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 starts from rest and accelerates uniformly at 1.5 m/s² along a runway. Find its speed v after 1900 m.

v₀ = 0a = 1.5 m/s²s = 1900 m

v = m/s

2.KinematicsHard

A ball is thrown straight up at 20 m/s from the edge of a roof 25 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height h_max above the ground.

v₀ = 20 m/sh₀ = 25 mg = 9.81 m/s²

h_max = m

3.Newton’s lawsMedium

A 9 kg block slides down a slope at 30° to the horizontal. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.

m = 9 kgθ = 30°μ = 0.25g = 9.81 m/s²

a = m/s²

4.Newton’s lawsHard

A 1 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 1.5 kg mass. The coefficient of kinetic friction between block and table is 0.3. Use g = 9.81 m/s². Find the acceleration a of the system.

m₁ = 1 kgm₂ = 1.5 kgμ = 0.3g = 9.81 m/s²

a = m/s²

5.Projectile motionMedium

A ball is launched from level ground at 65° above the horizontal and lands 100 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v₀.

θ = 65°R = 100 mg = 9.81 m/s²

v₀ = m/s

6.Projectile motionHard

A ball is launched at 15 m/s and 45° above the horizontal from the top of a 25 m cliff. Ignore air resistance and use g = 9.81 m/s². At what angle α below the horizontal does it hit the ground?

v₀ = 15 m/sθ = 45°h = 25 mg = 9.81 m/s²

α = °

7.Work & energyMedium

A 1800 kg car drives up a 3° slope at a constant 54 km/h. Air resistance and rolling resistance together amount to 750 N. Find the power P the wheels must deliver, in kW.

m = 1800 kgθ = 3°F_r = 750 Nv = 54 km/hg = 9.81 m/s²

P = kW

8.Work & energyHard

A 1 kg cart is released from rest at a height of 11.3 m on a smooth track that leads into a vertical circular loop of radius 2.5 m. Find the normal force N the track exerts on the cart at the top of the loop.

m = 1 kgh = 11.3 mR = 2.5 mg = 9.81 m/s²

N = N

9.Momentum & collisionsMedium

A 4.5 kg glider moving at 5 m/s makes a perfectly elastic head-on collision with a 5 kg glider at rest on a frictionless track. Find the velocity v′₂ of the second glider after the collision.

m₁ = 4.5 kgm₂ = 5 kgv₁ = 5 m/sv₂ = 0

v′₂ = m/s

10.Momentum & collisionsHard

A 12 g bullet is fired horizontally into a 1 kg block hanging from light strings 1.1 m long, and stays in it. The strings swing out to a largest angle of 35° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.

m = 12 gM = 1 kgL = 1.1 mθ = 35°g = 9.81 m/s²

v = m/s

11.Hydrostatic forceMedium

A vertical rectangular gate 1.5 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.

d = 1 mH = 2 mb = 1.5 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 glycerin, with its top edge 3 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 3 mH = 1.5 mb = 1.5 mρ = 1260 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 30 mm. The water level is kept 4 m above the orifice and the discharge coefficient is 0.62. Find the flow rate Q. Use g = 9.81 m/s².

d = 30 mmh = 4 mC_d = 0.62g = 9.81 m/s²

Q = L/s

14.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 200 mm to 100 mm while rising 3 m. At section 1 the pressure is 300 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 200 mmd₂ = 100 mmp₁ = 300 kPav₁ = 1.25 m/sΔz = 3 mρ = 1000 kg/m³

p₂ = kPa

15.Ideal gas lawMedium

What volume V does 190 g of CO₂ occupy at 100 kPa and 85 °C? Treat it as an ideal gas.

m = 190 gM = 44.01 g/molP = 100 kPaT = 85 °CR = 8.314 J/(mol·K)

V = L

16.Ideal gas lawHard

A fixed amount of He goes from 300 kPa, 5 L, 35 °C to 550 kPa and 250 °C. Find the new volume V₂.

