Physics Formulas and Laws for Classes 6 to 12

Motion

Speed

speed = distance ÷ time

In words: Speed equals distance divided by time.

Speed tells you how much distance something covers in each unit of time. A faster object covers more distance in the same time. If you know any two of speed, distance and time, you can find the third.

What the letters mean

speed

how fast something moves, in m/s or km/h

distance

length of the path travelled, in m or km

time

time taken, in s or h

Worked examples

1. Example 1. A bus travels 150 km in 3 hours. What is its average speed?

   1. speed = distance ÷ time
   2. speed = 150 ÷ 3

   Answer: 50 km/h

2. Example 2. A runner covers 100 m in 12.5 s. What is her speed?

   1. speed = 100 ÷ 12.5

   Answer: 8 m/s

Watch out: Keep the units matching. Metres with seconds give m/s; kilometres with hours give km/h. To change km/h into m/s, divide by 3.6.

Gravitation and pressure

Density

ρ = m ÷ V

In words: Density equals mass divided by volume.

Density measures how much mass is packed into a given volume. An iron ball is denser than a wooden ball of the same size. An object floats in a liquid if its density is less than the liquid's density.

What the letters mean

ρ

density, in kg/m³ or g/cm³

m

mass

V

volume

Worked examples

1. Example 1. A block has a mass of 540 g and a volume of 200 cm³. Find its density.

   1. ρ = m ÷ V = 540 ÷ 200

   Answer: 2.7 g/cm³

2. Example 2. An iron piece has a mass of 78 g and a volume of 10 cm³. What is its density?

   1. ρ = 78 ÷ 10

   Answer: 7.8 g/cm³

Watch out: Water has a density of 1 g/cm³, which is the same as 1000 kg/m³.

Gravitation and pressure

Pressure

p = F ÷ A

In words: Pressure equals force divided by area.

Pressure is the force pressing on each unit of area. The same force on a smaller area gives a bigger pressure, which is why a sharp knife cuts better than a blunt one and why high heels sink into soft ground.

What the letters mean

p

pressure, in pascal (Pa), where 1 Pa = 1 N/m²

F

force pressing at right angles to the surface, in newtons (N)

A

area of contact, in m²

Worked examples

1. Example 1. A force of 500 N acts on an area of 0.25 m². Find the pressure.

   1. p = F ÷ A = 500 ÷ 0.25

   Answer: 2000 Pa

2. Example 2. A heel presses down with a force of 600 N on an area of 0.002 m². What pressure does it exert?

   1. p = 600 ÷ 0.002

   Answer: 300000 Pa

Motion

First equation of motion

v = u + at

In words: Final velocity equals initial velocity plus acceleration times time.

This equation is for motion with constant acceleration. Acceleration is the change in velocity every second, so after t seconds the velocity has changed by a × t.

What the letters mean

v

final velocity, in m/s

u

initial velocity, in m/s

a

acceleration, in m/s²

t

time, in s

Worked examples

1. Example 1. A car moving at 10 m/s accelerates at 2 m/s² for 5 s. Find its final speed.

   1. v = u + at
   2. v = 10 + 2 × 5

   Answer: 20 m/s

2. Example 2. A ball is dropped from rest. Taking g = 10 m/s², find its speed after 3 s.

   1. u = 0, a = g = 10 m/s²
   2. v = 0 + 10 × 3

   Answer: 30 m/s

Motion

Second equation of motion

s = ut + ½at²

In words: Distance equals initial velocity times time, plus half acceleration times time squared.

It gives the distance covered in time t when acceleration is constant. The first part (ut) is the distance if there were no acceleration; the second part is the extra distance gained because of acceleration.

What the letters mean

s

distance covered, in m

u

initial velocity, in m/s

t

time, in s

a

acceleration, in m/s²

Worked examples

1. Example 1. A body starts at 5 m/s and accelerates at 2 m/s² for 4 s. How far does it travel?

   1. s = ut + ½at²
   2. s = 5 × 4 + ½ × 2 × 4²
   3. s = 20 + 16

   Answer: 36 m

2. Example 2. A stone falls from rest for 2 s. Taking g = 10 m/s², how far does it fall?

   1. u = 0, so s = ½gt²
   2. s = ½ × 10 × 2²

   Answer: 20 m

Motion

Third equation of motion

v² = u² + 2as

In words: Final velocity squared equals initial velocity squared plus two times acceleration times distance.

