LECTURES 1-10 INTRODUCTION TO CLASSICAL MECHANICS. Prof. N. Harnew University of Oxford MT 2016

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1 LECTURES 1-10 INTRODUCTION TO CLASSICAL MECHANICS Prof. N. Harnew University of Oxford MT

2 OUTLINE : INTRODUCTION TO MECHANICS LECTURES Outline of lectures 1.2 Book list 1.3 What is Classical Mechanics? 1.4 Vectors in mechanics Vector components in 3D Unit vectors Vector algebra 2.1 Scalar and vector products The scalar (dot) product The vector (cross) product Examples of scalar & vector products in mechanics 2.2 Differentiation of vectors Vector velocity Vector acceleration 3.1 Dimensional analysis The period of a pendulum

3 3.1.2 Kepler s third law The range of a cannon ball Example of limitations of the method 3.2 Newton s Laws of motion 3.3 Frames of reference 3.4 The Principle of Equivalence 4.1 Newton s Second Law 4.2 Newton s Third Law 4.3 Energy conservation in one dimension 5.1 Conservative forces Examples 5.2 Potential with turning points Oscillation about stable equilibrium Bounded and unbounded potentials 6.1 Lab & CM frames of reference 6.2 Internal forces and reduced mass 6.3 The Centre of Mass CM of a continuous volume Velocity in the Centre of Mass frame

4 6.3.3 Momentum in the CM frame Motion of CM under external forces 6.4 Kinetic energy and the CM 7.1 Two-body collisions - general concepts Momentum exchange and impulse An off-axis collision in 2D 7.2 Elastic collisions in the Lab frame Elastic collisions in 1D in the Lab frame Special case in 1D where target particle is at rest Collision in 2D : equal masses, target at rest 8.1 Elastic collisions in the CM frame 8.2 Lab to CM : 2-body 1D elastic collision Collision in 1D : numerical example 8.3 Relationship between speeds in CM in 2D 8.4 Lab to CM : 2-body 2D elastic collision 9.1 Examples of 2D elastic collisions Example 1: Equal masses, target at rest Example 2: Elastic collision, m 2 = 2m 1, θ 1 = Inelastic collisions in the Lab frame in 1D (u 2 = 0)

5 9.2.1 Coefficient of restitution 9.3 Inelastic collisions viewed in the CM frame Kinetic energy in the CM : alternative treatment Coefficient of restitution in the CM Example of inelastic process 10.1 Resisted motion and limiting speed 10.2 Air resistance 10.3 Example 1 : Resistive force, F R v 10.4 Example 2: Resistive force, F R v Work done on the body by the force for F R v 2 5

6 Two groups of lectures 1.1 Outline of lectures 10 in MT - mostly 1D & 2D linear motion. 19 in HT - 3D full vector treatment of Newtonian mechanics, rotational dynamics, orbits, introduction to Lagrangian dynamics Info on the course is on the web: Synopsis and suggested reading list Problem sets Lecture slides 6

7 7 1.2 Book list Introduction to Classical Mechanics A P French & M G Ebison (Chapman & Hall) Introduction to Classical Mechanics D. Morin (CUP) (good for Lagrangian Dynamics and many examples). Classical Mechanics : a Modern Introduction, M W McCall (Wiley 2001) Mechanics Berkeley Physics Course Vol I C Kittel et al. (McGraw Hill) Fundamentals of Physics Halliday, Resnick & Walker (Wiley) Analytical Mechanics 6th ed, Fowles & Cassidy (Harcourt) Physics for Scientists & Engineers, (Chapters on Mechanics) P.A Tipler & G. Mosca (W H Freeman) Classical Mechanics T W B Kibble & F H H Berkshire (Imperial College Press)

8 1.3 What is Classical Mechanics? Classical mechanics is the study of the motion of bodies in accordance with the general principles first enunciated by Sir Isaac Newton in his Philosophiae Naturalis Principia Mathematica (1687). Classical mechanics is the foundation upon which all other branches of Physics are built. It has many important applications in many areas of science: I Astronomy (motion of stars and planets) I Molecular and nuclear physics (collisions of atomic and subatomic particles) I Geology (e.g., the propagation of seismic waves) I Engineering (eg structures of bridges and buildings) Classical Mechanics covers: 8 I The case in which bodies remain at rest I Translational motion by which a body shifts from one point in space to another I Oscillatory motion e.g., the motion of a pendulum or spring I Circular motion motion by which a body executes a circular orbit about another fixed body [e.g., the (approximate) motion of the earth about the sun] I More general rotational motion orbits of planets or bodies that are spinning I Particle collisions (elastic and inelastic)

