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 120 km/h brakes with a constant deceleration of 6.5 m/s² until it stops. Find the braking distance s.

v₀ = 120 km/ha = −6.5 m/s²v = 0

s = m

2.KinematicsHard

A tram starts from rest and accelerates at 2.5 m/s² for 5 s, then runs at constant speed for 9 s, and finally brakes at 2 m/s² until it stops. Find the total distance s between the two stops.

a₁ = 2.5 m/s²t₁ = 5 st₂ = 9 sa₃ = −2 m/s²

s = m

3.Newton’s lawsMedium

A 13 kg block slides down a slope at 25° 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 = 13 kgθ = 25°μ = 0.3g = 9.81 m/s²

a = m/s²

4.Newton’s lawsHard

A 1.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 6 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 = 1.5 kgm₂ = 6 kgμ = 0.2g = 9.81 m/s²

a = m/s²

5.Projectile motionMedium

A ball is kicked from level ground at 20 m/s and 25° above the horizontal towards a wall 9 m away. Ignore air resistance and use g = 9.81 m/s². At what height y does the ball pass the wall?

v₀ = 20 m/sθ = 25°x = 9 mg = 9.81 m/s²

y = m

6.Projectile motionHard

A ball is launched at 30° above horizontal from the top of a 15 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.

v₀ = 31 m/sθ = 30°h = 15 mg = 9.81 m/s²

x = m

7.Work & energyMedium

A 10 kg block slides 10 m down a slope at 40° to the horizontal, with a coefficient of kinetic friction of 0.35. Find the work W_f done against friction.

m = 10 kgθ = 40°μ = 0.35d = 10 mg = 9.81 m/s²

W_f = J

8.Work & energyHard

A 2.5 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 = 0.1) and then runs onto a spring (k = 2000 N/m) on a smooth surface. Find the maximum compression x of the spring.

m = 2.5 kgh = 4 mL = 2.5 mμ = 0.1k = 2000 N/mg = 9.81 m/s²

x = cm

9.Momentum & collisionsMedium

A 15 g bullet is fired horizontally into a 5 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 6 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.

m = 15 gM = 5 kgh = 6 cmg = 9.81 m/s²

v = m/s

10.Momentum & collisionsHard

A 0.5 kg ball hits a wall at 21 m/s and bounces straight back at 4 m/s. The contact lasts 40 ms. Find the average force F_avg of the wall on the ball.

m = 0.5 kgv_in = 21 m/sv_out = 4 m/sΔt = 40 ms

F_avg = N

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

F = kN

12.Hydrostatic forceHard

A vertical rectangular gate 2 m wide sits in the wall of a tank of seawater, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 2.5 mH = 1.5 mb = 2 mρ = 1025 kg/m³g = 9.81 m/s²

F = kN

13.Continuity & BernoulliMedium

A pitot-static tube in a wind tunnel reads a difference of 1100 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.

p₀ − p = 1100 Paρ = 1.2 kg/m³

v = m/s

14.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 100 mm to 50 mm while rising 10 m. At section 1 the pressure is 240 kPa and the velocity is 1.75 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 100 mmd₂ = 50 mmp₁ = 240 kPav₁ = 1.75 m/sΔz = 10 mρ = 1000 kg/m³

p₂ = kPa

15.Ideal gas lawMedium

What volume V does 260 g of He occupy at 700 kPa and 20 °C? Treat it as an ideal gas.

m = 260 gM = 4.003 g/molP = 700 kPaT = 20 °CR = 8.314 J/(mol·K)

V = L

16.Ideal gas lawHard

Find the density ρ of Ar at 1000 kPa and 30 °C, treating it as an ideal gas.

M = 39.95 g/molP = 1000 kPaT = 30 °CR = 8.314 J/(mol·K)

ρ = kg/m³

17.Heat engines & refrigeratorsMedium

A reversible refrigerator keeps a cold room at 0 °C by removing 5 kW of heat, and rejects heat to the surroundings at 35 °C. Find the rate Q̇_H at which it rejects heat.

T_L = 0 °CT_H = 35 °CQ̇_L = 5 kW

Q̇_H = kW

18.Heat engines & refrigeratorsHard

A refrigerator keeps its cold space at -10 °C in a kitchen at 30 °C, removing heat from the cold space at 5 kW. Find the highest possible coefficient of performance COP_max.

T_H = 30 °CT_L = -10 °CQ̇_L = 5 kW

COP_max =

19.Waves, sound & opticsMedium

A string fixed at both ends is 0.6 m long; waves travel on it at 200 m/s. Find the frequency f₂ of its harmonic number 2.

v = 200 m/sn = 2L = 0.6 m

f₂ = Hz

20.Waves, sound & opticsHard

A parking sensor at rest sends ultrasound of 35 kHz towards a car that approaches at 30 m/s. Find the frequency f″ of the echo received back at the sensor (speed of sound 343 m/s).

f = 35 kHzu = 30 m/sv = 343 m/s

f″ = Hz

3Solutions

1.Kinematics

Answer: s = 85.47 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.

