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 260 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 = 260 km/hs = 1900 m

a = m/s²

2.KinematicsHard

A ball is thrown straight up at 10 m/s from the edge of a roof 30 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height h_max above the ground.

v₀ = 10 m/sh₀ = 30 mg = 9.81 m/s²

h_max = m

3.Newton’s lawsMedium

A 30 kg sled is pulled across level snow by a rope at 35° above the horizontal with a force of 200 N. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the acceleration a of the sled.

m = 30 kgF = 200 Nθ = 35°μ = 0.1g = 9.81 m/s²

a = m/s²

4.Newton’s lawsHard

A 10 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 3.5 kg mass. The coefficient of kinetic friction between block and table is 0.2. Use g = 9.81 m/s². Find the acceleration a of the system.

m₁ = 10 kgm₂ = 3.5 kgμ = 0.2g = 9.81 m/s²

a = m/s²

5.Projectile motionMedium

A ball is kicked from level ground at 34 m/s and 70° above the horizontal towards a wall 30 m away. Ignore air resistance and use g = 9.81 m/s². At what height y does the ball pass the wall?

v₀ = 34 m/sθ = 70°x = 30 mg = 9.81 m/s²

y = m

6.Projectile motionHard

A ball is thrown from ground level at 30 m/s and 70° above the horizontal. It lands on the flat roof of a building 11 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 = 30 m/sθ = 70°h_roof = 11 mg = 9.81 m/s²

x = m

7.Work & energyMedium

A block slides from rest down a slope at 45° to the horizontal. The coefficient of kinetic friction is 0.1. Find its speed v after it has slid 18 m along the slope.

θ = 45°μ = 0.1d = 18 mg = 9.81 m/s²

v = m/s

8.Work & energyHard

A 4 kg block is released from rest at height 4 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 2.5 m long (μ = 0.2) and then runs onto a spring (k = 1000 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.

m = 4 kgh = 4 mL = 2.5 mμ = 0.2k = 1000 N/mg = 9.81 m/s²

v = m/s

9.Momentum & collisionsMedium

A 5 kg glider moving at 10 m/s makes a perfectly elastic head-on collision with a 3 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 = 3 kgv₁ = 10 m/sv₂ = 0

v′₁ = m/s

10.Momentum & collisionsHard

A 0.45 kg ball hits a wall at 13 m/s and bounces straight back at 9 m/s. The contact lasts 50 ms. Find the magnitude of the impulse J the wall gives the ball.

m = 0.45 kgv_in = 13 m/sv_out = 9 m/sΔt = 50 ms

J = N·s

11.Hydrostatic forceMedium

A vertical rectangular gate 1 m wide sits in the wall of a tank of water, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 2.5 mH = 1 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

12.Hydrostatic forceHard

A vertical rectangular gate 1.5 m wide sits in the wall of a tank of oil, with its top edge 0.5 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 0.5 mH = 1.5 mb = 1.5 mρ = 880 kg/m³g = 9.81 m/s²

y_cp = m

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 2 m above the orifice and the discharge coefficient is 0.63. Find the flow rate Q. Use g = 9.81 m/s².

d = 20 mmh = 2 mC_d = 0.63g = 9.81 m/s²

Q = L/s

14.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 75 mm to 40 mm while rising 8 m. At section 1 the pressure is 250 kPa and the velocity is 3 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 75 mmd₂ = 40 mmp₁ = 250 kPav₁ = 3 m/sΔz = 8 mρ = 1000 kg/m³

p₂ = kPa

15.Ideal gas lawMedium

A tank of 20 L contains 440 g of O₂ at 45 °C. Treat it as an ideal gas and find the pressure P.

m = 440 gM = 32 g/molV = 20 LT = 45 °CR = 8.314 J/(mol·K)

P = kPa

16.Ideal gas lawHard

Find the density ρ of O₂ at 350 kPa and 0 °C, treating it as an ideal gas.