P₁ = 300 kPaV₁ = 5 LT₁ = 35 °CP₂ = 550 kPaT₂ = 250 °C

V₂ = L

17.Heat engines & refrigeratorsMedium

A heat engine works between a hot reservoir at 725 °C and a cold reservoir at 5 °C. It takes in 1200 kJ of heat. Find the most work W_max it could possibly produce.

T_H = 725 °CT_L = 5 °CQ_H = 1200 kJ

W_max = kJ

18.Heat engines & refrigeratorsHard

A heat pump delivers 8 kW of heat to a heating system at 35 °C, taking heat from a source at 5 °C. Its actual COP is 40% of the Carnot COP between these temperatures. Find the electrical power Ẇ it uses.

T_H = 35 °CT_L = 5 °Cη_II = 40%Q̇_H = 8 kW

Ẇ = kW

19.Waves, sound & opticsMedium

An observer moves at 5 m/s towards a stationary source of 440 Hz. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.

f = 440 Hzv = 343 m/sv_o = 5 m/s

f′ = Hz

20.Waves, sound & opticsHard

Light of wavelength 633 nm falls at right angles on a grating with 400 lines per millimetre. Find the angle θ₁ of the order-1 maximum.

N = 400 mm⁻¹λ = 633 nmm = 1

θ₁ = °

3Solutions

1.Kinematics

Answer: v = 75.5 m/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.

Speed after the distance

v = √(2·a·s) = √(2·1.5·1900) = 75.5 m/s

Answer

v = 75.5 m/s

(271.8 km/h)

2.Kinematics

Answer: h_max = 45.39 m

Rise above the roof

0 = v₀² − 2·g·Δh → Δh = 20²/(2·9.81) = 20.39 m

Height above the ground

h_max = 25 + 20.39 = 45.39 m

Answer

h_max = 45.39 m

3.Newton’s laws

Answer: a = 2.781 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 30° − 0.25·cos 30°) = 9.81·(0.5 − 0.2165) = 2.781 m/s²

Answer

a = 2.781 m/s²

4.Newton’s laws

Answer: a = 4.709 m/s²

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (14.71 − 2.943)/2.5 = 4.709 m/s²

Answer

a = 4.709 m/s²

5.Projectile motion

Answer: v₀ = 35.79 m/s

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.

Solve the range formula for v₀

v₀ = √(g·R/sin 2θ) = √(9.81·100/sin 130°) = 35.79 m/s

Answer

v₀ = 35.79 m/s

6.Projectile motion

Answer: α = 66.64 °

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 = 15·cos 45° = 10.61 m/s

v_y0 = 15·sin 45° = 10.61 m/s

Vertical speed at impact

Energy or v² = v₀² + 2as in the vertical direction.

v_y = √(10.61² + 2·9.81·25) = 24.56 m/s

Impact angle

tan α = v_y/v_x = 24.56/10.61 → α = 66.64°

Answer

α = 66.64°

7.Work & energy

Answer: P = 25.11 kW

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). Power is force times speed: P = F·v.

Driving force

m·g = 1800·9.81 = 17658 N

mg·sin θ = 17658·sin 3° = 924.1 N

F = 924.1 + 750 = 1674 N

Power

v = 54/3.6 = 15 m/s

P = F·v = 1674·15 = 25112 W = 25.11 kW

Answer

P = 25.11 kW

8.Work & energy

Answer: N = 39.63 N

Speed at the top

v² = 2g(h − 2R) = 2·9.81·(11.3 − 5) = 123.6 m²/s²

Forces at the top

m·g = 1·9.81 = 9.81 N

N = mv²/R − mg = 1·123.6/2.5 − 9.81 = 39.63 N

Answer

N = 39.63 N

The cart just makes it round when h = 2.5R; then N = 0 at the top.