This equation links velocity and distance directly, without time. It is the quickest choice when the problem does not mention time, for example stopping distances.

What the letters mean

v

final velocity, in m/s

u

initial velocity, in m/s

a

acceleration, in m/s² (negative when slowing down)

s

distance, in m

Worked examples

1. Example 1. A vehicle starts from rest and accelerates at 5 m/s² over 40 m. What is its final speed?

   1. v² = 0 + 2 × 5 × 40 = 400
   2. v = √400

   Answer: 20 m/s

2. Example 2. A car moving at 20 m/s brakes with a deceleration of 5 m/s². How far does it travel before stopping?

   1. v = 0, so 0 = 20² + 2 × (−5) × s
   2. 10s = 400

   Answer: 40 m

Watch out: Deceleration is acceleration with a negative sign.

Force and laws of motion

Newton's first law (inertia)

ΣF = 0 ⇒ velocity stays constant

In words: If the net force is zero, the velocity does not change.

An object at rest stays at rest, and an object moving in a straight line keeps moving at a steady speed, unless a net force acts on it. This tendency to resist a change in motion is called inertia. A heavier object has more inertia.

What the letters mean

ΣF

net force: all the forces added together, with direction

N

a force of one newton

Worked examples

1. Example 1. A 5 kg box rests on a table. Its weight pulls down with 50 N and the table pushes up with 50 N. What is the net force?

   1. Up and down forces are in opposite directions
   2. Net force = 50 − 50

   Answer: 0 N, so the box stays at rest

Watch out: No net force does not mean no force. Forces can be present and balanced.

Force and laws of motion

Newton's second law

F = ma

In words: Force equals mass times acceleration.

A bigger force gives a bigger acceleration, and a heavier object accelerates less for the same force. One newton is the force that gives a mass of 1 kg an acceleration of 1 m/s².

What the letters mean

F

net force, in newtons (N)

m

mass, in kg

a

acceleration, in m/s²

Worked examples

1. Example 1. What force gives a 5 kg object an acceleration of 3 m/s²?

   1. F = ma = 5 × 3

   Answer: 15 N

2. Example 2. A force of 60 N acts on a 12 kg trolley. Find its acceleration.

   1. a = F ÷ m = 60 ÷ 12

   Answer: 5 m/s²

Force and laws of motion

Newton's third law and conservation of momentum

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

In words: Total momentum before a collision equals total momentum after it.

Forces come in pairs: if A pushes B, then B pushes A equally hard in the opposite direction. Because of this, the total momentum of objects that only interact with each other does not change.

What the letters mean

m₁, m₂

masses of the two objects, in kg

u₁, u₂

velocities before the collision, in m/s

v₁, v₂

velocities after the collision, in m/s

Worked examples

1. Example 1. A 2 kg trolley moving at 3 m/s hits a 1 kg trolley at rest. They stick together. Find their common speed.

   1. Momentum before = 2 × 3 + 1 × 0 = 6
   2. Momentum after = (2 + 1) × v
   3. v = 6 ÷ 3

   Answer: 2 m/s

Force and laws of motion

Momentum

p = mv

In words: Momentum equals mass times velocity.

Momentum measures how hard it is to stop a moving object. A slow truck and a fast bullet can both have a lot of momentum. Force is also the rate of change of momentum, which is Newton's second law in its original form.

What the letters mean

p

momentum, in kg m/s

m

mass, in kg

v

velocity, in m/s

Worked examples

1. Example 1. Find the momentum of a 1000 kg car moving at 15 m/s.

   1. p = mv = 1000 × 15

   Answer: 15000 kg m/s

2. Example 2. A 0.05 kg ball moves at 40 m/s. What is its momentum?

   1. p = 0.05 × 40

   Answer: 2 kg m/s

Gravitation and pressure

Weight

W = mg

In words: Weight equals mass times gravitational acceleration.

Mass is the amount of matter in an object and stays the same everywhere. Weight is the pull of gravity on that mass, so it changes from place to place. A person weighs less on the Moon because g there is smaller.

