Download 12. MATHEMATICAL PHYSICS

The IB Physics Compendium 2005: Mathematical physics
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12.* MATHEMATICAL PHYSICS
12.1.* Mathematical physics
This section is intended for those who want to explore the applications in physics of the calculus
commonly learned in high school. Calculus = the mathematics of derivatives and integrals.
12.2.* The one-hour calculus course
1. Derivative = gradient function
The gradient shows how steeply inclined a graph is. For a straight line, the gradient has one specific
value. For y = x its 1, for y = 2x it is 2, for y = ½x it is ½; for a horizontal line the gradient is 0.
Fig. calc1: Graphs of y = 0, y = ½x, y = x, y = 2x,
For a nonlinear curve, the gradient is changing. Below is the graph of y = x2 :
Fig. calc2 : y = x2 with tangents at x = 0, x =1 and x =2
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The IB Physics Compendium 2005: Mathematical physics
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The gradient in any point shows how the curve is inclined right there; it can be found by making a
very large graph millimeter graph paper and drawing a tangent at the point in question; that is a line
showing roughly how the curve is going very near the point. The gradient of this tangent is the
gradient of the parabola at this point. For the y = x2 curve above, one can find that the gradient is 0
when x = 0, it is 2 when x = 1 and it is 4 when x = 2.
Fig. calc3: The points (0,0) , (1,2) , (2,4) making the graph of y = 2x
We notice that the gradients make the graph of a new function, y = 2x. This - another function
which shows the gradient in the first function - is the "derivative".


A function takes a value gives a new (or the same) value; y = x 2 changes 1 to 1, 2 to 4, 3
to 9,....
The derivative takes a function and gives another function: it changes y = x2 to y = 2x
2. Some derivative rules
We found that y = 2x is the derivative of y = x2. Instead of finding the gradient function = derivative
graphically it can be proved that certain rules give the derivative of any function we have. (This will
be left to later mathematics lessons).
When we take the derivative of a function it can be symbolised with "D" or "(d/dx)". So for
example:
Dx2 = 2x or (d/dx)x2 = 2x
Rule A:
Dxn = nx(n-1)

Dx2 = 2x is an example with n = 2 where Dx2 = 2x(2-1) = 2x1 = 2x

Dx3 = 3x(3-1) = 3x2

Dx4 = 4x(4-1) = 4x3
Note that

Dx = Dx1 = 1x(1-1) = x0 = 1
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The IB Physics Compendium 2005: Mathematical physics
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Rule B: Constants are not involved in the derivation


D5x2 = 5Dx2 = 5*2x = 10x
D7x4 = 7Dx4 = 7*4x3 = 28x3
Rule C: If we have several terms, we take the derivative of one at a time


D(x3 + x2) = Dx3 + Dx2 = 3x2 + 2x
D(2x5 - 7x) = 2Dx5 - 7Dx = 2*5x4 - 7*x0 = 10x4 - 7
3. Derivatives in physics
The derivative tells what the gradient of something else is. This is common in physics; for example
the velocity is the gradient of the displacement when both are graphed with the time on the
horisontal axis. From Mechanics we know that
s = ut +½at2 and v = u + at
Here a is a constant (it is uniformly accelerated motion) and u is a constant (we can reach higher
final velocities by accelerating for a longer time, but that does not change what the initial velocity
was). We can take the derivative of the s-function with t instead of x as the variable:

D(ut + ½at2) = Dut + D(½at2) = uDt + ½aDt2 = u*t0 + ½a*2t = u + at as expexted
4. Integral = area function or antiderivative
Some physical quantities are the gradient of something else, like the velocity is the gradient of the
displacement. Others are the area under a graph of the other, for example the displacement is the
area under the velocity graph. Since from a velocity graph we get back the same quantity
(displacement) as we used when taking the derivative we may accept that the integral is the
"opposite" of the derivative.
Fig. calc 4: Graphs of y = 2x with areas from x = 0 to x = 0, 1 and 2 shadowed
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The IB Physics Compendium 2005: Mathematical physics
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If we make a function to show how much area we have under the graph we can use the triangles
above.

from x = 0 to x = 0 no area is yet found

from x = 0 to x = 1 we have the area (1*2)/2 = 1

from x = 0 to x = 2 we have the area (2*4)/2 = 4
If we now make a graph of the accumulated area under the y = 2x graph we get:
Fig. calc5: Graph of (0,0) , (1,1), (2,4) or y = x2
The function we get in this way is, as expected, y = x2. We can write this as
2xdx = x2
This looks a bit confusing, but just note that the integral symbol consists of two parts, a  and a dx
which tells that x is the variable (may be useful if we have several letters involved). Whatever is in
between the  and the dx is the function to integrate.
Summary:


derivative: D(functiontoderive) or (d/dx)functiontoderive
integral: (functiontointegrate)dx
Like the derivative, the integral takes one function and returns another function.
5. Some integration rules
Rule A: xndx= x(n+1)/(n+1)

xdx = x1dx = x(1+1)/(1+1) = ½x2
Rule B: Constants are not involved in integration


