Physics for Engineers: cram pack

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

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 4 m/s² until it stops. Find the braking distance s.

v₀ = 90 km/ha = −4 m/s²v = 0

s = m

2.KinematicsHard

A ball is thrown straight up at 18 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 = 18 m/sh₀ = 25 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 30° above the horizontal with a force of 120 N. The coefficient of kinetic friction is 0.4. Use g = 9.81 m/s². Find the acceleration a of the sled.

m = 30 kgF = 120 Nθ = 30°μ = 0.4g = 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 5 kg mass. The coefficient of kinetic friction between block and table is 0.25. Use g = 9.81 m/s². Find the acceleration a of the system.

m₁ = 10 kgm₂ = 5 kgμ = 0.25g = 9.81 m/s²

a = m/s²

5.Projectile motionMedium

A ball is launched from level ground at 40 m/s and must land 70 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.

v₀ = 40 m/sR = 70 mg = 9.81 m/s²

θ = °

6.Projectile motionHard

A ball is launched at 13 m/s and 50° above the horizontal from the top of a 10 m cliff. Ignore air resistance and use g = 9.81 m/s². Find the speed v with which it hits the ground.

v₀ = 13 m/sθ = 50°h = 10 mg = 9.81 m/s²

v = m/s

7.Work & energyMedium

A 1000 kg car drives up a 3° slope at a constant 54 km/h. Air resistance and rolling resistance together amount to 800 N. Find the power P the wheels must deliver, in kW.

m = 1000 kgθ = 3°F_r = 800 Nv = 54 km/hg = 9.81 m/s²

P = kW

8.Work & energyHard

A 3 kg block rests on a rough horizontal table (μ = 0.1). A light string runs from it over a smooth pulley at the edge of the table to a hanging 1 kg block. The system is released from rest. Find the speed v of the blocks after the hanging block has dropped 2 m.

m₁ = 3 kgm₂ = 1 kgμ = 0.1d = 2 mg = 9.81 m/s²

v = m/s

9.Momentum & collisionsMedium

A 5 g bullet is fired horizontally into a 2 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 11 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.

m = 5 gM = 2 kgh = 11 cmg = 9.81 m/s²

v = m/s

10.Momentum & collisionsHard

At a crossing a 1500 kg car travelling east at 19 m/s collides with a 1300 kg car travelling north at 23 m/s. The cars lock together. Find the direction θ of the wreck just after the collision, measured north of east.

m₁ = 1500 kgv₁ = 19 m/s (E)m₂ = 1300 kgv₂ = 23 m/s (N)

θ = °

11.Hydrostatic forceMedium

A vertical rectangular gate 2 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 = 2.5 mb = 2 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 resultant hydrostatic force F on the gate.

d = 0.5 mH = 2.5 mb = 1.5 mρ = 880 kg/m³g = 9.81 m/s²

F = kN

13.Continuity & BernoulliMedium

A pitot-static tube in a water pipe reads a difference of 21 kPa between the stagnation pressure and the static pressure. Find the water velocity v at the tip.

p₀ − p = 21 kPaρ = 1000 kg/m³

v = m/s

14.Continuity & BernoulliHard

A horizontal venturi meter (inlet 250 mm, throat 125 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 400 mm. The discharge coefficient is 0.99. 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 = 250 mmd₂ = 125 mmΔh = 400 mmC_d = 0.99ρ_Hg = 13 600 kg/m³g = 9.81 m/s²

Q = L/s

15.Ideal gas lawMedium

A tank of 20 L contains 100 g of N₂ at 35 °C. Treat it as an ideal gas and find the pressure P.

m = 100 gM = 28.01 g/molV = 20 LT = 35 °CR = 8.314 J/(mol·K)

P = kPa

16.Ideal gas lawHard

A fixed amount of Ar goes from 150 kPa, 2 L, 15 °C to 600 kPa and 90 °C. Find the new volume V₂.