What the letters mean

W

weight, a force in newtons (N)

m

mass, in kg

g

gravitational acceleration: about 9.8 m/s² on Earth and about 1.6 m/s² on the Moon

Worked examples

1. Example 1. Find the weight of a 60 kg person on Earth. Take g = 10 m/s².

   1. W = mg = 60 × 10

   Answer: 600 N

2. Example 2. What is the weight of the same person on the Moon, where g = 1.6 m/s²?

   1. W = 60 × 1.6

   Answer: 96 N

Gravitation and pressure

Universal law of gravitation

F = G m₁ m₂ ÷ r²

In words: Gravitational force equals G times the two masses, divided by the square of the distance between them.

Every object attracts every other object. The pull is bigger for larger masses and gets weaker quickly as the distance grows: doubling the distance makes the force four times smaller. G is a very small number, so the pull is only noticeable when one mass is huge, like the Earth.

What the letters mean

F

gravitational force, in N

G

gravitational constant, 6.67 × 10^{-11} N m² kg^{-2}

m₁, m₂

the two masses, in kg

r

distance between their centres, in m

Worked examples

1. Example 1. Two people of mass 60 kg and 70 kg stand 1 m apart. Find the pull between them.

   1. F = 6.67 × 10^-11 × 60 × 70 ÷ 1²

   Answer: about 2.8 × 10^-7 N (far too small to feel)

2. Example 2. Find g at the Earth's surface. Mass of Earth = 5.97 × 10^24 kg, radius = 6.37 × 10^6 m.

   1. g = GM ÷ R²
   2. g = 6.67 × 10^-11 × 5.97 × 10^24 ÷ (6.37 × 10^6)²

   Answer: about 9.8 m/s²

Gravitation and pressure

Archimedes' principle (buoyant force)

F_b = ρ V g

In words: Buoyant force equals liquid density times submerged volume times g.

A liquid pushes up on any object placed in it. The upward force equals the weight of the liquid the object pushes aside. That is why a ship made of steel floats: it pushes aside a large volume of water.

What the letters mean

F_b

upward buoyant force, in N

ρ

density of the liquid, in kg/m³

V

volume of liquid displaced (the submerged volume), in m³

g

gravitational acceleration

Worked examples

1. Example 1. A stone of volume 0.002 m³ is fully under water (density 1000 kg/m³). Taking g = 10 m/s², find the buoyant force.

   1. F = ρVg = 1000 × 0.002 × 10

   Answer: 20 N

Work, energy and power

Work done

W = F × s

In words: Work equals force times distance moved in the direction of the force.

In physics, work is done only when a force moves something. Pushing a wall that does not move is hard, but it does no work. The unit of work is the joule (J): 1 J = 1 N × 1 m.

What the letters mean

W

work done, in joules (J)

F

force, in N

s

distance moved in the direction of the force, in m

Worked examples

1. Example 1. A force of 20 N moves a box 5 m along the floor. How much work is done?

   1. W = F × s = 20 × 5

   Answer: 100 J

Work, energy and power

Kinetic energy

KE = ½mv²

In words: Kinetic energy equals half mass times velocity squared.

Kinetic energy is the energy of motion. Because velocity is squared, doubling the speed makes the kinetic energy four times bigger. That is why high speeds are so dangerous.

What the letters mean

KE

kinetic energy, in J

m

mass, in kg

v

speed, in m/s

Worked examples

1. Example 1. Find the kinetic energy of a 2 kg ball moving at 6 m/s.

   1. KE = ½ × 2 × 6²
   2. KE = 1 × 36

   Answer: 36 J

2. Example 2. A 1000 kg car moves at 20 m/s. Find its kinetic energy.

   1. KE = ½ × 1000 × 20²

   Answer: 200000 J

Work, energy and power

Gravitational potential energy

PE = mgh

In words: Potential energy equals mass times g times height.

An object lifted to a height stores energy because gravity can pull it down again. The higher it is, or the heavier it is, the more energy it stores. When it falls, this energy turns into kinetic energy.