9 Forces in mechanics Relative magnitude of forces: Strong force - nuclear : 1 Electromagnetism - charged particles : Weak force - β decay : 10 5 Gravitational - important for masses, relative strength :

10 Not too fast! Classical Mechanics valid on scales which are: Not too fast eg. high energy particle tracks from CERN v << c [speed of light in vacuo] If too fast, time is no longer absolute - need special relativity. 10

11 Classical Mechanics valid on scales which are: Not too small! Images of atom planes in a lattice by scanning tunneling electron microscope Particles actually have wave-like properties : λ = h p (h = Js) Hence for scales >> λ, wave properties can be ignored Not too small! 11

12 Not too large! Classical Mechanics valid on scales which are: Not too large! Gravitational lens produced by a cluster of galaxies Space is flat in classical mechanics - curvature of space is ignored Also in Newtonian mechanics, time is absolute 12

13 1.4 Vectors in mechanics The use of vectors is essential in the formalization of classical mechanics. Notation: A scalar is characterised by magnitude only: energy, temperature. A vector is a quantity characterised by magnitude and direction: eg. Force, momentum, velocity. Vector: a (bold); in components a = (a x, a y, a z ) Magnitude of a is a or simply a. Two vectors are equal if they have the same magnitude and direction (i.e. parallel) a = b gives a x = b x, a y = b y, a z = b z 13

14 1.4.1 Vector components in 3D Projecting the components: p = (p x, p y, p z ) x-component p x = p sin(θ) cos(φ) y-component p y = p sin(θ) sin(φ) z-component p z = p cos(θ) Magnitude p = (p 2 x + p 2 y + p 2 z) Direction tan(φ) = (py /p x ) cos(θ) = (p z / p ) 14

15 1.4.2 Unit vectors A unit vector is a vector with magnitude equal to one. e.g. Three unit vectors defined by orthogonal components of the Cartesian coordinate system: i = (1,0,0), obviously i = 1 j = (0,1,0), j = 1 k = (0,0,1), k = 1 A unit vector in the direction of general vector a is written â = a/ a a is written in terms of unit vectors a = a x i + a y j + a z k x O i z k j y 15

16 1.4.3 Vector algebra Sum of two vectors To calculate the sum of two vectors c = a + b Triangle rule: Put the second vector nose to tail with the first and the resultant is the vector sum. c = a + b : in (x, y, z) components (c x, c y, c z ) = (a x +b x, a y +b y, a z +b z ) Alternatively c = a + b c x i + c y j + c z k = (a x + b x )i + (a y + b y )j + (a z + b z )k 16

17 Vector algebra laws a + b = b + a : commutative law a + (b + c) = (a + b) + c: associative law Can treat vector equations in same way as ordinary algebra a + b = c a = c b Note that vector b is equal in magnitude to b but in the opposite direction. so a b = a + ( b) 17

18 Multiplication of a vector by a scalar This gives a vector in the same direction as the original but of proportional magnitude. For any scalars α and β and vectors a and b (αβ) a = α(β a ) = β(α a ) = a (αβ) : associative & commutative (α + β)a = αa + βa : distributive α(a + b) = αa + αb 18

19 2.1.1 The scalar (dot) product Scalar (or dot) product definition: a.b = a. b cos θ ab cos θ (write shorthand a = a ). Scalar product is the magnitude of a multiplied by the projection of b onto a. Obviously if a is perpendicular to b then a.b = 0 Also a.a = a 2 (since θ =0 ) Hence a = (a.a) 19

20 Properties of scalar product (i) i.i = j.j = k.k = 1 and i.j = j.k = k.i = 0 (ii) This leads to a.b = (a x i + a y j + a z k).(b x i + b y j + b z k) = a x b x + a y b y + a z b z iii) a.b = b.a : commutative a.(b + c) = a.b + a.c : distributive (iv) Parentheses are important Note (u.v) w u (v.w) because one is a vector along ŵ, the other is along û. 20

21 2.1.2 The vector (cross) product Vector (or cross) product of two vectors, definition: a b = a b sinθ ˆn where ˆn is a unit vector in a direction perpendicular to both a and b. To get direction of a b use right hand rule: 21 i) Make a set of directions with your right hand thumb & first index finger, and with middle finger positioned perpendicular to plane of both ii) Point your thumb along the first vector a iii) Point your 1st index finger along b, making the smallest possible angle to a iv) The direction of the middle finger gives the direction of a b. b c=axb a