M = 32 g/molP = 350 kPaT = 0 °CR = 8.314 J/(mol·K)

ρ = kg/m³

17.Heat engines & refrigeratorsMedium

A heat engine works between a hot reservoir at 500 °C and a cold reservoir at 5 °C. Find the highest possible thermal efficiency η_max.

T_H = 500 °CT_L = 5 °C

η_max = %

18.Heat engines & refrigeratorsHard

A heat pump keeps a house at 20 °C while it is 0 °C outside, delivering 10 kW of heat to the house. Find the highest possible coefficient of performance COP_max.

T_H = 20 °CT_L = 0 °CQ̇_H = 10 kW

COP_max =

19.Waves, sound & opticsMedium

A pipe open at both ends is 0.8 m long. The speed of sound is 343 m/s. Find the frequency f₂ of its harmonic number 2.

v = 343 m/sn = 2L = 0.8 m

f₂ = Hz

20.Waves, sound & opticsHard

Light of wavelength 650 nm passes through a single slit 0.15 mm wide. Find the width w of the central bright maximum on a screen 1.5 m away (between the first dark fringes on either side).

λ = 650 nma = 0.15 mmL = 1.5 m

w = mm

3Solutions

1.Kinematics

Answer: a = 1.373 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 = 260/3.6 = 72.22 m/s

Acceleration

a = v²/(2s) = 72.22²/(2·1900) = 1.373 m/s²

Answer

a = 1.373 m/s²

2.Kinematics

Answer: h_max = 35.1 m

Rise above the roof

0 = v₀² − 2·g·Δh → Δh = 10²/(2·9.81) = 5.097 m

Height above the ground

h_max = 30 + 5.097 = 35.1 m

Answer

h_max = 35.1 m

3.Newton’s laws

Answer: a = 4.862 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.

Normal force

N = m·g − F·sin θ = 294.3 − 200·sin 35° = 179.6 N

Horizontal forces

m·a = F·cos θ − μ·N = 163.8 − 17.96

a = 145.9/30 = 4.862 m/s²

Answer

a = 4.862 m/s²

4.Newton’s laws

Answer: a = 1.09 m/s²

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (34.34 − 19.62)/13.5 = 1.09 m/s²

Answer

a = 1.09 m/s²

5.Projectile motion

Answer: y = 49.78 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.

Time to the wall

v_x = 34·cos 70° = 11.63 m/s

t = x/v_x = 30/11.63 = 2.58 s

Height at that time

y = 31.95·2.58 − ½·9.81·2.58² = 49.78 m

Answer

y = 49.78 m

The ball is still rising at the wall.

6.Projectile motion

Answer: x = 54.65 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 = 10.26 m/s, v_y = 28.19 m/s

Time to land on the roof

Larger root: the ball is descending.

11 = 28.19·t − 4.905·t²

t = (28.19 + √(28.19² − 2·9.81·11))/9.81 = 5.326 s

Horizontal distance

x = v_x·t = 10.26·5.326 = 54.65 m

Answer

x = 54.65 m

7.Work & energy

Answer: v = 14.99 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).

Energy balance along the slope

The block drops a height d·sin θ while friction removes μmg·cos θ·d. The mass cancels.

½v² = g·d·(sin θ − μ·cos θ)

v = √(2·9.81·18·(0.7071 − 0.1·0.7071)) = 14.99 m/s

Answer

v = 14.99 m/s

8.Work & energy

Answer: v = 8.287 m/s

Energy left after the rough strip

The ramp is smooth, so only the strip removes energy.

mgh − μmgL = 39.24·(4 − 0.2·2.5) = 137.3 J

Speed at the spring

½mv² = mg(h − μL) → v = √(2g(h − μL)) = √(2·9.81·3.5) = 8.287 m/s

Answer

v = 8.287 m/s

9.Momentum & collisions

Answer: v′₁ = 2.5 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.