What the letters mean

PE

potential energy, in J

m

mass, in kg

g

gravitational acceleration

h

height above the reference level, in m

Worked examples

1. Example 1. A 5 kg object is held 10 m above the ground. Take g = 10 m/s². Find its potential energy.

   1. PE = mgh = 5 × 10 × 10

   Answer: 500 J

2. Example 2. How much potential energy does a 2 kg book gain when lifted 3 m? Take g = 9.8 m/s².

   1. PE = 2 × 9.8 × 3

   Answer: 58.8 J

Work, energy and power

Power

P = W ÷ t

In words: Power equals work done divided by time taken.

Power is how fast work is done, or how fast energy is used. Two machines can do the same work, but the more powerful one does it in less time. The unit is the watt (W): 1 W = 1 J/s.

What the letters mean

P

power, in watts (W)

W

work done or energy used, in J

t

time taken, in s

Worked examples

1. Example 1. A pump does 18000 J of work in 30 s. What is its power?

   1. P = W ÷ t = 18000 ÷ 30

   Answer: 600 W

2. Example 2. A boy does 3000 J of work in 60 s. Find his power.

   1. P = 3000 ÷ 60

   Answer: 50 W

Heat

Heat needed to change temperature

Q = mcΔT

In words: Heat equals mass times specific heat capacity times change in temperature.

Different materials need different amounts of heat to warm up. Specific heat capacity (c) is the heat needed to raise 1 kg of the material by 1 °C. Water has a high value, so it warms and cools slowly.

What the letters mean

Q

heat energy, in J

m

mass, in kg

c

specific heat capacity, in J/(kg °C); water is about 4200

ΔT

rise in temperature, in °C

Worked examples

1. Example 1. How much heat is needed to warm 2 kg of water by 10 °C? (c = 4200 J/kg °C)

   1. Q = mcΔT = 2 × 4200 × 10

   Answer: 84000 J

Heat

Latent heat

Q = mL

In words: Heat equals mass times latent heat.

While a substance melts or boils, it takes in heat but its temperature does not rise. That hidden heat is called latent heat. The latent heat of fusion is the heat needed to change 1 kg from solid to liquid at its melting point.

What the letters mean

Q

heat energy, in J

m

mass, in kg

L

latent heat in J/kg; for melting ice it is 3.34 × 10^{5} J/kg

Worked examples

1. Example 1. How much heat melts 0.5 kg of ice at 0 °C into water at 0 °C? (L = 3.34 × 10^5 J/kg)

   1. Q = mL = 0.5 × 334000

   Answer: 167000 J

Electricity

Ohm's law

V = IR

In words: Voltage equals current times resistance.

For a conductor at a steady temperature, the current through it is proportional to the voltage across it. Resistance is the ratio V ÷ I. The unit of resistance is the ohm (Ω).

What the letters mean

V

potential difference, in volts (V)

I

current, in amperes (A)

R

resistance, in ohms (Ω)

Worked examples

1. Example 1. A 12 V battery drives current through a 4 Ω resistor. Find the current.

   1. I = V ÷ R = 12 ÷ 4

   Answer: 3 A

2. Example 2. A current of 0.5 A flows through a 20 Ω resistor. What is the voltage across it?

   1. V = IR = 0.5 × 20

   Answer: 10 V

Electricity

Resistors in series and in parallel

Series: R = R₁ + R₂ + R₃ Parallel: 1/R = 1/R₁ + 1/R₂ + 1/R₃

In words: In series the resistances add; in parallel the reciprocals add.

In series the same current flows through each resistor, so the total resistance is the sum. In parallel the current has several paths, so the total resistance is smaller than the smallest one.

What the letters mean

R

total (equivalent) resistance, in Ω

R₁, R₂, R₃

individual resistances, in Ω

Worked examples

1. Example 1. Find the total resistance of 2 Ω, 3 Ω and 5 Ω in series.

   1. R = 2 + 3 + 5

   Answer: 10 Ω

2. Example 2. Find the total resistance of 6 Ω and 3 Ω in parallel.

   1. 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2
   2. R = 2

   Answer: 2 Ω

Electricity

Electric power and energy

P = VI E = Pt

In words: Power equals voltage times current; energy equals power times time.

Electric power tells you how fast a device uses energy. Household bills use the unit kilowatt-hour (kWh), also called a "unit": the energy used by a 1 kW device running for 1 hour.