2xdx = 2x1dx = 2*½x2 = x2 as expected
5x4dx = 5x4dx = 5*x(4+1)/(4+1) = x5
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The IB Physics Compendium 2005: Mathematical physics
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Rule C: If several terms are integrated, we can take one at a time

(3x2 - 7x3)dx = 3x2dx - 7x3dx = 3x2dx - 7x3dx =
(2+1)
3*x /(2+1) - 7*x(3+1)/(3+1) = x3 - 7x4/4
6. Integrals in physics
We expect the integral of the velocity as a function of time (v = u +at) to be the displacement as a
function of time (s = ut + ½at2)

(u + at)dt = udt + atdt = (u*1)dt + atdt = ut0dt + at1dt
=u*t(0+1)/(0+1) + a*t(1+1)/(1+1) = ut + ½at2 as expected
We may also use integrals in other cases where an area under a graph is used. The work W = Fs
when the force F is constant, but if it is not, then W = the area under the graph of F as a function of
time. For example, the work done in stretching a spring against the force F = kx requires the work
given by the area of the triangle under the F-graph; streching it from s = 0 to s = x gives W = kx*x/2
= ½kx2, which is then the elastic potential energy stored in the stretched spring. This could also
have been found with an integral:
W = kxdx = kxdx = kx1dx = k*x(1+1)/(1+1) = kx2/2 = ½kx2 = Ep,elas
In Mechanics, we had for objects far out in space that the force of gravity is
F = Gm1m2/r2
where r is the variable, the distance from the center of the planet, and the others are constants. We
also had that
Ep = - Gm1m2/r
We can sort of understand this if we think of Ep as the amount of work done when an object is
falling or rising in the graviational field; so it should have something to do with the integral of the
force function. Here Gm1m2 is a constant, call it k. Now

(k/r2)dr = kr-2dr = k*r(-2+1)/(-2+1) = -kr-1 = -k/x
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The IB Physics Compendium 2005: Mathematical physics
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which may explain the minus sign and the change from r2 to r. Note that we use r = infinity as the
zero level since that is where the force function is zero; at r = 0 we would have F = infinite. (When
we integrated y = 2x we could start from x = 0 as a "zero level" since y = 2x = 0 when x = 0).
7. Some other rules
Integration constants
If we take the derivative of a function y = k where k is a constant, then
Dk = D(k*1) = D(k*x0) = kDx0 = k*0*x0-1 = 0 independent of x
Therefore any constant will disappear in derivation, for example
Dx2 = 2x, D(x2 + 1) = 2x, D(x2 + 2) = 2x, D(x2 -127) = 2x etc.
On the other hand, if we integrate 2xdx = x2, the result could as well have been x2 +1 or x2 +2 or x2
- 127 or anything similar. We could then write x 2 + C where C is an unknown (positive or negative)
integration constant. In physics we can mostly set C = 0, for example as in: (u + at)dt = ... = ut +
½at2 but we should really have ... = s0 + ut + ½at2 where s0 is the displacement at the time t = 0.
But this can usually be set to zero by defining s = 0 when t = 0; that is we start measuring the time
when an object passes a chosen starting line.
The integral of y = 1/x
We would get x-1dx = x(-1+1)/(-1+1) which involves division by zero. It can be shown in
mathematics that x-1dx = ln x (the natural logarithm of x). This is useful in thermodynamics where
we find the work done in an isothermal process as the area under a graph in a diagram with
pressure P as a function of volume V:
PV = nRT gives P = nRT/V so the work is found using W = (nRT/V)dV = nRTV-1dV = nRTlnV
when n, R and T are constants.
Trigonometric functions
Most useful are :
Dsin x = cos x
cos x dx = sin x
Dcos x = - sin x
sin x dx = - cos x
We can also differentiate (= take the derivative of) or integrate a function more than once. If we
have a displacement function (with the time as a variable) and differentiate it once we get a velocity
function. If we differentiate that again, we get an acceleration function. If we try trigonometric
functions, then Dsinx = cosx, and then Dcosx = -sin x. We got the original function back, only with
an extra minus sign! Now think of a mass oscillating on a spring. Then F = ma and F = kx or -ks
(with directions) for the force on it. Then
ma = -ks gives a = -(k/m)s
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The IB Physics Compendium 2005: Mathematical physics
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that is, the acceleration function is the displacement function, give or take a negative constant. This
may help explain why sine-functions are important for oscillating things, like the water molecules
in an ocean wave. More about this in the Waves topic.
12.3* Further calculus in physics (and why E = mc2)
Using the "dx": Gradient (slope) = derivative
For constant velocity v = s / t and for a changing velocity, the average velocity is s/t. If t gets
smaller and smaller, we reach the instantaneous velocity which is the gradient of the displacementas-function-of-time curve or: the time derivative of displacement. When s and t become
infinetely small we write v = ds/dt or velocity is the "rate of change" in the displacement. We will
when necessary calculate with ds and dt in the same way as with any other variable. Note that "ds"
then is not "d times s" but one variable.
Area under curve = integral
For constant velocity, s = vt which is the area of a rectangle in a velocity-as-function-of-time curve.
If the velocity is not constant but the acceleration is then the v-curve is a straight line, and the area
under it can be found as a trapeze (here a triangle on top of a rectangle). From this the usual
formulas v = u +at, s = ut + ½at2 etc. were found. If the acceleration is not constant either, the area
under the curve is found by splitting the t-axis into short intervals with the length t and using the
v-value in the beginning or end of the interval to find an approximate area for it. Summing up all
the thin vertical slices roughly gives the area under the curve as s =  vt. If the t becomes
infinitely small and the slices infinitely many, this gives s =  vdt.
Changing variables in integration
Here we will assume that you learned the usual rules of deriving and integrating functions in maths,
like that if y = x2 then dy/dx = 2x or that x2dx = x3/3. In physics it is sometimes convenient to
change one variable in integration to another. Let us take (using something else than x as a
variable!)
a2da = a3/3 (plus an integration constant, if you like)
We can find the same result in a different way (which here is unnecessary, but shows the steps to
take).