Elastic collision with a target at rest

Solving momentum and energy conservation together gives these two results.

v′₁ = (5 − 3)/(5 + 3)·10 = 2.5 m/s

v′₂ = 2·5/(5 + 3)·10 = 12.5 m/s

Answer

v′₁ = 2.5 m/s

10.Momentum & collisions

Answer: J = 9.9 N·s

Change in momentum

J = m·Δv = 0.45·(9 + 13) = 9.9 N·s

Answer

J = 9.9 N·s

11.Hydrostatic force

Answer: F = 29.43 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 = 2.5 + 1/2 = 3 m

A = H·b = 1·1 = 1 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3·1 = 29.43 kN

Answer

F = 29.43 kN

12.Hydrostatic force

Answer: y_cp = 1.4 m

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 = 0.5 + 1.5/2 = 1.25 m

A = H·b = 1.5·1.5 = 2.25 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·1.25·2.25 = 24.28 kN

Centre of pressure

Pressure grows with depth, so the resultant acts below the centroid.

I_c = b·H³/12 = 1.5·1.5³/12 = 0.4219 m⁴

y_cp = y_c + I_c/(y_c·A) = 1.25 + 0.4219/(1.25·2.25) = 1.4 m

Answer

y_cp = 1.4 m

13.Continuity & Bernoulli

Answer: Q = 1.24 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) = √(2·9.81·2) = 6.264 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) = 0.63·0.0003142·6.264 = 0.00124 m³/s = 1.24 L/s

Answer

Q = 1.24 L/s

14.Continuity & Bernoulli

Answer: p₂ = 120.4 kPa

Continuity and Bernoulli

Work in Pa, then divide by 1000 to get kPa.

v₂ = v₁·(d₁/d₂)² = 3·(75/40)² = 10.55 m/s

½ρ(v₁² − v₂²) = ½·1000·(3² − 10.55²)/1000 = −51.12 kPa

ρgΔz = 1000·9.81·8/1000 = 78.48 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 250 + (−51.12) − 78.48 = 120.4 kPa

Answer

p₂ = 120.4 kPa

15.Ideal gas law

Answer: P = 1819 kPa

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

Moles

n = m/M = 440/32 = 13.75 mol

T = 45 + 273.15 = 318.1 K

Pressure

P = nRT/V = 13.75·8.314·318.1/0.02 = 1819 kPa

Answer

P = 1819 kPa

16.Ideal gas law

Answer: ρ = 4.932 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)

T = 0 + 273.15 = 273.1 K

Density

ρ = 350·32/(8.314·273.1) = 4.932 kg/m³

Answer

ρ = 4.932 kg/m³

17.Heat engines & refrigerators

Answer: η_max = 64.02 %

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 = 500 + 273.15 = 773.1 K

T_L = 5 + 273.15 = 278.1 K

Carnot efficiency

η_max = 1 − T_L/T_H = 1 − 278.1/773.1 = 64.02 %

Answer

η_max = 64.02 %

18.Heat engines & refrigerators

Answer: COP_max = 14.66

Absolute temperatures

T_H = 20 + 273.15 = 293.1 K

T_L = 0 + 273.15 = 273.1 K

Best possible COP

COP_HP,max = T_H/(T_H − T_L) = 293.1/(293.1 − 273.1) = 14.66

Answer

COP_max = 14.66

19.Waves, sound & optics

Answer: f₂ = 428.8 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, f_n = 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, f_n = n·v/(4L). When source or observer move along the line between them, f′ = f·(v ± v_o)/(v ∓ v_s): use the upper signs for motion towards each other. Two waves of nearly equal frequency beat at |f₁ − f₂|.

Wavelength of this mode

λ = 2L/n = 2·0.8/2 = 0.8 m

Frequency

f = v/λ = 343/0.8 = 428.8 Hz

Answer

f₂ = 428.8 Hz

20.Waves, sound & optics

Answer: w = 13 mm

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, …

First dark fringe

y₁ = λL/a = 650 × 10⁻⁹·1.5/(0.15 × 10⁻³) = 6.5 mm

Width of the central maximum

w = 2y₁ = 13 mm

Answer

w = 13 mm

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