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 car driving at 90 km/h brakes with a constant deceleration of 8 m/s² until it stops. How long does braking take? Find t.

v₀ = 90 km/ha = −8 m/s²v = 0

t = s

2.KinematicsHard

A ball is thrown straight up at 10 m/s from the edge of a roof 15 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 = 15 mg = 9.81 m/s²

h_max = m

3.Newton’s lawsMedium

A 15 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.

m = 15 kgθ = 25°μ = 0.2g = 9.81 m/s²

a = m/s²

4.Newton’s lawsHard

A 6.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 3 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 = 6.5 kgm₂ = 3 kgμ = 0.3g = 9.81 m/s²

a = m/s²

5.Projectile motionMedium

A ball is launched at 70° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.

v₀ = 18 m/sθ = 70°g = 9.81 m/s²

x = m

6.Projectile motionHard

A ball is launched at 16 m/s and 70° above the horizontal from the top of a 40 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.

v₀ = 16 m/sθ = 70°h = 40 mg = 9.81 m/s²

H = m

7.Work & energyMedium

A 0.5 kg block slides 14 m down a slope at 35° to the horizontal, with a coefficient of kinetic friction of 0.3. Find the work W_f done against friction.

m = 0.5 kgθ = 35°μ = 0.3d = 14 mg = 9.81 m/s²

W_f = J

8.Work & energyHard

A 10 kg block is released from rest at height 3.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3.5 m long (μ = 0.3) and then runs onto a spring (k = 1500 N/m) on a smooth surface. Find the maximum compression x of the spring.

m = 10 kgh = 3.5 mL = 3.5 mμ = 0.3k = 1500 N/mg = 9.81 m/s²

x = cm

9.Momentum & collisionsMedium

A 3 kg glider moving at 12 m/s makes a perfectly elastic head-on collision with a 4.5 kg glider at rest on a frictionless track. Find the velocity v′₂ of the second glider after the collision.

m₁ = 3 kgm₂ = 4.5 kgv₁ = 12 m/sv₂ = 0

v′₂ = m/s

10.Momentum & collisionsHard

A 2000 kg car runs into a parked 1100 kg car. The wrecks lock together and skid 24.5 m before stopping. The coefficient of kinetic friction between tyres and road is 0.5. Use g = 9.81 m/s². Find the speed v₁ of the first car just before the collision.

m₁ = 2000 kgm₂ = 1100 kgd = 24.5 mμ = 0.5g = 9.81 m/s²

v₁ = 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 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 2.5 mH = 1 mb = 1.5 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 water, 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 = 2 mb = 3 mρ = 1000 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 30 mm. The water level is kept 6 m above the orifice and the discharge coefficient is 0.6. Find the flow rate Q. Use g = 9.81 m/s².

d = 30 mmh = 6 mC_d = 0.6g = 9.81 m/s²

Q = L/s

14.Continuity & BernoulliHard

A horizontal venturi meter (inlet 100 mm, throat 50 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 340 mm. The discharge coefficient is 0.96. 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 = 100 mmd₂ = 50 mmΔh = 340 mmC_d = 0.96ρ_Hg = 13 600 kg/m³g = 9.81 m/s²

Q = L/s

15.Ideal gas lawMedium

What volume V does 200 g of He occupy at 700 kPa and 90 °C? Treat it as an ideal gas.

m = 200 gM = 4.003 g/molP = 700 kPaT = 90 °CR = 8.314 J/(mol·K)

V = L

16.Ideal gas lawHard

A fixed amount of He goes from 100 kPa, 20 L, 35 °C to 100 kPa and 50 °C. Find the new volume V₂.

P₁ = 100 kPaV₁ = 20 LT₁ = 35 °CP₂ = 100 kPaT₂ = 50 °C

V₂ = L

17.Heat engines & refrigeratorsMedium

A heat engine works between a hot reservoir at 350 °C and a cold reservoir at 20 °C. Find the highest possible thermal efficiency η_max.

T_H = 350 °CT_L = 20 °C

η_max = %

18.Heat engines & refrigeratorsHard

A heat pump keeps a house at 22 °C while it is 7 °C outside, delivering 9 kW of heat to the house. Find the minimum power input Ẇ_min.

T_H = 22 °CT_L = 7 °CQ̇_H = 9 kW

Ẇ_min = kW

19.Waves, sound & opticsMedium

A pipe open at both ends must have its harmonic number 4 at 1143 Hz. The speed of sound is 343 m/s. Find the length L of the pipe.

v = 343 m/sn = 4f₄ = 1143 Hz

L = m

20.Waves, sound & opticsHard

A sensor at rest sends ultrasound of 20 kHz towards a car that approaches at 5 m/s and mixes the echo with the emitted signal. Find the beat frequency f_b (speed of sound 343 m/s).

f = 20 kHzu = 5 m/sv = 343 m/s

f_b = Hz

3Solutions

1.Kinematics

Answer: t = 3.125 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.

Units

v₀ = 90/3.6 = 25 m/s

Braking time

0 = v₀ − a·t → t = v₀/a = 25/8 = 3.125 s

Answer

t = 3.125 s

2.Kinematics

Answer: h_max = 20.1 m

Rise above the roof

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

Height above the ground

h_max = 15 + 5.097 = 20.1 m

Answer

h_max = 20.1 m

3.Newton’s laws

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.