Units

v₀ = 120/3.6 = 33.33 m/s

Braking distance

0 = v₀² − 2·a·s → s = v₀²/(2a) = 33.33²/(2·6.5) = 85.47 m

Answer

s = 85.47 m

2.Kinematics

Answer: s = 182.8 m

Top speed

v = a₁·t₁ = 2.5·5 = 12.5 m/s

Distance of each phase

s₁ = ½·2.5·5² = 31.25 m

s₂ = 12.5·9 = 112.5 m

s₃ = 12.5²/(2·2) = 39.06 m

Total distance

s = 31.25 + 112.5 + 39.06 = 182.8 m

Answer

s = 182.8 m

3.Newton’s laws

Answer: a = 1.479 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° − 0.3·cos 25°) = 9.81·(0.4226 − 0.2719) = 1.479 m/s²

Answer

a = 1.479 m/s²

4.Newton’s laws

Answer: a = 7.456 m/s²

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (58.86 − 2.943)/7.5 = 7.456 m/s²

Answer

a = 7.456 m/s²

5.Projectile motion

Answer: y = 2.988 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 = 20·cos 25° = 18.13 m/s

t = x/v_x = 9/18.13 = 0.4965 s

Height at that time

y = 8.452·0.4965 − ½·9.81·0.4965² = 2.988 m

Answer

y = 2.988 m

The ball is still rising at the wall.

6.Projectile motion

Answer: x = 105.7 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 = 31·cos 30° = 26.85 m/s

v_y = v₀·sin θ = 31·sin 30° = 15.5 m/s

Time in the air

Solve y(t) = h + v_y·t − ½gt² = 0 for the positive root.

0 = 15 + 15.5·t − 4.905·t²

t = (v_y + √(v_y² + 2gh))/g = (15.5 + √(240.2 + 294.3))/9.81 = 3.937 s

Horizontal distance

x = v_x·t = 26.85·3.937 = 105.7 m

Answer

x = 105.7 m

7.Work & energy

Answer: W_f = 263 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 = 98.1·0.766 = 75.15 N

F = μN = 0.35·75.15 = 26.3 N

Work against friction

W_f = F·d = 26.3·10 = 263 J

Answer

W_f = 263 J

8.Work & energy

Answer: x = 30.33 cm

Energy left after the rough strip

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

mgh − μmgL = 24.53·(4 − 0.1·2.5) = 91.97 J

Maximum compression

½kx² = 91.97 → x = √(2·91.97/2000) = 0.3033 m = 30.33 cm

Answer

x = 30.33 cm

9.Momentum & collisions

Answer: v = 362.7 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.

Swing after impact

V = √(2·g·h) = √(2·9.81·0.06) = 1.085 m/s

Collision

m·v = (m + M)·V → v = (0.015 + 5)·1.085/0.015 = 362.7 m/s

Answer

v = 362.7 m/s

10.Momentum & collisions

Answer: F_avg = 312.5 N

Change in momentum

J = m·Δv = 0.5·(4 + 21) = 12.5 N·s

Average force

F_avg = J/Δt = 12.5/0.04 = 312.5 N

Answer

F_avg = 312.5 N

11.Hydrostatic force

Answer: F = 176.6 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 + 3/2 = 4 m

A = H·b = 3·1.5 = 4.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4·4.5 = 176.6 kN

Answer

F = 176.6 kN

12.Hydrostatic force

Answer: F = 98.04 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 + 1.5/2 = 3.25 m

A = H·b = 1.5·2 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1025·9.81·3.25·3 = 98.04 kN

Answer

F = 98.04 kN

13.Continuity & Bernoulli

Answer: v = 42.82 m/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.

Bernoulli to the stagnation point

The flow comes to rest at the tip, so its kinetic energy shows up as extra pressure.

p₀ − p = ½ρv²

v = √(2·1100/1.2) = 42.82 m/s

Answer

v = 42.82 m/s

14.Continuity & Bernoulli

Answer: p₂ = 118.9 kPa

Continuity and Bernoulli

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

v₂ = v₁·(d₁/d₂)² = 1.75·(100/50)² = 7 m/s

½ρ(v₁² − v₂²) = ½·1000·(1.75² − 7²)/1000 = −22.97 kPa

ρgΔz = 1000·9.81·10/1000 = 98.1 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 240 + (−22.97) − 98.1 = 118.9 kPa

Answer

p₂ = 118.9 kPa

15.Ideal gas law

Answer: V = 226.1 L

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

Moles

n = 260/4.003 = 64.95 mol

T = 20 + 273.15 = 293.1 K

Volume

V = nRT/P = 64.95·8.314·293.1/700 = 226.1 L

Answer

V = 226.1 L

16.Ideal gas law

Answer: ρ = 15.85 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)

T = 30 + 273.15 = 303.1 K

Density

ρ = 1000·39.95/(8.314·303.1) = 15.85 kg/m³

Answer

ρ = 15.85 kg/m³

17.Heat engines & refrigerators

Answer: Q̇_H = 5.641 kW

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 = 35 + 273.15 = 308.1 K

T_L = 0 + 273.15 = 273.1 K

Reversible heat ratio

Q̇_H = Q̇_L·T_H/T_L = 5·308.1/273.1 = 5.641 kW

Answer

Q̇_H = 5.641 kW

Power input: Ẇ = 0.6407 kW.

18.Heat engines & refrigerators

Answer: COP_max = 6.579

Absolute temperatures

T_H = 30 + 273.15 = 303.1 K

T_L = -10 + 273.15 = 263.1 K

Best possible COP

COP_R,max = T_L/(T_H − T_L) = 263.1/(303.1 − 263.1) = 6.579

Answer

COP_max = 6.579

19.Waves, sound & optics

Answer: f₂ = 333.3 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 = 2·0.6/2 = 0.6 m

Frequency

f = v/λ = 200/0.6 = 333.3 Hz

Answer

f₂ = 333.3 Hz

20.Waves, sound & optics

Answer: f″ = 41709 Hz

Frequency at the car

f′ = 35000·(343 + 30)/343 = 38061 Hz

Frequency of the echo

f″ = f′·343/(343 − 30) = 41709 Hz

Answer

f″ = 41709 Hz

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