22 Properties of vector product i j = k ; j k = i ; k i = j ; i i = 0 etc. (a + b) c = (a c) + (b c) : distributive a b = b a : NON-commutative (a b) c a (b c) : NON-associative If m is a scalar, m(a b) = (ma) b = a (mb) = (a b)m a b = 0 if vectors are parallel (0 o ) i.e a a = 0 22

23 Vector product in components Cross product written out in components: a b = (ax, a y, a z ) (b x, b y, b z ) = (a x i + a y j + a z k) (b x i + b y j + b z k) Since i i = j j = k k = 0 and i j = k etc. a b = (ay b z a z b y )i (a x b z a z b x )j + (a x b y a y b x )k This is much easier when we write in determinant form: a b = i j k a x a y a z b x b y b z (1) 23

24 2.1.3 Examples of scalar & vector products in mechanics a) Scalar product Work done on a body by a force through distance dr from position 1 to 2 W 12 = 2 1 F.dr Only the component of force parallel to the line of displacement does work. b) Vector product A torque about O due to a force F acting at B : τ = r F Torque is a vector with direction perpendicular to both r and F, magnitude of r F sin θ. 24

25 2.2 Differentiation of vectors Notation: a dot above the function indicates derivative wrt time. A dash indicates derivative wrt a spatial coordinate. ẏ dy dt y dy dx a(t + t) a(t) a ȧ = lim = lim t 0 t t 0 t a(t) = a x (t)i + a y (t)j + a z (t)k ( ) ax (t + t) a x (t) ȧ = lim i +... t 0 t Hence ȧ = ȧ x i + ȧ y j + ȧ z k 25

26 2.2.1 Vector velocity r = r2 r 1 r v = lim t 0 t Velocity at any point is tangent to the path at that point v = dr dt = ṙ In one dimension: Abandon vector notation and simply write v = dx dt = ẋ, (+v in +x direction, v in x direction). 26

27 Example - 1D motion A body has velocity v 0 = 15 ms 1 at position x 0 = 20m and has a time-dependent acceleration a(t) = 6t 4 [ms 2 ]. Find the value of x for which the body instantaneously comes to rest. a(t) = 6t 4 [ms 2 ] ; x 0 = 20 m ; v 0 = 15 ms 1 v = 6t 4 v = a(t)dt = 3t 2 4t + c At t = 0, v = 15 ms 1 c = 15 ms 1 v = 3t 2 4t 15 v = 0 for 3t 2 4t 15 = 3(t 3)(t ) = 0 27 t = 3s (also 5 3 s) x = v(t)dt = t 3 2t 2 15t + c x = 20 m at t = 0, c = 20 m x(t) = = 16 m

28 2.2.2 Vector acceleration v = v2 v 1 v a = lim = v = r t 0 t In one dimension: Abandon vector notation and simply write a = dv dt = v = ẍ, (+a in +x direction, a in x direction). 28

29 Example: constant acceleration - projectile motion in 2D a = dv dt = constant r = 0 at t = 0 v v 0 dv = t 0 adt v = v 0 + at v = dr dt r 0 dr = t 0 (v 0 + at)dt r = v 0 t at2 Under gravity: a = gŷ a x = 0; a y = g vx = v 0 cos θ x = (v0 cos θ)t vy = v 0 sin θ gt y = (v0 sin θ)t 1 2 gt2 Trajectory: y = (tan θ)x g (sec 2 θ)x 2 2v0 2 29

30 30 The monkey and the hunter

31 3.1 Dimensional analysis A useful method for determining the units of a variable in an equation Useful for checking the correctness of an equation which you have derived after some algebraic manipulation. Dimensions need to be correct! Determining the form of an equation itself Most physical quantities can be expressed in terms of combinations of basic dimensions. These are certainly not unique : 31 mass (M) length (L) time (T) electric charge (Q) temperature (θ)

32 Note: The term "dimension" is not quite the same as "unit", but obviously closely related. Quantity Unit Dimension Frequency Hertz (Hz) = (cycles) s 1 T 1 Force Newton (N) = kg m s 2 MLT 2 Energy Joule (J) = N m = kg m 2 s 2 ML 2 T 2 Power Watt (W) = J s 1 = kg m 2 s 3 ML 2 T 3 Current Ampere (A) = Cs 1 QT 1 EMF Volt (V) = Nm C 1 = kg m 2 s 2 C 1 ML 2 T 2 Q 1 Dimensional analysis is best illustrated with examples. 32