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 25° − 0.2·cos 25°) = 9.81·(0.4226 − 0.1813) = 2.368 m/s²

Answer

a = 2.368 m/s²

4.Newton’s laws

Answer: a = 1.084 m/s²

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (29.43 − 19.13)/9.5 = 1.084 m/s²

Answer

a = 1.084 m/s²

5.Projectile motion

Answer: x = 21.23 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 = 18·cos 70° = 6.156 m/s

v_y = v₀·sin θ = 18·sin 70° = 16.91 m/s

Time in the air

Starting and landing at the same height: t = 2v_y/g.

t = 2·16.91/9.81 = 3.448 s

Horizontal distance

x = v_x·t = 6.156·3.448 = 21.23 m

Answer

x = 21.23 m

6.Projectile motion

Answer: H = 51.52 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.

Vertical launch speed

v_y = 16·sin 70° = 15.04 m/s

Rise above the cliff

Δh = v_y²/(2g) = 15.04²/(2·9.81) = 11.52 m

Height above the ground

H = 40 + 11.52 = 51.52 m

Answer

H = 51.52 m

7.Work & energy

Answer: W_f = 16.88 J

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).

Friction force

N = mg·cos θ = 4.905·0.8192 = 4.018 N

F = μN = 0.3·4.018 = 1.205 N

Work against friction

W_f = F·d = 1.205·14 = 16.88 J

Answer

W_f = 16.88 J

8.Work & energy

Answer: x = 56.61 cm

Energy left after the rough strip

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

mgh − μmgL = 98.1·(3.5 − 0.3·3.5) = 240.3 J

Maximum compression

½kx² = 240.3 → x = √(2·240.3/1500) = 0.5661 m = 56.61 cm

Answer

x = 56.61 cm

9.Momentum & collisions

Answer: v′₂ = 9.6 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 = (3 − 4.5)/(3 + 4.5)·12 = −2.4 m/s

v′₂ = 2·3/(3 + 4.5)·12 = 9.6 m/s

Answer

v′₂ = 9.6 m/s

10.Momentum & collisions

Answer: v₁ = 24.03 m/s

Speed after the collision

V = √(2·μ·g·d) = √(2·0.5·9.81·24.5) = 15.5 m/s

Momentum in the collision

m₁·v₁ = (m₁ + m₂)·V → v₁ = 3100·15.5/2000 = 24.03 m/s

Answer

v₁ = 24.03 m/s

(86.51 km/h)

11.Hydrostatic force

Answer: F = 44.15 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 = 1·1.5 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3·1.5 = 44.15 kN

Answer

F = 44.15 kN

12.Hydrostatic force

Answer: y_cp = 1.722 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 + 2/2 = 1.5 m

A = H·b = 2·3 = 6 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·1.5·6 = 88.29 kN

Centre of pressure

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

I_c = b·H³/12 = 3·2³/12 = 2 m⁴

y_cp = y_c + I_c/(y_c·A) = 1.5 + 2/(1.5·6) = 1.722 m

Answer

y_cp = 1.722 m

13.Continuity & Bernoulli

Answer: Q = 4.602 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) = √(2·9.81·6) = 10.85 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) = 0.6·0.0007069·10.85 = 0.004602 m³/s = 4.602 L/s

Answer

Q = 4.602 L/s

14.Continuity & Bernoulli

Answer: Q = 17.85 L/s

Pressure difference from the manometer

Δp = (13600 − 1000)·9.81·0.34 = 42026 Pa

Throat velocity

Continuity and Bernoulli together give the ideal throat velocity.

β = 50/100 = 0.5

v₂ = √(2·42026/(1000·(1 − 0.0625))) = 9.469 m/s

Actual flow rate

Q = C_d·A₂·v₂ = 0.96·0.001963·9.469 = 0.01785 m³/s = 17.85 L/s

Answer

Q = 17.85 L/s

15.Ideal gas law

Answer: V = 215.5 L

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

Moles

n = 200/4.003 = 49.96 mol

T = 90 + 273.15 = 363.1 K

Volume

V = nRT/P = 49.96·8.314·363.1/700 = 215.5 L

Answer

V = 215.5 L

16.Ideal gas law

Answer: V₂ = 20.97 L

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

Absolute temperatures

T₁ = 308.1 K, T₂ = 323.1 K

Combined gas law

V₂ = V₁·(P₁/P₂)·(T₂/T₁) = 20·(100/100)·(323.1/308.1) = 20.97 L

Answer

V₂ = 20.97 L

17.Heat engines & refrigerators

Answer: η_max = 52.96 %

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 = 350 + 273.15 = 623.1 K

T_L = 20 + 273.15 = 293.1 K

Carnot efficiency

η_max = 1 − T_L/T_H = 1 − 293.1/623.1 = 52.96 %

Answer

η_max = 52.96 %

18.Heat engines & refrigerators

Answer: Ẇ_min = 0.4574 kW

Absolute temperatures

T_H = 22 + 273.15 = 295.1 K

T_L = 7 + 273.15 = 280.1 K

Best possible COP

COP_HP,max = T_H/(T_H − T_L) = 295.1/(295.1 − 280.1) = 19.68

Minimum power

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

Ẇ_min = Q̇_H/COP = 9/19.68 = 0.4574 kW

Answer

Ẇ_min = 0.4574 kW

19.Waves, sound & optics

Answer: L = 0.6002 m

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

λ = v/f = 343/1143 = 0.3001 m

Length

L = nλ/2 = 4·0.3001/2 = 0.6002 m

Answer

L = 0.6002 m

20.Waves, sound & optics

Answer: f_b = 591.7 Hz

Frequency at the car

f′ = 20000·(343 + 5)/343 = 20292 Hz

Frequency of the echo

f″ = f′·343/(343 − 5) = 20592 Hz

Beat frequency

f_b = |f″ − f| = 591.7 Hz

Answer

f_b = 591.7 Hz

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