What the letters mean

P

power, in watts (W)

V

voltage, in V

I

current, in A

E

energy, in joules or in kWh

t

time, in s or in hours

Worked examples

1. Example 1. An iron on 220 V draws 5 A. What is its power?

   1. P = VI = 220 × 5

   Answer: 1100 W

2. Example 2. A 2 kW heater runs for 5 hours. How many units are used, and what is the cost at ₹6 per unit?

   1. E = Pt = 2 × 5 = 10 kWh
   2. Cost = 10 × 6

   Answer: 10 units, costing ₹60

Light

Mirror formula and magnification

1/v + 1/u = 1/f m = −v/u

In words: One over image distance plus one over object distance equals one over focal length.

Use the Cartesian sign convention: distances are measured from the mirror, and distances in the direction of the incoming light are positive. For an object in front of a mirror, u is negative. A concave mirror has a negative f. A negative magnification means the image is inverted.

What the letters mean

u

object distance

v

image distance

f

focal length

m

magnification: image height ÷ object height

Worked examples

1. Example 1. An object is 30 cm in front of a concave mirror of focal length 10 cm. Find the image distance.

   1. u = −30 cm, f = −10 cm
   2. 1/v = 1/f − 1/u = −1/10 + 1/30 = −2/30
   3. v = −15 cm

   Answer: 15 cm in front of the mirror (v = −15 cm)

2. Example 2. For the same mirror, find the magnification.

   1. m = −v/u = −(−15)/(−30)

   Answer: −0.5: the image is real, inverted and half the size

Light

Lens formula and magnification

1/v − 1/u = 1/f m = v/u

In words: One over image distance minus one over object distance equals one over focal length.

The lens formula works the same way as the mirror formula, with distances measured from the centre of the lens and the same sign convention. A convex lens has a positive focal length and a concave lens a negative one.

What the letters mean

u

object distance (negative on the object side)

v

image distance

f

focal length

m

magnification

Worked examples

1. Example 1. An object is placed 30 cm from a convex lens of focal length 10 cm. Find the image distance.

   1. u = −30 cm, f = +10 cm
   2. 1/v = 1/f + 1/u = 1/10 − 1/30 = 2/30
   3. v = 15 cm

   Answer: 15 cm on the other side of the lens

2. Example 2. Find the magnification for this lens.

   1. m = v/u = 15 ÷ (−30)

   Answer: −0.5: the image is real, inverted and half the size

Light

Refractive index and Snell's law

n = c ÷ v = sin i ÷ sin r

In words: Refractive index equals the speed of light in air divided by its speed in the medium, which also equals sine of the angle of incidence over sine of the angle of refraction.

Light slows down when it enters glass or water, and bends towards the normal. The refractive index n tells you how much it slows down. A larger n means more bending.

What the letters mean

n

refractive index of the medium (no unit)

c

speed of light in air or vacuum, 3 × 10^{8} m/s

v

speed of light in the medium

i, r

angles of incidence and refraction, measured from the normal

Worked examples

1. Example 1. The refractive index of glass is 1.5. Find the speed of light in glass (in units of 10^8 m/s).

   1. v = c ÷ n = 3 ÷ 1.5

   Answer: 2 × 10^8 m/s

2. Example 2. Light enters glass (n = 1.5) from air at an angle of incidence of 30°. Find the angle of refraction.

   1. sin r = sin i ÷ n = 0.5 ÷ 1.5 = 0.333
   2. r = sin⁻¹ 0.333

   Answer: about 19.5°

Waves

Wave speed

v = fλ

In words: Wave speed equals frequency times wavelength.

Frequency (f) is the number of waves passing a point each second, and wavelength (λ) is the length of one wave. Together they give how fast the wave travels. It works for sound, light and water waves.

What the letters mean

v

wave speed, in m/s

f

frequency, in hertz (Hz)

λ

wavelength, in m

Worked examples

1. Example 1. A sound wave has a frequency of 500 Hz and a wavelength of 0.68 m. Find its speed.

   1. v = fλ = 500 × 0.68

   Answer: 340 m/s

2. Example 2. A radio wave travels at 3 × 10^8 m/s and has a frequency of 100 MHz (1 × 10^8 Hz). Find its wavelength.

   1. λ = v ÷ f = 3 × 10^8 ÷ 1 × 10^8

   Answer: 3 m

Motion

Projectile motion

R = u² sin 2θ ÷ g H = u² sin²θ ÷ 2g

In words: Range equals launch speed squared times sine of twice the angle, divided by g.