Let b = a2 and therefore a = b = b½.

One would think that we simply get bdb, but this is ½b2 and since b = a2 it would give
½b2 = ½a4 which is not the correct result a3/3
To get the correct result we must find out what to replace da with like this: Start with b = a2 We
take the derivative of both sides (with respect to something, we don't care what):

db = 2ada where 2a is multiplied with the "inner" derivative of a; then
(e.g. if we derived for time then we would have db/dt = 2ada/dt, but multiply the whole with dt
and we have the db = 2ada left)

da = db/2a and since a = b we get da = db/2b so

a2da = b(db/2b) = (b/2)db = ½(b)db and since (x)dx = (2x3/2)/3 we get
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The IB Physics Compendium 2005: Mathematical physics

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½(2b3/2)/3 = b3/2/3 but since a = b½ this = a3/3 as it should
Here the whole substitution of b = a2 was unnecessary, but in some cases a substitution would give
a function easier to integrate.
Example 1: Simple derivative
Since v = ds/dt and s =  vdt we can for a constant acceleration take v = u +at and get s = (u+at)dt
which with conventional integrating rules gives the familiar s = ut + ½at2.
Example 2: Inner derivative
We have seen the formula P = Fv and used it for situations of constant velocity where F is e.g. the
force pulling a train, and other forces (friction) keep the velocity constant. But if no other forces act
(a rocket force F accelerates a spaceship) the power = the rate of work done which goes to
increasing the kinetic energy.
E = ½mv2 => dE = d(½mv2) = ½(2mvdv) = mdv so dE = mvdv and dividing with dt:
dE/dt = mvdv/dt = mva = mav and with F = ma we get dE/dt = P = Fv.
Example 3 : Newton's II law in relativity
Instead of F = ma we will for linear acceleration have

F = dp/dt = (d/dt)mv = (d/dt)(m0v/(1 - v2/c2)½ , that is we need to derive the ratio of two
functions, and need to use the derivation rule
D(f/g) = (f'g - g'f)/g2
where now f = m0v and g = (1 - v2/c2)½
 so f' = m0*dv/dt = m0a remembering the inner derivative dv/dt = a
 and g' = ½(1 - v2/c2)½-1 *( (- 2/c2)v*dv/dt) where two layers of inner derivatives are needed;
we get
 g' = ½(1 - v2/c2)-½(- 2va/c2) then giving the expression f'g - g'f as
 m0a(1 - v2/c2)½ - (1 - v2/c2)-½(- va/c2)m0v since ½*2 = 1
 the downstairs part is g2 = (1 - v2/c2)
We will now multiply both the upstairs and downstairs part with (1 - v2/c2)½ as a result of which
the downstairs becomes (1 - v2/c2)3/2 and in the second term of the upstairs (1 - v2/c2)½(1 - v2/c2)-½ =
1 so that the whole upstairs now is



m0a(1 - v2/c2)1 - ( - va/c2)m0v which discarding the parentheses gives
m0a - m0av2/c2 + m0av2/c2 = m0a so the whole result is
m0a / (1 - v2/c2)3/2 or using γ = 1 / (1 - v2/c2)1/2 finally
F = γ3m0a
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The IB Physics Compendium 2005: Mathematical physics
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Example 4: The ideal angle for pulling a sleigh
When a an object is pulled on a horisontal surface by a force F acting in a direction at an angle  up
from the horison one will find that:







the normal force is now mg - Fsin
the resultant horisontal force is Fcos - Ffr = Fcos - (mg-Fsin)
so Ftot = F(cos + sin)-mg
which has a maximum when cos + sin has a maximum
the derivative is sin - cos
let sin - cos = 0 to find the maximum when
= tan or  = arctan 
Example 5 : deriving Ek = mc2 - m0c2 in relativity
(Note that the word "derive" can mean both "take the derivative of" and "show why a formula is
true"!). In relativity we must change Newton's II law F = ma since m is not constant. We can use the
momentum instead : F = ma = m(v-u) = mv - mu and then impulse = Ft = change in momentum =
p = mv - mu which for an infinitely short time dt becomes:
Fdt = dp and dividing with dt then: F = dp/dt (force is the rate of change in momentum).
We also know from relativity that mass increases as m = m0/(1- v2/c2)½ . If a resultant force F acts
on an object accelerating it from rest we will assume that all the work done goes into increasing its
kinetic energy. Before, the work W = Fs. Here we use