33 3.1.1 The period of a pendulum How does the period of a pendulum depend on its length? Variables: period P, mass m, length l, acceleration due to gravity g Guess the form: let P = k m a l b g c (k is a dimensionless constant) T 1 = M a L b (LT 2 ) c = M a L b+c T 2c Compare terms: a = 0, b + c = 0, 2c = 1 c = 1/2, b = 1/2 P = k We know that P = 2π l g l g : we obtained this form using dimensions and without using equation of motion: IMPRESSIVE! 33

34 3.1.2 Kepler s third law How does the period of an orbiting mass depend on its radius? Variables: period P, central mass M0, orbit radius r, Gravitational constant G Guess the form: let P = k M a 0 r b G c (k is a dimensionless constant) Dimensions of G (MLT 2 ).L 2 M 2 T 1 = M a L b (M 1 L 3 T 2 ) c = M (a c) L b+3c T 2c Compare terms: a c = 0, b + 3c = 0, 2c = 1 a = 1/2, c = 1/2, b = 3/2 P = k M 1/2 0 r 3/2 G 1/2 P 2 = k 2 GM 0 r 3 GmM 0 = mv 2 r 2 r v = 2πr P P 2 = 4π2 GM 0 r 3 k 2 = 4π 2 34

35 3.1.3 The range of a cannon ball A cannon ball is fired with V y upwards and V x horizontally, assume no air resistance. Variables: V x, V y, distance travelled along x (range) R, acceleration due to gravity g First with no use of directed length dimensions Let R = kv a x V b y g c. 35 (k is a dimensionless constant) Dimensionally L = (L/T ) a+b (L/T 2 ) c Compare terms: a + b + c = 1 and a + b + 2c = 0, which leaves one exponent undetermined. Now use directed length dimensions, then V x will be dimensioned as L x /T, V y as L y /T, R as L x and g as L y /T 2 The dimensional equation becomes: L x = (L x /T ) a (L y /T ) b (L y /T 2 ) c a = 1, b = 1 and c = 1. vx vy R = k g x = vx t y = vy t 1 2 gt 2 = 0 t = 2vy g x = 2v x v y g

36 3.1.4 Example of limitations of the method Let y = f (x1, x 2,... x n ) where x 1, x 2,... x n have independent dimensions However in general y = (x a 1 x b 2... x n ) φ(u 1,... u k ) where u i are dimensionless variables Extend to how the period of a rigid pendulum depends on length pivot to CM. In actual fact P P(g, l, m, I) where I is the moment of inertia [I] = ML 2 can define u = I T = l g φ(u) i.e. Equation is not reproduced ml 2 T = 2π I mgl 36

37 3.2 Newton s Laws of motion NI: Every body continues in a state of rest or in uniform motion (constant velocity in straight line) unless acted upon by an external force. NII: The rate of change of momentum is equal to the applied force; where the momentum is defined as the product of mass and velocity (p = mv). [i.e. the applied force F on a body is equal to its mass m multiplied by its acceleration a.] NIII: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body [i.e. action and reaction forces are equal in magnitude and opposite in direction.] 37

38 3.3 Frames of reference A frame of reference is an environment which is used to observe an event or the motion of a particle. A coordinate system is associated with the frame to observe the event (eg the body s location over time). The observer is equipped with measuring tools (eg rulers and clocks) to measure the positions and times of events. In classical mechanics, time intervals between events is the same in all reference frames (time is absolute). In relativity, we will need to use space-time frames. A reference frame in which NI is satisfied is called an inertial reference frame. 38

39 Inertial reference frames A frame in which Newton s first law is satisfied: Deep space The Earth? [Only in circumstances where we can ignore gravity & the spin of the Earth.] Principle of Relativity : The laws of Physics are the same in all inertial frames of reference. At t = 0, x = 0, x = 0 and S and S are coincident. Galilean Transformation of coordinates: x = x v 0 t, y = y, z = z, t = t Velocity of a body v in S; velocity measured in S v = v v 0 Acceleration measured in S a = a Hence F = F (consistent with the principle of relativity) 39

40 3.4 The Principle of Equivalence The Principle of Equivalence dictates that m = m. Inertial mass = Gravitational mass This may seem obvious, but it was not an original postulate of Newton 40