A projectile is thrown at an angle θ and then moves under gravity alone. Its horizontal speed stays constant while its vertical speed changes. The range R is the horizontal distance covered on level ground, and H is the maximum height. For a given speed, the range is greatest at 45°.

What the letters mean

u

launch speed, in m/s

θ

angle above the horizontal

g

gravitational acceleration

R

horizontal range, in m

H

maximum height, in m

Worked examples

1. Example 1. A ball is thrown at 20 m/s at 30° above the horizontal. Take g = 10 m/s². Find the range.

   1. R = u² sin 2θ ÷ g
   2. R = 400 × sin 60° ÷ 10 = 40 × 0.866

   Answer: about 34.6 m

2. Example 2. Find the maximum height for the same throw.

   1. H = u² sin²θ ÷ 2g = 400 × 0.25 ÷ 20

   Answer: 5 m

Force and laws of motion

Circular motion

a = v² ÷ r F = mv² ÷ r

In words: Centripetal acceleration equals speed squared divided by the radius.

An object moving in a circle at a steady speed is still accelerating, because its direction keeps changing. The acceleration points towards the centre. The force that causes it is called the centripetal force; it can be tension, friction or gravity.

What the letters mean

a

centripetal acceleration, in m/s²

v

speed along the circle, in m/s

r

radius of the circle, in m

F

centripetal force, in N

m

mass, in kg

Worked examples

1. Example 1. A car moves at 10 m/s round a bend of radius 50 m. Find its centripetal acceleration.

   1. a = v² ÷ r = 100 ÷ 50

   Answer: 2 m/s²

2. Example 2. A 1000 kg car takes a curve of radius 100 m at 20 m/s. What force towards the centre is needed?

   1. F = mv² ÷ r = 1000 × 400 ÷ 100

   Answer: 4000 N

Electricity

Coulomb's law

F = k q₁ q₂ ÷ r²

In words: The force between two charges equals k times the product of the charges, divided by the square of the distance.

Like charges repel and unlike charges attract. The force is bigger for larger charges and falls quickly with distance. In air, the constant k is about 9 × 10^9 N m²/C².

What the letters mean

F

force between the charges, in N

k

Coulomb constant, 9 × 10^{9} N m² C^{-2}

q₁, q₂

the charges, in coulombs (C)

r

distance between them, in m

Worked examples

1. Example 1. Two charges of 1 µC (1 × 10^-6 C) each are 0.1 m apart in air. Find the force between them.

   1. F = 9 × 10^9 × 10^-6 × 10^-6 ÷ 0.1²
   2. F = 9 × 10^-3 ÷ 0.01

   Answer: 0.9 N

Electricity

Capacitance and stored energy

C = Q ÷ V U = ½CV²

In words: Capacitance equals charge divided by voltage; stored energy equals half C times V squared.

A capacitor stores electric charge. Its capacitance C says how much charge it holds for each volt across it. The unit is the farad (F); common values are in microfarads (µF), where 1 µF = 10^-6 F.

What the letters mean

C

capacitance, in farads (F)

Q

charge stored, in coulombs (C)

V

voltage across the capacitor, in volts

U

energy stored, in joules

Worked examples

1. Example 1. A capacitor stores 10 µC of charge at 5 V. Find its capacitance (in µF).

   1. C = Q ÷ V = 10 ÷ 5

   Answer: 2 µF

2. Example 2. Find the energy stored in a 2 µF capacitor charged to 10 V (answer in µJ).

   1. U = ½CV² = ½ × 2 × 100

   Answer: 100 µJ

Oscillations

Hooke's law and the spring

F = −kx T = 2π √(m ÷ k)

In words: The restoring force equals minus the spring constant times the stretch; the period is 2 pi times the square root of mass over spring constant.

A spring pulls or pushes back with a force proportional to how far it is stretched or compressed. The minus sign shows the force acts against the stretch. A mass on a spring then swings to and fro with a period T that depends only on the mass and the stiffness.