Then


W =Fds = (dp/dt)ds d(mv)ds/dt which gives W =  d(mv)v since ds/dt = v
d(mv) = mdv + vdm (the usual rule for deriving the product of two functions, note that
both m and v are variables here)
multiplying both sides with v gives d(mv)v = mvdv + v2dm
So instead of finding out what W = Ek =  d(mv)v is, we can find out what (mvdv + v2dm) is. Now
we first use the formula for mass increase:

m = m0/(1- v2/c2)½ square both sides so

m2 = m02/(1- v2/c2) ; mult. with parenthesis and div. with m2 :

m02 / m2 = (1- v2/c2) divide with m02

1 / m2 = (1- v2/c2) / m02; this expression for 1/m2 will be useful later
Then we differentiate (take the derivative of) both sides, where m0 and c are constants:

start with m02 / m2 = (1- v2/c2)

on the left we get: - 2m02dm/m3 since the derivative of 1/x2 is -2/x3 (remember the inner
dm!)

on the right the 1 is lost and we get: -2vdv/c2 (remember the inner dv!)

so - 2vdv/c2 = - 2m02dm/m3 ; cancel -2 and multiply with mc2 :

mvdv = m02c2dm/m2 = m02c2dm(1/m2)
From the equation before the differentiation we take our expression for (1/m2):

mvdv = m02c2dm(1/m2) now gives
 Thomas Illman and Vasa övningsskola
The IB Physics Compendium 2005: Mathematical physics


10
mvdv = m02c2dm((1- v2/c2) / m02) which gives
mvdv = dmc2(1- v2/c2) = dm(c2 - v2)
Now we can return to the mvdv + v2dm we had in the beginning and get

mvdv + v2dm = dm(c2 - v2) + v2dm = (c2 - v2 + v2)dm = c2dm
And this finally means that the integral simply is:


