· web viewexample 5 : deriving ek = mc2 - m0c2 in relativity (note that the word...

The IB Physics Compendium 2005: Mathematical physics

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

Thomas Illman and Vasa övningsskola

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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 = x2 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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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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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

ò2xdx = x2

This looks a bit confusing, but just note that the integral symbol consists of two parts, a ò and a dx which tells that x is the variable (may be useful if we have several letters involved). Whatever is in between the ò and the dx is the function to integrate. Summary:

derivative: D(functiontoderive) or (d/dx)functiontoderive integral: ò(functiontointegrate)dx

Like the derivative, the integral takes one function and returns another function. 5. Some integration rules Rule A: òxndx= x(n+1)/(n+1)

òxdx = òx1dx = x(1+1)/(1+1) = ½x2

Rule B: Constants are not involved in integration

ò2xdx = 2òx1dx = 2*½x2 = x2 as expected ò5x4dx = 5òx4dx = 5*x(4+1)/(4+1) = x5

Rule C: If several terms are integrated, we can take one at a time


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ò(3x2 - 7x3)dx = ò3x2dx - ò7x3dx = 3òx2dx - 7òx3dx = 3*x(2+1)/(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)

ò(u + at)dt = òudt + òatdt = ò(u*1)dt + aòtdt = uòt0dt + aòt1dt =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 = òkxdx = kòxdx = kòx1dx = 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

ò(k/r2)dr = kòr-2dr = k*r(-2+1)/(-2+1) = -kr-1 = -k/x 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).


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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 ò2xdx = x2, the result could as well have been x2 +1 or x2 +2 or x2

- 127 or anything similar. We could then write x2 + C where C is an unknown (positive or negative) integration constant. In physics we can mostly set C = 0, for example as in: ò(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 òx-1dx = x(-1+1)/(-1+1) which involves division by zero. It can be shown in mathematics that òx-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 = ò(nRT/V)dV = nRTòV-1dV = nRTlnV when n, R and T are constants. Trigonometric functions Most useful are : Dsin x = cos x Dcos x = - sin x òcos x dx = sin x òsin 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 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.


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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 Ds/Dt. If Dt gets smaller and smaller, we reach the instantaneous velocity which is the gradient of the displacement-as-function-of-time curve or: the time derivative of displacement. When Ds and Dt 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 Dt 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 = S vDt. If the Dt becomes infinitely small and the slices infinitely many, this gives s = ò 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 òx2dx = 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!) òa2da = 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 = Öb = b½. One would think that we simply get òbdb, 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 = Öb we get da = db/2Öb so òa2da = òb(db/2Öb) = ò(Öb/2)db = ½ò(Öb)db and since ò(Öx)dx = (2x3/2)/3 we get 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.


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Example 1: Simple derivative Since v = ds/dt and s = ò vdt we can for a constant acceleration take v = u +at and get s = ò(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

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:


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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 = FDt = change in momentum = Dp = 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

W =òFds = ò(dp/dt)ds òd(mv)ds/dt which gives W = ò d(mv)v since ds/dt = vThen

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 = ò d(mv)v is, we can find out what ò(mvdv + v2dm) is. Now we first use the formula for mass increase:

m = m0/(1- v2/c2)½ square both sides so m2 = m0

2/(1- v2/c2) ; mult. with parenthesis and div. with m2 : m0

2 / 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 = - 2m0

2dm/m3 ; cancel -2 and multiply with mc2 : mvdv = m0

2c2dm/m2 = m02 c 2 dm(1/m 2 )

From the equation before the differentiation we take our expression for (1/m2):

mvdv = m02c2dm(1/m2) now gives

mvdv = m02c2dm((1- v2/c2) / m0

2) 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


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And this finally means that the integral simply is:

W = òc2dm = c2òdm 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 ...

12.4 Constants and formulas available in IB examinations

Page 1: Fundamental constants Gravity acceleration g 9.81 ms-2 Gravitational constant G 6.67 x 10-11 Nm2kg-2 Avogadro's constant NA 6.02 x 1023 mol-1


