# Derivations of Formula

### Below is a full list of all required derivations for the Advanced Higher Physics course :-

## Unit 1 : Rotational Motion and Astrophysics

### Equation 1 - v = u + at

### Integration of Acceleration with respect to time :-

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### Equation 2 - s = ut + ^{1}/_{2}at^{2}

^{1}/

_{2}at

^{2}

### Integration of Velocity with respect to time :-

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### Equation 3 - v^{2} = u^{2} +2as

^{2}= u

^{2}+2as

### Square both sides of Equation of Motion 1, then substitute in Equation of Motion 2 :-

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### Equation 4 - Radial Acceleration

### In the above diagram, a particle moves along the Arc AB in a time of Δt, with a velocity of v.

### The time (Δt) can be found using the following formula :-

### The Change in Velocity (Δv) can be found by the following approximation :-

### The Average Acceleration of the Particle can then be found by the following :-

### The above formula gives the Acceleration over a large time period. If the value of θ is made to tend towards zero, the value of sinθ ≈ θ and the Instantaneous Acceleration at a point can be found :-

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### Equation 5 - Centripetal Force

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### Equation 6 - Angular Momentum

### The diagram below shows an object of mass m rotating around a central point P at a distance of r.

### The object has a Linear Momentum equal to :-

### p = m v

### The Moment of Momentum (L) is therefore equal to :-

### L = m v r

### And by substituting for Angular Velocity :-

### L = mr^{2}ω

^{2}ω

### Finally by substituting for Moment of Inertia :-

### L = I ω

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### Equation 7 - Orbital Period

### By combining the above formula for the Gravitational Force Between objects and the Centripetal Force formula, the Orbital period of any orbiting body can be found :-

### Gravitational Force :-

### Centripetal Force :-

### By combining the above formulae together and by then substituting Velocity, the Period can be found as follows :-

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### Equation 8 - Gravitational Potential Energy

### Gravitational Potential Energy can be found using:-

### E_{p} = mg h

_{p}= mg h

### However, by substituting the Gravitational Force (mg) formula from earlier within the Advanced Higher course, the Gravitational Potential Energy can be shown as :-

### mg = ( GMm ) / r^{2}

^{2}

### E_{p} = ( ( GMm ) / r^{2} ) x h

_{p}= ( ( GMm ) / r

^{2}) x h

### If the Value of h is taken as the height above a point mass then h = r , and the Gravitational Potential Energy can be found as :-

### E_{p} = ( GMm ) / r

_{p}= ( GMm ) / r

### However, by convention, we state that Gravitational Potential Energy is equal to 0 J at infinity, and as such , the formula is given as :-

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### Equation 9 - Kinetic Energy of a Satellite

### The Kinetic Energy of a satellite can be found be deriving a form of E_{k} = ^{1}/_{2}mv^{2 }, based upon a Centripetal Force calculation :-

_{k}=

^{1}/

_{2}mv

^{2 }, based upon a Centripetal Force calculation :-

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### Equation 10 - Combined Energy of a satellite

### The combined Energy of a satellite can be found by the addition of the Kinetic Energy and the Gravitational Potential Energy :-

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### Equation 11 - Escape Velocity

### The Potential Energy of an Object in a Gravitational field is given by :-

### As the Gravitational Potential Energy is defined as equal to 0 J at Infinity, therefore to escape the Gravitational Potential Well, the object must be given Energy equal to :-

### The lower limit to Escape completely would be to provide the object with an initial Kinetic Energy such that at infinity, all Kinetic Energy has been converted to Gravitational Potential Energy :-

### E_{k} + E_{p} = 0

_{k}+ E

_{p}= 0

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### Equation 12 - Schwartzchild Radius (Event Horizon of a Black Hole)

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## Unit 2 : Quanta and Waves

### Equation 13 - De Broglie Wavelength

### For a Photon, as it has no rest mass, it can be shown that the Energy can be given as :-

### E = p x C

### Also the Energy of a Photon is related to its Frequency. By combining these two formulae together, the following can be derived :-

### E = p x v and E = h x f

### p x v = h x f

### p = ^{hf }/ _{v }

^{hf }/

_{v }

### p = ^{h} / _{λ}

^{h}/

_{λ}

### λ = ^{h} / _{p }

^{h}/

_{p }

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### Equation 14 - SHM Velocity

### The displacement on the vertical axis is equal to :-

### By differentiation, an equation of the velocity can be obtained :-

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### Equation 15 - SHM Accleration

### By differentiation of SHM Velocity, the acceleration can also be found :-

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### Equation 16 - SHM Velocity (2nd method)

### By combining the two equations for SHM Velocity and Acceleration and by using the trig identity below, the following can be derived :-

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### Equation 17 - SHM Kinetic Energy

### The velocity of a particle undergoing SHM was found by :-

### If this formula is combined with the Kinetic Energy formula, a method to find the Kinetic Energy of the particle at any given point in its motion can be derived :-

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### Equation 18 - SHM Potential Energy