W = c2dm = c2dm which if we intgrate from the rest mass m0 to the mass m after
acceleration gives
W = c2(m - m0) and since we assumed that the work goes to kinetic energy,
Ek = mc2 - m0c2
That is, terms of the form mass*c2 express some kind of energy that the particle has by virtue of
having a mass. Even at rest there is some energy E0 = m0c2 which can be released (the commonly
known E = mc2 really refers to this) if, in any reaction, the mass would decrease. Which it in some
nuclear reactions does ...
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The IB Physics Compendium 2005: Mathematical physics
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12.4 Constants and formulas available in IB examinations
Page 1: Fundamental constants
Gravity acceleration
g
Gravitational constant
G
Avogadro's constant
NA
Gas constant
R
Boltzmann's constant
k
Stefan-Boltzmann constant 
Coulomb constant
k
Permittivity of free space
0
Permeability of free space
0
Speed of light in vacuum
c
Planck's constant
h
Charge on electron
e
Electron rest mass
me
Proton rest mass
mp
Neutron rest mass
mn
Unified atomic mass unit
u
9.81 ms-2
6.67 x 10-11 Nm2kg-2
6.02 x 1023 mol-1
8.31 JK-1 mol-1
1.38 x 10-23 JK-1
5.67 x 10-8 Wm-2K-4
8.99 x 109 Nm2C-2
8.85 x 10-12 C2N-1m-2
4 x 10-7 TmA-1
3.00 x 108 ms-1
6.63 x 10-34 Js
1.60 x 10-19 C
9.11 x 10-31 kg = 0.000549u = 0.511 MeV/c2
1.673 x 10-27 kg = 1.007276 u = 938 MeV/c2
1.675 x 10-27 kg = 1.008665 u = 940 MeV/c2
1.661 x 10-27 kg = 931.5 MeV/c2
Page 2 : SI prefixes and unit conversions
tera = T = 1012, giga = G = 109, mega = M = 106, kilo = k =103, hecto = h = 102, deca = da = 101, deci = d =
10-1, centi = c = 10-2, milli = m = 10-3, micro =  = 10-6, nano = n = 10-9, pico = p = 10-12, femto = f = 10-15
1 light year (ly) = 9.46 x 1015 m
1 astronomical unit(AU) = 1.50 x 1011 m
1 kilowatt-hour (kWh) = 3.60 x 106 J
1 parsec = 3.26 ly
1 radian (rad) = 180o/
1 atm = 1.01 x 105 Nm-2 =101 kPa = 760 mmHg
Page 3 : Electrical circuit symbols
Symbols are given for: cell, battery, lamp, ac supply, switch, ammeter, voltmeter, galvanometer, resistor,
potentiometer, transformer, heating element
Page 4 : Measurement and mechanics
Horizontal,vertical components vector A:
AH = Acos
Uncertainties: If y = a  b then y = a + b
If y = ab/c then y/y = a/a + b/b + c/c
AV = Asin
v = s/t
a = v/t
g = F/m
g =Gm/r2
Page 5 : Mechanics
v = u + at
s = ((u+v)/2)t
s = ut + ½at2
v2 = u2 + 2as
s : displacement, t : time, u : initial speed (?), v : final speed (?), a : acceleration
F = ma
p =mv
F = p/t
Ek =p2/2m
Impulse = Ft = p
W = (Fss =) Fs cos
Ek = ½mv2
Ep= mgh
F = (-) kx
Eelas=½kx2
P (= E/t or W/t) = work/time = Fv
a = v2/r = 42r/T2
 = Frsin
Ffr = kN
Ffr < or = sN
F = Gm1m2/r2
Ep = -Gm1m2/r
V=-Gm/r
T2/R3=const.
Page 6
Q = mcT
Thermal physics and Waves
Q = mL
p = F/A
pV = nRT
Q = U + W
W = pV
 = = (TH - TC)/ TH(Carnot efficiency)
 Thomas Illman and Vasa övningsskola
efficiency = (QH - QC)/ QH
QC/QH = TC/TH(Carnot cycle)
The IB Physics Compendium 2005: Mathematical physics
f = 1/T
v = f
fbeat = f1 - f2 
Moving source: f' = f ( 1 / (1  vs/v))
dsin = n
s = D/d
Moving observer: f' = f (1  vo/v)
sin1/sin 2 = v1/v2
n=c/v
Page 7 : Electricity and magnetism, electromagnetism
F = kq1q2/r2 k = 1/40
E = F/q
E = kq / r2
E = - V/x
V = kq/r
I = q/t
1/R = 1/R1 + 1/R2
F = qvBsin
n1sin1 = n2sin2
E = V/d
R=V/I
P = VI = I2R = V2/R
R=R1+R2
B = 0I/2r
B = 0NI/L = 0nI
F = ILBsin
 = BLv
 = - N/t
F/L = 0I1I2/2r
=BAcos
Irms = I0/2
Vrms = V0/2 Vp/Vs = Np/Ns
Page 8 : Atomic, nuclear and quantum physics
E = mc2
E = hf
p = h/
hf = φ + Ek,max
N = N0e-t
12
hf = hf0 + eVs
T½=ln 2/
Page 9-10 : SL options
Page 11 : Biomedical physics and The history and development of physics
 = 10 log ( I / I0 ) where I0 = 10-12 Wm-2
I = I0e-x
x½ = ln2/
Mechanical Advantage = load/effort;Velocity Ratio = distance moved by effort/ distance moved by load;
Absorbed dose = Absorbed Energy / mass; Exposure = total charge / mass;
Dose equivalent = quality factor x Absorbed dose
1/TE = 1/TB + 1/TR
1/ = RH (1/n2 - 1/m2)
xp  h/2
Et  h/2
Page 12 : Astrophysics and Relativity
L = AT4
max = 2.90 x 10-3/T
d(parsec) = 1/p(arc-second) b = L/4d2
2
2
v = Hd
/  v/c
 = 1/( 1 - v /c )½
t = t0
L = L0 / 
ux' =(ux - v)/(1 - uxv/c2)
m = m0
p =m0u
E2 = p2c2 + m02c4
f/f = gh/c2
RSch = 2GM / c2
Page 13 : Optics
n1sin1 = n2sin2
n = 1/sinc
1/f = 1/u + 1/v
M = i / o
 = 1.22/b
dsin = n
 = /b
 Thomas Illman and Vasa övningsskola
E0= m0c2
E = mc2
m = hi / ho = v / u
The IB Physics Compendium 2005: Mathematical physics
13
12.5 Exercise problems
1.1.1. Can you think of a physical quantity that could not be used to describe some aspect of living
beings?
1.2.1. Change a) 50 cg to kg b) 0.034 kg to ng c) 23400 kN to MN
1.3.1. A lifeboat drifts 14 km east one day and 12 km north the following. How far from the starting
point is it then? In what direction (angle from the direction due north from the starting place). [18
km, 49o]
1.4.1. Manipulate the values on the axis until these (x,y)-data points give approximately a straight
line. What function do you think describes them best?
(1.01, 0.98), (2.04, 0.26), (2.99, 0.11), (4.87, 0.05), (6.91, 0.02)
1.5.1. A plane drops from the altitude 5500 m to 2700 m. If the uncertainty of the altimeter is 40 m,
how much altitude was lost (with uncertainty?) [280080m]