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Gas constant R 8.31 JK-1 mol-1 Boltzmann's constant k 1.38 x 10-23 JK-1 Stefan-Boltzmann constant s 5.67 x 10-8 Wm-2K-4 Coulomb constant k 8.99 x 109 Nm2C-2 Permittivity of free space e0 8.85 x 10-12 C2N-1m-2 Permeability of free space 0 4p x 10-7 TmA-1 Speed of light in vacuum c 3.00 x 108 ms-1 Planck's constant h 6.63 x 10-34 JsCharge on electron e 1.60 x 10-19 CElectron rest mass me 9.11 x 10-31 kg = 0.000549u = 0.511 MeV/c2 Proton rest mass mp 1.673 x 10-27 kg = 1.007276 u = 938 MeV/c2 Neutron rest mass mn 1.675 x 10-27 kg = 1.008665 u = 940 MeV/c2 Unified atomic mass unit u 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 parsec = 3.26 ly1 astronomical unit(AU) = 1.50 x 1011 m 1 radian (rad) = 180o/p1 kilowatt-hour (kWh) = 3.60 x 106 J 1 atm = 1.01 x 105 Nm-2 =101 kPa = 760 mmHg Page 3 : Electrical circuit symbolsSymbols are given for: cell, battery, lamp, ac supply, switch, ammeter, voltmeter, galvanometer, resistor, potentiometer, transformer, heating element Page 4 : Measurement and mechanicsHorizontal,vertical components vector A: AH = Acos AV = AsinUncertainties: If y = a ± b then Dy = Da + DbIf y = ab/c then Dy/y = Da/a + Db/b + Dc/c v = Ds/Dt a = Dv/Dt g = F/m g =Gm/r2

Page 5 : Mechanicsv = u + at s = ((u+v)/2)t s = ut + ½at2 v2 = u2 + 2ass : displacement, t : time, u : initial speed (?), v : final speed (?), a : acceleration F = ma p =mv F = Dp/Dt Ek =p2/2m Impulse = FDt = Dp W = (Fss =) Fs cos Ek = ½mv2 Ep= mgh P (= E/t or W/t) = work/time = Fv F = (-) kx Eelas=½kx2

a = v2/r = 4p2r/T2 t = FrsinQ Ffr = kN Ffr < or = sN F = Gm1m2/r2 Ep = -Gm1m2/r V=-Gm/r T2/R3=const.

Page 6 Thermal physics and WavesQ = mcDT Q = mL p = F/A pV = nRT DQ = DU + DW DW = pDV efficiency = (QH - QC)/ QH h = = (TH - TC)/ TH(Carnot efficiency) QC/QH = TC/TH(Carnot cycle) f = 1/T v = fl fbeat = ½f1 - f2 ½ Moving source: f' = f ( 1 / (1 ± vs/v)) Moving observer: f' = f (1 ± vo/v) dsinQ = nl s = lD/d sinQ1/sin Q2 = v1/v2 n = c / v n1sinQ1 = n2sinQ2


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Page 7 : Electricity and magnetism, electromagnetismF = kq1q2/r2 k = 1/4pe0 E = F/q E = kq / r2 E = - DV/Dx E = V/d

V = kq/r I = Dq/Dt R = V / I P = VI = I2R = V2/R R=R1+R2

1/R = 1/R1 + 1/R2 B = 0I/2pr B = 0NI/L = 0nI F = ILBsin

F = qvBsin F/L = 0I1I2/2pr e = BLv F=BAcos e = - NDF/Dt

Irms = I0/Ö2 Vrms = V0/Ö2 Vp/Vs = Np/Ns

Page 8 : Atomic, nuclear and quantum physicsE = mc2 E = hf p = h/l hf = φ + Ek,max hf = hf0 + eVs

N = N0e-lt T½=ln 2/l Page 9-10 : SL optionsPage 11 : Biomedical physics and The history and development of physicsb = 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/l = RH (1/n2 - 1/m2) DxDp ³ h/2p DEDt ³ h/2p : Astrophysics and RelativityL = sAT4 lmax = 2.90 x 10-3/T d(parsec) = 1/p(arc-second) b = L/4pd2 v = Hd Dl/l » v/c g = 1/( 1 - v2/c2 )½ Dt = gDt0 L = L0 / g

ux' =(ux - v)/(1 - uxv/c2) m = gm0 p =gm0u E0= m0c2 E = mc2

E2 = p2c2 + m02c4 Df/f = gDh/c2 RSch = 2GM / c2

Page 13 : Optics n1sin1 = n2sin2 n = 1/sinc 1/f = 1/u + 1/v m = hi / ho = v / u M = i / o = l/b = 1.22l/b dsin = nl

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


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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?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]


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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? [67o 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]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]]


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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 process3.9-1. 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%]


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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 W? [7.5 W]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 W] b) the resistance of the external circuit [1.44 W]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] 6.2-1. What is the frequency of a photon emitted when an electron drops from the P-shell to the O-shell 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]


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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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