What the letters mean

F

restoring force, in N

k

spring constant, in N/m (stiffer spring: larger k)

x

stretch or compression from the natural length, in m

T

time for one full oscillation, in s

m

mass, in kg

Worked examples

1. Example 1. A spring with k = 200 N/m is stretched by 0.05 m. What force does it exert?

   1. |F| = kx = 200 × 0.05

   Answer: 10 N

2. Example 2. A 0.5 kg mass hangs on that spring and oscillates. Find the period.

   1. T = 2π √(m ÷ k) = 2π √(0.5 ÷ 200)
   2. T = 2π × 0.05

   Answer: about 0.314 s

Oscillations

Simple pendulum

T = 2π √(l ÷ g)

In words: The period equals 2 pi times the square root of length over g.

For small swings, the time for one swing depends only on the length of the string and on g, not on the mass of the bob or on how wide the swing is. A longer pendulum swings more slowly.

What the letters mean

T

time period, in s

l

length of the pendulum, in m

g

gravitational acceleration, 9.8 m/s² on Earth

Worked examples

1. Example 1. Find the period of a 1 m pendulum on Earth. Take g = 9.8 m/s².

   1. T = 2π √(1 ÷ 9.8) = 2π × 0.319

   Answer: about 2.01 s

Modern physics

Mass–energy equivalence

E = mc²

In words: Energy equals mass times the speed of light squared.

Mass and energy are two forms of the same thing. Because c² is enormous, a tiny amount of mass corresponds to a huge amount of energy. This is the source of the energy released in nuclear reactions and in the Sun.

What the letters mean

E

energy, in J

m

mass converted, in kg

c

speed of light, 3 × 10^{8} m/s

Worked examples

1. Example 1. How much energy is equivalent to a mass of 1 g (0.001 kg)?

   1. E = mc² = 0.001 × (3 × 10^8)²
   2. E = 0.001 × 9 × 10^16

   Answer: 9 × 10^13 J

Modern physics

Photoelectric effect

KE_max = hf − φ

In words: The maximum kinetic energy of an emitted electron equals the photon energy minus the work function.

When light of high enough frequency falls on a metal, it knocks out electrons. Each photon carries energy hf. Part of it (the work function φ) is needed to free the electron and the rest becomes its kinetic energy. Below a threshold frequency no electrons come out, however bright the light.

What the letters mean

KE_max

maximum kinetic energy of the electron

h

Planck's constant, 6.63 × 10^{-34} J s = 4.14 × 10^{-15} eV s

f

frequency of the light, in Hz

φ

work function of the metal, in eV

Worked examples

1. Example 1. Light of frequency 1.0 × 10^15 Hz falls on a metal with work function 2.3 eV. Find the maximum kinetic energy of the electrons. (h = 4.14 × 10^-15 eV s)

   1. hf = 4.14 × 10^-15 × 1.0 × 10^15 = 4.14 eV
   2. KE = 4.14 − 2.3

   Answer: 1.84 eV

Modern physics

de Broglie wavelength

λ = h ÷ mv

In words: Wavelength equals Planck's constant divided by momentum.

Every moving object has a wave nature. The wavelength is inversely proportional to the momentum. For everyday objects it is far too small to notice; for electrons it is comparable to the size of atoms, which is why electron microscopes work.

What the letters mean

λ

de Broglie wavelength, in m

h

Planck's constant, 6.63 × 10^{-34} J s

m

mass, in kg

v

speed, in m/s

Worked examples

1. Example 1. Find the wavelength of a 0.1 kg ball moving at 10 m/s.

   1. λ = h ÷ mv = 6.63 × 10^-34 ÷ (0.1 × 10)

   Answer: 6.63 × 10^-34 m (far too small to detect)

Modern physics

Radioactive decay and half-life

N = N₀ × (½)^t/T

In words: The amount left equals the starting amount times one half raised to the number of half-lives.

The half-life T of a radioactive substance is the time in which half of its nuclei decay. After each half-life the remaining amount is halved again: ½, ¼, ⅛ and so on.

What the letters mean

N

amount remaining

N₀

amount at the start

t

time elapsed

T

half-life (in the same unit as t)

Worked examples

1. Example 1. A sample of 80 g has a half-life of 10 days. How much remains after 30 days?

   1. Number of half-lives = 30 ÷ 10 = 3
   2. N = 80 × (½)³ = 80 ÷ 8

   Answer: 10 g