1.5.2. A cube with the side 857 cm has the mass 42030 kg. What is its density, with uncertainty?
[684218 kgm-3 or approx. 700200 kgm-3 ]
2.2.1. If you move 45 meters forwards and 20 backwards, what is the a) distance b) displacement ?
2.2.2. If you move in a half-circle with the radius 40 m, what was the a) distance b) displacement ?
[a) 130m, b) 80m]
2.2.3. Draw a qualitative graph of a) the velocity b) speed c) displacement d) the distance as a
function of time when a billiard ball moves towards a wall, bounces straight back without losing
energy and comes to rest in a "hole" some distance from the starting point in the opposite direction
to where it first moved.
2.3.1. What was the (average) speed and velocity of the boat in problem 1.3.1, if the "days" are 24
hours ? [0.54 kmh-1, 0.38 kmh-1]
2.4.1. Draw two possible graphs each of a) acceleration b) velocity as a function of time for a ball
thrown upwards and falling back down to the ground.
2.4.2. What is the acceleration of a) a car that speeds up from 50 kmh-1 to 80 kmh-1 in 4.5 s b) of
one that slows down from 90 kmh-1 to 30 kmh-1 in 7.1 s? [a) 1.9 ms-2 b) -2.3 ms-2]
2.5.1. If an object is uniformly decelarated to rest in 2.34 s while covering a displacement of 5.05
m, what was its initial velocity? [4.3 ms-1]
2.5.2. Draw a graph of the velocity as a function of time for an object which accelerates from rest
with 4.0 ms-2, for t = 1, 2, 3 and 4 seconds.
2.5.3. Calculate the displacement covered in after 1, 2, 3 and 4 seconds and make a graph of the
displacement as a function of time.
2.6.1. A ball is thrown up with 5.0 ms-1. After what time is it moving down with 3.0 ms-1? [0.82 s]
2.6.2. A train is decelerated from 108 kmh-1 to 18 kmh-1 while moving 400 m forwards. Find the
acceleration. [-1.1 ms-2]
2.6.3. A car accelerates uniformly from 20 ms-1 for 8.5 s while moving 230 m forwards. Find the
acceleration. [1.7 ms-2]
2.6.4. A ball rolls from the initial velocity u to rest, decelerating uniformly. Derive an expression
for the initial velocity as a function of the displacement and time to come to rest. [u = 2s/t]
2.6.5. A ball is thrown from 1.8 m above the ground with the initial velocity 14 ms -1 up and falls
back to it. Find the time. [3.0 s]
2.7.1. A car with the mass 850 kg is decelerated from 40 ms -1 by the net force 200 N for 2.3 seconds. How
far will it move during this? [91 m]
2.7.2. On planet X, a forcemeter shows FG = 55 N when a 20 kg object hangs from it. If the object is
released, after what time will it have fallen 10 m down?
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The IB Physics Compendium 2005: Mathematical physics
14
2.8.1. A 1.5 kg object is on a horisontal surface. The force needed to overcome static friction is 3.8 N. Find
the coefficient of static friction.
2.8.2. A 300g object slides to a stop from the initial velocity 2.0 ms -1 while moving 68 cm forwards. Find the
coefficient of kinetic (dynamic) friction.
2.8.3. An object is given the initial speed 14 ms-1 up a slope inclined 33o where the friction coefficient is
0.12. How far will it have moved when the velocity has dropped to 5.0 ms-1 ? [13.5 m]
2.8.4. Estimate the order of magnitude of the kinetic (dynamic) friction coefficient for car with summer tires
on newly fallen snow, using relevant values you find reasonable.[hint: think of a car decelerating to rest,
estimate reasonable values for relevant quantities]
2.9.1. A 70 kg box is pulled 15 m along a horisontal floor by an 80 N force doing the work 350 J in this. At
what angle up from the horisontal level is the force acting? [73o]
2.9.2. A bomb is dropped from a hovering helicopter at the altitude 4400 m. If we ignore the energy lost to
air resistance, what will the ground impact speed be? [294 ms-1]
2.9.3. A pump moves 50 liters of water per minute 3.70 meters up. Find its power if the efficiency is 71 %.
[43 W]
2.10.1. When a "massless" spring is compressed 7.5 cm it acts with a force of 3.4 N on a 500 g object
attached to it. If the same spring with the same object is stretched out 13 cm from its equilibrium and then
released, what speed will the object have when it has been pulled back to 4.0 cm from the equilibrium, still
on the same side of the equilibrium as when it was released? [1.2 ms-1]
2.11.1. Find the period of a 150g mass oscillating on a spring with the spring constant 10 Nm-1.
2.11.2. Find the spring constant of a spring where a 75 g mass moves from one extreme position to the other
in 0.12 s.
2.11.3. Find the time period of a 150 g mass oscillating on a 90 cm long pendulum.
2.11.4. Find the gravity acceleration on a planet where the mass on the pendulum in problem 2.11.3.
completes 5 oscillations in 4.80 seconds.
2.12.1. Show that the kinetic energy depends on momentum and mass as Ek = p2/2m.
2.12.2. A rolling ball A moving at 5.0 ms-1 hits another ball B with twice its mass being at rest. B is given the
velocity 2.3 ms-1 in a direction 25 degrees to the right from the one A was moving in before the collision.
What is the magnitude and direction (angle to original direction) of A's velocity after the collision? [67 o to
the left of A's original direction, speed 2.1 ms-1]
2.12.3. A 900 kg car coming from the north at 20 ms-1 collides head-on with one coming from the south at 28
