m = ±1, ±2, …

Line spacing

d = 1/100 mm = 1 × 10⁻⁵ m

Grating equation

sin θ = mλ/d = 2·532 × 10⁻⁹/(1 × 10⁻⁵) = 0.1064

θ = 6.108°

Answer

θ₂ = 6.108°

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