### If an Ideal system (no Energy loss)

### Therefore at the equilibrium point of the oscillation, E_{tot} = E_{k} + 0 it can be shown that :-

_{tot}= E

_{k}+ 0 it can be shown that :-

### By substituting into the general equation above, the Potential Energy can be shown by :-

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### Equation 19 - Travelling Wave

### Since wave motion is repetitive after each wavelength, the formula describing the wave can be derived as follows :-

### At point i ) x = 0, t = x/v

### At point ii ) x = x, t = t

### Motion repetitive, therefore motion of particles at i ) and ii ) are identical, allowing the replacement of t within the calculation by :-

### t = t - ^{x}/_{v}

^{x}/

_{v}

### The general formula can then be derived as follows :-

### y = a sin ωt substitute in t = t - ^{x}/_{v }

^{x}/

_{v }

### y = a sin ω(t - ^{x}/_{v})

^{x}/

_{v})

### y = a sin ω(t - ^{x}/_{v}) substitute in ω = 2 π f and v = f λ

^{x}/

_{v}) substitute in ω = 2 π f and v = f λ

### y = a sin 2 π ( ft - ^{x}/_{λ})

^{x}/

_{λ})

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### Equation 20 - Young's Double Slit

### A bright maxima is found at point P.

### N is a point on the diagram such that the length NP is given by the following :-

### NP = S_{1}P

_{1}P

### This gives an overall path difference to point P as :-

### S_{2}P - S_{1}P = S_{2}N = mλ

_{2}P - S

_{1}P = S

_{2}N = mλ

### As the line PM >> S_{1}S_{2}, the line S_{1}N crosses S_{2}P at (approximately) a right angle and S_{1}S_{2}N is a right angled triangle. For this triangle :-

_{1}S

_{2}, the line S

_{1}N crosses S

_{2}P at (approximately) a right angle and S

_{1}S

_{2}N is a right angled triangle. For this triangle :-

### sinθ = ( S_{2}N / S_{1}S_{2} ) = ^{λ}/_{d }

_{2}N / S

_{1}S

_{2}) =

^{λ}/

_{d }

### Also looking at the triangle MPO :-

### tanθ = ^{OP}/_{MO} = ^{Δx}/_{D}

^{OP}/

_{MO}=

^{Δx}/

_{D}

### But as stated earlier sinθ ~ tanθ ~ θ, therefore :-

### Δx = ^{λD}/_{d}

^{λD}/

_{d}

### The above formula is normally stated in terms of wavelength :-

### λ = ^{Δxd}/_{D}

^{Δxd}/

_{D}

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### Equation 21 - Brewster's Angle

### The angle of incidence (i_{p}) that this occurs at is known as Brewster's Angle.

_{p}) that this occurs at is known as Brewster's Angle.

### If we apply Snell's Law a formula for the Brewster's Angle can be derived :-

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## Unit 3 : Electromagnetism

### Equation 22 - The Electric field strength

### The Electric Field Strength (E) at a given point is defined as the force experienced by a unit positive charge at that point.

### Where :-

### E - Electric Field Strength (NC^{-1})

^{-1})

### F - Force experienced by the charge (N)

### Q_{t} - Charge of the Charged particle (C)

_{t}- Charge of the Charged particle (C)

### As direction of the force is dependant on charge type, Electric field strength is a vector quantity.

### By substituting the value for force gained using Coulomb's Law, the following formula for Electric field strength can be derived :-

### Therefore by substituting into the above Electric field strength formula :-

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### Equation 23 - Electrostatic Potential due to point charges

### As the Electric field strength around a point charge varies, so does the Electrostatic Potential.

### E = - dV/dr

### Also previously, the Electric field strength could also be found using :-

### By combining the above formulae an expression for Electrostatic Potential (V) can be derived :-

### Integrating the above gives :-

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### Equation 24 - Closest Approach of Charged Particles

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### Equation 25 - Force per Unit Length

### The Magnetic field generated by wire 1 is given by :-

### By substitution of the Force calculation above, the Force applied to wire two can be derived :-

### As these fields are not based upon a point source, an extended description as a base unit is required - the Force per unit length :-

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### Equation 26 - Magnetic Force on a Moving Charge

### The Force experienced by the Charge can be found using :-

### F = BILsinθ

### The time taken for the charge to travel the distance L and the Current this movement generates can be found using the following formulae and combined :-

### t = ^{L} / _{v} and I = ^{q} / _{t }

^{L}/

_{v}and I =

^{q}/

_{t }

### IL = qv

### By substituting into the above formula and assuming the Charge is moving perpendicularly to the Magnetic field , the following can be derived :-

### F = BILsinθ

### F = BILsin90

### F = BIL

### F = qvB

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### Equation 27 - Frequency of Circular Motion in a B Field

### By using the Centripetal Force calculations from Unit 1, the radius of curvature of the Charge can be found :-

### Also, by applying calculations for Angular Velocity from the Circular Motion section of Unit 1, the Period and Frequency of the rotation can be derived :-

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