ms-1 with the mass 750 kg. If the cars stay together after collision, find the velocity of the wreck. [1.8 ms -1 to
the north]
2.12.4. A 120 g ball moving at +2.1 ms-1 hits a 190 g ball at rest in an elastic collision. Find the velocities of
the balls after the collision, assuming that they are parallel to the first ball's initial velocity. [- 0.47 ms-1 and
+1.63 ms-1]
2.12.5. The net force acting on a 5.0 kg ball is described by a graph of force as a function of time with lines
through the points (0s, 0N), (3s, 140N), (8s, 0N). The velocity of the ball at t = 0s is 15 ms -1 in a direction
opposite to that of the force at t = 3s. Find the velocity of the ball at t = 8 s. [97 ms-1]
2.12.6. Estimate the order of magnitude of the a) momentum b) kinetic energy of a running cat.
2.12.7. Find a general formula for the velocities after an elastic collision of objects with the masses m 1 and
m2 and the initial velocities u1 and u2.
2.13.1. An object is shot at the initial velocity 44 ms-1 in a direction 35o upwards from a cliff 80 m
above sea level. Find a) the maximum altitude of the object [112 m above sea] b) the time it takes to
reach it [2.57 s] c) the range (how far it travels horizontally) [265 m] d) the speed it hits the water
with [59.2 ms-1]e) the angle it hits the water with [52.5o down from the horizon]
2.14.1. A wrench exerts the torque 100 Nm on a nut when it is pulled with the force 1600 N acting
at a distance of 32 cm from the pivot point in the nut. Which two possible angles between the force
and the wrench arm can you find? [11.3o and 168.7o]
2.14.2. A 7.0 m long seesaw is in balance when a 50 kg person is sitting at one end and a 70 kg is
sitting - where? [2.5 m from pivot]
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The IB Physics Compendium 2005: Mathematical physics
15
2.14.3. A 4.00 m long 33 kg uniform beam is held horisontally in place by vertical ropes attached to
each end. A 15 kg object is hanging from a point 1.20 m from one end. Find the (forces of) tension
in the ropes.[206 N and 265 N]
2.14.4. A 500 kg load is hanging from the end of a 5.50 m long 64 kg uniform beam attached to a
wall at a 40o angle to the vertical direction. A horisontal rope is attached to the midpoint of the
beam and the wall. Find a) the tension in the rope [8758 N] b) the direction and magnitude of the
reaction force of the wall on the beam. [58o from vertical direction, magnitude 10359 N]]
2.15.1. The planet Venus revolves around the sun at a distance of 108 million km in 228 earth days.
Find its a) speed [34400 ms-1]b) centripetal acceleration. [0.011 ms-2]
2.15.2. A horisontal curve has the radius 140 m and the static friction coefficient is 0.09. At what
maximum speed can a car take the curve without skidding? [11 ms-1]
2.16.1. At what altitude above the surface of the earth (radius 6370 km) will the force of gravity on
a 400 kg satellite be 20.0 N? [82855 km]
2.17.1. Find the mass of earth if its radius is 6370 km and surface gravity acceleration 9.81 ms -2.
[5.97 x 1024 kg]
2.17.2. A rocket is launched from the south pole to a stable orbit 600 km above the earth's surface.
Find the energy that must be supplied by its rocket engine (assuming, unrealistically, that its mass
of 15500 kg remains constant). [5.26 x 1011 J]
2.17.3. Find a) the gravitational potential at the surface of a planet with the radius 3500 km and the
gravity acceleration 7.70 ms-2. [3.003 x 107 Jkg-1] b) the gravitational potential energy of a 15.5 ton
flying saucer at that point. [-4.65 x 1011 J]
2.18.1. If the planet in m16-3 had the escape speed 8.81 kms-1, could it then have the surface gravity
constant 7.70 ms-2? [No, since the surface gravity acceleration should then be 11.1 ms-2]
2.18.2. Estimate the size of an asteroid made of rock with the density 3000 kgm -3, from whose
surface you could throw a rock away into space. Estimate approximate values for necessary
quantities. [Radius ca 15 km for a spherical asteriod if one can throw with 20 ms-1, may vary
depending on assumptions and methods]
2.19.1. Planet I has two moons, B and O. If B revolves around I in 55 earth days at a distance of 280
000 km, and O revolves around I in 68 earth days, find its distance from I. [ca 320 000 km]
3.2-1. a)What is the difference between temperature and heat? b) between heat and internal energy?
3.3-1. Which are the four states of matter?
3.3-2. Draw the Maxwell-Boltzmann speed distribution diagram and how it will change if the
temperature is increased.
3.3-3. Draw a graph of a) potential energy b) the force between two atoms as a function of the
distance between them.
3.3-4. In what way can these diagrams explain why objects generally expand when the temperature
increases?
3.4-1. Heating 200 g of a material from +10oC to 300 K requires 50 kJ of energy. Find the specific
heat capacity of this material.[14.7 kJkg-1K-1]
3.4-2. Find the heat capacity of the same object.[2.9 kJK-1]
3.4-3. The latent heats of a) fusion b) vaporisation for water are 334 kJkg-1 and 2260 kJkg-1. How
much a) ice could be melted b) water at 373K boiled with 100 000 J of energy? [a) 299g b) 44 g]
3.6-1. Which are the 3 ways of transferring thermal energy?
3.7-1. What pressure does a cube of stone with the side length 0.50 m and the density 3000 kgm-3