P₁ = 150 kPaV₁ = 2 LT₁ = 15 °CP₂ = 600 kPaT₂ = 90 °C

V₂ = L

17.Heat engines & refrigeratorsMedium

A reversible (Carnot) heat engine takes in 2500 kJ of heat from a reservoir at 575 °C and rejects 849 kJ per cycle. Find the temperature T_L of the cold reservoir in °C.

T_H = 575 °CQ_H = 2500 kJQ_L = 849 kJ

T_L = °C

18.Heat engines & refrigeratorsHard

A heat engine takes in 4100 kJ of heat from a reservoir at 850 °C, produces 1517 kJ of work and rejects the rest to a reservoir at 25 °C, per cycle. Find the total entropy generated S_gen per cycle.

T_H = 850 °CT_L = 25 °CQ_H = 4100 kJW = 1517 kJ

S_gen = kJ/K

19.Waves, sound & opticsMedium

An object stands 15 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 = 15 cmf = 5 cm

d_i = cm

20.Waves, sound & opticsHard

An object is 10 cm in front of a concave mirror with a radius of curvature of 30 cm. Find the magnification m (negative for an inverted image).

d_o = 10 cmR = 30 cm

m =

3Solutions

1.Kinematics

Answer: s = 78.13 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 = 90/3.6 = 25 m/s

Braking distance

0 = v₀² − 2·a·s → s = v₀²/(2a) = 25²/(2·4) = 78.13 m

Answer

s = 78.13 m

2.Kinematics

Answer: h_max = 41.51 m

Rise above the roof

0 = v₀² − 2·g·Δh → Δh = 18²/(2·9.81) = 16.51 m

Height above the ground

h_max = 25 + 16.51 = 41.51 m

Answer

h_max = 41.51 m

3.Newton’s laws

Answer: a = 0.3401 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 − 120·sin 30° = 234.3 N

Horizontal forces

m·a = F·cos θ − μ·N = 103.9 − 93.72

a = 10.2/30 = 0.3401 m/s²

Answer

a = 0.3401 m/s²

4.Newton’s laws

Answer: a = 1.635 m/s²

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (49.05 − 24.53)/15 = 1.635 m/s²

Answer

a = 1.635 m/s²

5.Projectile motion

Answer: θ = 12.71 °

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.

Range formula

R = v₀²·sin 2θ/g → sin 2θ = 9.81·70/40² = 0.4292

Angle

2θ = 25.42° → θ = 12.71°

Answer

θ = 12.71°

The other solution is 90° − θ = 77.29°.

6.Projectile motion

Answer: v = 19.11 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.

Split the launch velocity

v_x = 13·cos 50° = 8.356 m/s

v_y0 = 13·sin 50° = 9.959 m/s

Vertical speed at impact

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

v_y = √(9.959² + 2·9.81·10) = 17.19 m/s

Impact speed

v = √(8.356² + 17.19²) = 19.11 m/s

Answer

v = 19.11 m/s

Check with energy: v² = v₀² + 2gh.

7.Work & energy

Answer: P = 19.7 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 = 1000·9.81 = 9810 N

mg·sin θ = 9810·sin 3° = 513.4 N

F = 513.4 + 800 = 1313 N

Power

v = 54/3.6 = 15 m/s

P = F·v = 1313·15 = 19701 W = 19.7 kW

Answer

P = 19.7 kW

8.Work & energy

Answer: v = 2.62 m/s

Energy released and lost

m₂g·d = 9.81·2 = 19.62 J

μ·m₁g·d = 0.1·29.43·2 = 5.886 J

Kinetic energy of both blocks

½(m₁ + m₂)v² = 13.73 → v = √(2·13.73/4) = 2.62 m/s

Answer

v = 2.62 m/s

9.Momentum & collisions

Answer: v = 589.1 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.