9.Momentum & collisions

Answer: v′₂ = 4.737 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 = (4.5 − 5)/(4.5 + 5)·5 = −0.2632 m/s

v′₂ = 2·4.5/(4.5 + 5)·5 = 4.737 m/s

Answer

v′₂ = 4.737 m/s

10.Momentum & collisions

Answer: v = 166.6 m/s

Rise of the block

h = L·(1 − cos θ) = 1.1·(1 − cos 35°) = 0.1989 m

Speed after impact

V = √(2·9.81·0.1989) = 1.976 m/s

Collision

v = (m + M)·V/m = 1.012·1.976/0.012 = 166.6 m/s

Answer

v = 166.6 m/s

11.Hydrostatic force

Answer: F = 58.86 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 + 2/2 = 2 m

A = H·b = 2·1.5 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2·3 = 58.86 kN

Answer

F = 58.86 kN

12.Hydrostatic force

Answer: F = 104.3 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 = 3 + 1.5/2 = 3.75 m

A = H·b = 1.5·1.5 = 2.25 m²

Resultant force

F = ρ·g·y_c·A = 1260·9.81·3.75·2.25 = 104.3 kN

Answer

F = 104.3 kN

13.Continuity & Bernoulli

Answer: Q = 3.882 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) = √(2·9.81·4) = 8.859 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) = 0.62·0.0007069·8.859 = 0.003882 m³/s = 3.882 L/s

Answer

Q = 3.882 L/s

14.Continuity & Bernoulli

Answer: p₂ = 258.9 kPa

Continuity and Bernoulli

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

v₂ = v₁·(d₁/d₂)² = 1.25·(200/100)² = 5 m/s

½ρ(v₁² − v₂²) = ½·1000·(1.25² − 5²)/1000 = −11.72 kPa

ρgΔz = 1000·9.81·3/1000 = 29.43 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 300 + (−11.72) − 29.43 = 258.9 kPa

Answer

p₂ = 258.9 kPa

15.Ideal gas law

Answer: V = 128.6 L

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

Moles

n = 190/44.01 = 4.317 mol

T = 85 + 273.15 = 358.1 K

Volume

V = nRT/P = 4.317·8.314·358.1/100 = 128.6 L

Answer

V = 128.6 L

16.Ideal gas law

Answer: V₂ = 4.63 L

Combined gas law: for a fixed amount of ideal gas, P₁V₁/T₁ = P₂V₂/T₂.

Absolute temperatures

T₁ = 308.1 K, T₂ = 523.1 K

Combined gas law

V₂ = V₁·(P₁/P₂)·(T₂/T₁) = 5·(300/550)·(523.1/308.1) = 4.63 L

Answer

V₂ = 4.63 L

17.Heat engines & refrigerators

Answer: W_max = 865.6 kJ

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 = 725 + 273.15 = 998.1 K

T_L = 5 + 273.15 = 278.1 K

Carnot efficiency

η_max = 1 − T_L/T_H = 1 − 278.1/998.1 = 72.13 %

Maximum work

W_max = η_max·Q_H = 0.7213·1200 = 865.6 kJ

Answer

W_max = 865.6 kJ

18.Heat engines & refrigerators

Answer: Ẇ = 1.947 kW

Absolute temperatures

T_H = 35 + 273.15 = 308.1 K

T_L = 5 + 273.15 = 278.1 K

Carnot and actual COP

COP_max = 308.1/(308.1 − 278.1) = 10.27

COP = 0.4·10.27 = 4.109

Power input

Ẇ = Q̇_H/COP = 8/4.109 = 1.947 kW

Answer

Ẇ = 1.947 kW

19.Waves, sound & optics

Answer: f′ = 446.4 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|.

Doppler formula

f′ = f·(v + v_o)/(v − v_s) = 440·(343 + 5)/(343 − 0)

f′ = 446.4 Hz

Answer

f′ = 446.4 Hz

20.Waves, sound & optics

Answer: θ₁ = 14.67 °

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 = 1/400 mm = 2.5 × 10⁻⁶ m

Grating equation

sin θ = mλ/d = 1·633 × 10⁻⁹/(2.5 × 10⁻⁶) = 0.2532

θ = 14.67°

Answer

θ₁ = 14.67°

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