cause on the floor it stands on? [15 kPa]
3.7-2. 5.00 m3 of an ideal gas has the temperature 400 K. What will its volume be if the pressure is
doubled and the temperature increased to 500 K?[3.13m3]
3.7-3. Draw in a diagram of pressure as a function of volume a graph for a) isobaric expansion b)
isothermal expansion c) adiabatic expansion d) any isochoric process
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The IB Physics Compendium 2005: Mathematical physics
16
3.9-1. A gas at 50 kPa expands isobarically from 2.00 m3 to 3.20 m3. Find the work done by the
gas.[60 kJ]
3.8-1. If 20 kJ of heat flowed into the gas during this, find its change in internal energy.[-40 kJ]
3.11-1. Find the efficiency of a heat engine which does 300 J of work when 800 J of heat flows out
of it. [27%]
3.12-1. The Carnot efficiency of an ideal heat engine is 40% when it releases "waste" heat at a
temperature of 300K. What minimum temperature must be found somewhere in the engine? [500
K]
3.13-1. In what way is a refrigerator relevant to the second law of thermodynamics?
4.1-1. Draw a graph of a) a wave with the wavelength 3.00 m and amplitude 1.50 mm b) the same
wave if its frequency is 10 Hz.
4.5-1. If the speed of sound is 340 ms-1, what would be the freqencies of the first and second
harmonics (stationary waves) in a) a 2.00 m long tube open at both ends [85 Hz, 170 Hz] b) a 2.00
m long tube open only at one end [42.5 Hz, 127.5 Hz] ?
4.6-1. A train moves towards us sending out a 500 Hz sound which we hear as 510 Hz. How fast is
the train moving (speed of sound in air same as before)?. [24kmh-1]
4.8-1. Laser light at 632 nm hits a diffraction grating with 500 lines per mm. At what angle to the
original direction of the light will we find the second order (n=2) maximum? [39.2o]
4.10-1. Light from air hits water, where it moves at 200 000 kms-1 at an angle of 38o. Where will it
be refracted? [Angle of refraction: 24o]
5.2-1. At what distance in vacuum from each other will two protons repel each other with a force of
1.00 pN? [15 nm]
5.2-2. If at that distance they repel each other with the force 0.25 pN, what is a) the permittivity
[3.54 x 10-11 C2N-1m-1] b) the relative permittivity of the medium they are in? [4]
5.2-3. Two electrons are 1 mm from each other. What is the resultant electric field in a point on a
line through them, 0.20 mm from one of them? [34 nVm-1]
5.3-3. Two parallel metal plates are connected to the terminals of a 4.5 V battery. The homogenous
electric field between them is 23 NC-1. What is the distance between them? [20 cm]
5.4-1. When a DC power source is connected to a 5.00 ohm resistor, the dissipated power is 200
mW. What is a) the current through the resistor [200 mA]b) the potential difference over it ? [1.00
V]
5.4-2. What resistor must be connected in parallel with the one in e4-1 to give a combined
resistance of 3.00 ? [7.5 ]
5.4-3. What is then the a) potential differences over the two resistors [What do you think?] b) the
currents through them? [0.20 A and 0.13 A]
5.5-1. A battery with the emf 12.00 V gives a terminal voltage of 11.50 V when the current 8.00 A
is drawn from it. Find a) its internal resistance [0.06 ] b) the resistance of the external circuit [1.44
]
5.6-1. How far from a straight wire carrying the current 2.00 A is the magnetic field intensity 2.00
% of that inside a solenoid carrying 1.00 A in 300 loops, and being 7.40 cm long? [3.93 mm]
5.7-1. What force is acting on a 3.00 cm long wire carrying 2.00 nA at an angle of 45 degrees to a
1.20 mT magnetic field? [510 pN]
5.7-2. If the current in e7-1 is carried by electrons moving at 5 % of the speed of light, how many
are needed? [25]
5.7-3. A piece of wire has the linear density (mass per length) µ and is falling downwards away
from another stationary but identical wire, both carrying the current I. Find an expression for the
resultant acceleration of the falling wire as a function of the distance it has fallen away from the
upper wire. State at least 3 different assumptions you make. [Depends on assumptions]
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The IB Physics Compendium 2005: Mathematical physics
17
6.2-1. What is the frequency of a photon emitted when an electron drops from the P-shell to the Oshell in a hydrogen atom? The Rydberg constant is 1.097 x 107 m-1. [40.2 THz]
6.2-2. What is the de Broglie wavelength of a neutron moving at 0.5 % of the speed of light? [0.26
pm]
6.5-1. If the relative uncertainties of the mass and velocity of the neutron in a2-2 both are 3.0 %,
what is the minimum absolute uncertainty in their position? [0.7 pm]
6.6-1. The work function of an unknown metal is 1.30 eV. What is the maximum kinetic energy of
photoelectrons emitted when ultraviolet light of the wavelength 100 nm is shining on the metal?
[10.7 eV]
6.7-1. What is the maximum frequency of photons emitted by an X-ray apparatus employing a 15.0
kV potential difference? [3620000 THz]
6.10-1. A new element, internationalbaccalaureatium, has been discovered. One isotope of it is
390
150Ib. Find its a) mass defect in kg b) binding energy per nucleon in MeV. The mass of a neutral
Ib-390 atom is 386.314159 u. [a) 1.15 x 10-26 kg b) 16.6 MeV/nucleon]
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