Swing after impact

V = √(2·g·h) = √(2·9.81·0.11) = 1.469 m/s

Collision

m·v = (m + M)·V → v = (0.005 + 2)·1.469/0.005 = 589.1 m/s

Answer

v = 589.1 m/s

10.Momentum & collisions

Answer: θ = 46.37 °

Momentum components

p_x = 1500·19 = 28500 kg·m/s

p_y = 1300·23 = 29900 kg·m/s

Direction

tan θ = p_y/p_x = 1.049 → θ = 46.37°

Answer

θ = 46.37°

11.Hydrostatic force

Answer: F = 257.5 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 + 2.5/2 = 5.25 m

A = H·b = 2.5·2 = 5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·5.25·5 = 257.5 kN

Answer

F = 257.5 kN

12.Hydrostatic force

Answer: F = 56.65 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 = 0.5 + 2.5/2 = 1.75 m

A = H·b = 2.5·1.5 = 3.75 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·1.75·3.75 = 56.65 kN

Answer

F = 56.65 kN

13.Continuity & Bernoulli

Answer: v = 6.481 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 = √(2·21000/1000) = 6.481 m/s

Answer

v = 6.481 m/s

14.Continuity & Bernoulli

Answer: Q = 124.8 L/s

Pressure difference from the manometer

Δp = (13600 − 1000)·9.81·0.4 = 49442 Pa

Throat velocity

Continuity and Bernoulli together give the ideal throat velocity.

β = 125/250 = 0.5

v₂ = √(2·49442/(1000·(1 − 0.0625))) = 10.27 m/s

Actual flow rate

Q = C_d·A₂·v₂ = 0.99·0.01227·10.27 = 0.1248 m³/s = 124.8 L/s

Answer

Q = 124.8 L/s

15.Ideal gas law

Answer: P = 457.3 kPa

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

Moles

n = m/M = 100/28.01 = 3.57 mol

T = 35 + 273.15 = 308.1 K

Pressure

P = nRT/V = 3.57·8.314·308.1/0.02 = 457.3 kPa

Answer

P = 457.3 kPa

16.Ideal gas law

Answer: V₂ = 0.6301 L

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

Absolute temperatures

T₁ = 288.1 K, T₂ = 363.1 K

Combined gas law

V₂ = V₁·(P₁/P₂)·(T₂/T₁) = 2·(150/600)·(363.1/288.1) = 0.6301 L

Answer

V₂ = 0.6301 L

17.Heat engines & refrigerators

Answer: T_L = 14.88 °C

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 temperature of the source

T_H = 575 + 273.15 = 848.1 K

Reversible heat ratio

T_L = T_H·Q_L/Q_H = 848.1·849/2500 = 288 K = 14.88 °C

Answer

T_L = 14.88 °C

18.Heat engines & refrigerators

Answer: S_gen = 5.013 kJ/K

Entropy generation: over a full cycle the working fluid has no net entropy change. The hot reservoir loses Q_H/T_H and the cold reservoir gains Q_L/T_L. Their sum is the entropy generated, which is zero for a reversible cycle and positive for every real one.

Heat rejected

Q_L = Q_H − W = 4100 − 1517 = 2583 kJ

Absolute temperatures

T_H = 850 + 273.15 = 1123 K

T_L = 25 + 273.15 = 298.1 K

Entropy changes of the reservoirs

ΔS_H = −4100/1123 = −3.65 kJ/K

ΔS_L = 2583/298.1 = 8.663 kJ/K

S_gen = 8.663 − 3.65 = 5.013 kJ/K

Answer

S_gen = 5.013 kJ/K

S_gen > 0, so the engine is possible but irreversible.

19.Waves, sound & optics

Answer: d_i = 7.5 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 − 1/15

d_i = 7.5 cm

Ray diagram

real, on the far side of the lens

Answer

d_i = 7.5 cm

20.Waves, sound & optics

Answer: m = 3

Focal length

f = R/2 = 15 cm

Mirror equation

1/d_i = 1/f − 1/d_o = 1/15 − 1/10

d_i = −30 cm

Magnification

m = −(−30)/10 = 3

Answer

m = 3

ExamProblems · examproblems.com/courses/physics-for-engineers