Physics Formulas: Derivations and Applications

Deriving Key Physics Formulas

Deriving v = u + at

    → v = u + at

Deriving s = ut + ½ at²

V_average =     But v = u + at       → V_average =  

V_average = s/t    → s = V_average(t)    → s = (t)     → s = ut + ½ at²

Deriving v² = u² + 2as

v = u + at   → v² = u² + 2uat + (at)²    {multiply out both sides}

We can rewrite this as v² = u² + 2a(ut + ½ at²)             {because 2a(ut + ½ at²) = 2uat + (at)²}

Now sub in s = ut + ½ at²    → v² = u² + 2as

Force and Motion

- F = Bqv: Consider a section of conductor of length l through which a current I is flowing.

If q is the charge which carries the current in this section of the conductor, then:

I = q/t, where t is the time it takes the charge q to travel a distance l.

The average velocity with which the charge flows is given by v = l/t, i.e. l = vt.

Substituting into the primary equation which we have for force (F = BIL),

we get F = B × q/t × vt i.e. F = Bqv

- F = ma *Force ∞ rate of change of momentum *F ∞ (mv – mu)/t *F ∞ m(v-u)/t *F ∞ ma *F = k (ma) *F = ma

- v = rω * θ = s/r *θ/t = s/tr *ω=v/r ( because w = q/t, v = s/t) *v=rω

Gravitational and Centripetal Force

We compare two formulae which we have for Force:

The first is the Universal Gravitational Force formula:                                        

The second is the Centripetal Force formula:                                                       

Equate both forces (because both equations apply to satellite motion)

Cancel one ‘m’ from both sides

Replace the d² in the first formula with r² and cancel one ‘r’ both sides

You now have                                                                              Equation (1)

Now  v = velocity = Distance/Time.

Distance in this case is the circumference of a circle (2pR for circular satellite orbits)

Þ 

                     Þ                       

                                    Equation (2)

Equating Equations (1) and (2) we get                        

To show that any object that obeys Hooke’s Law will also execute SHM

So we start with the equation for Hooke’s Law;                      F µ -s               Þ                    F = -k s

But F =  ma                                                      Þ                 ma = -k s

Now divide both sides by m                                                                Þ                  a = -

 s

This is equivalent to the equation for S.H.M. where the constant w² in this case is k/m.

Derive the diffraction grating formula             [2009]

From the diagram we can see that

(i)         For constructive interference to occur, the extra path length that the top ray travels must be an integer number of wavelengths (nl)                             {Eqn (1)}

(ii)        Using trigonometry, this extra path length is equal to d sin q, where d is the slit width                                                                                      {Eqn (2)}

 Equating (1) and (2) gives us nl = d Sin q

-  For resistors in parallel I_Total = I₁ + I₂

- For resistors in series VTotal = V1 + V2