41 4.1 Newton s Second Law The rate of change of momentum of a body is equal to the applied force on the body. F = dp dt = ma where p = mv In components: (Fx, F y, F z ) = m(a x, a y, a z ) Assuming constant mass, we can define the equation of motion in 1D : F = d(mv) dt = m d 2 x dt 2 We require two initial conditions for a unique solution: e.g. v = v 0 at t = 0 and x = x 0 at t = 0 We shall later solve the EOM for three examples: (i) F = constant, (ii) F v, (iii) F x 41

42 4.2 Newton s Third Law Action and reaction forces are equal in magnitude and opposite in direction. Electrostatic interaction F 12 = F 21 Compressed spring 42

43 Conservation of momentum Compressed spring F12 = m 1 a 1 = dp 1 dt and F 21 = m 2 a 2 = dp 2 dt F12 + F 21 = d dt (P 1 + P 2 ) = 0 (Newton III) Therefore (P1 + P 2 ) = constant In an isolated system, the total momentum is conserved. 43

44 Newton II : Example 1. E.O.M. under constant force How fast should we accelerate the triangular wedge to keep the block m stationary on the wedge? Forces on wedge: Forces on block: Horizontal: Fex F i sin θ = MA x Vertical: R F i cos θ Mg = 0 Horizontal: Fi sin θ = ma x Vertical: Fi cos θ mg = ma y For block to remain at the same place A x = a x and a y = 0 Fi = mg cos θ and a x = g tan θ = A x Hence Fex = Mg tan θ + mg tan θ = (m + M)g tan θ 44

45 Example constant force, continued What is the internal force that the blocks apply on each other and the reaction force by the ground on M? From before: Fi = mg cos θ R Fi cos θ Mg = 0 Hence: R = Fi cos θ + Mg = (m + M)g 45

46 Example 2. Force proportional to velocity Solve the equation of motion for the case F = βv (β > 0) with x = x 0 and v = v 0 at t = 0 m dv dv dt = βv dt = αv where α = β m v dv v 0 v = α t 0 dt v = v 0e αt v = dx dt x x 0 dx = t 0 vdt = t 0 v 0e αt dt x x0 = v 0 α e αt + v 0 α x = x0 + v 0 α (1 e αt ) When t, x x0 + v 0 α 46

47 Example 3. Force proportional to position: simple harmonic oscillator Solving the equation of motion for the case F = m d 2 x = kx dt 2 47 m d 2 x + k dt 2 mx = 0 ; trial solution x = A cos ωt + B sin ωt ẋ = Aω sin ωt + Bω cos ωt ; ẍ = Aω 2 cos ωt Bω 2 sin ωt ẍ = ω 2 x ω 2 = k m Alternatively x = x0 cos(ωt + φ) (or x = x 0 Re[e i(αt+φ) ] ) Expand : x = x0 (cos(ωt) cos φ sin(ωt) sin φ) A = x 0 cos φ ; B = x 0 sin φ x 2 0 = A2 + B 2 ; tan φ = B/A x0 = amplitude, φ = phase, ω = angular frequency (T = 2π ω )

48 x = x0 cos(ωt + φ) ω = k m φ = ω t 48

49 4.3 Energy conservation in one dimension Work done on a body by a force F 49 x2 W = x 1 F(x)dx = m x 2 dv x 1 dt dx We can write: dv dt dx dx = dt dv = vdv hence x 2 x 1 F(x)dx = m v 2 v 1 vdv = 1 2 m(v 2 2 v 1 2) = T 2 T 1 Now introduce an arbitrary reference point x0 x2 x 1 Fdx = x 2 x 0 Fdx x 1 x 0 Fdx defines a conservative force hence T 2 + [ x 2 x 0 Fdx] = T 1 + [ x 1 x 0 Fdx] We define the potential energy U(x) at a point x : U(x) U(x 0 ) = x x 0 Fdx and hence T 2 + U 2 = T 1 + U 1 (total energy PE + KE conserved) Note the minus sign. The potential energy (relative to a reference point) is always the negative of the work done by the force F(x) = du dx

50 5.1 Conservative forces W ab = b a F. dr = U(a) U(b) For a conservative field of force, the work done depends only on the initial and final positions of the particle independent of the path. The conditions for a conservative force (all equivalent) are: The force is derived from a (scalar) potential function: F(r) = U F(x) = du dx etc. There is zero net work by the force when moving a particle around any closed path: W = c F. dr = 0 In equivalent vector notation F = 0 For any force: W ab = 1 2 mv 2 b 1 2 mv 2 a Only for a conservative force: W ab = U(a) U(b) 50

51 Conservative force: example 1. Constant acceleration Consider a particle moving under constant force (in 1-D). F = ma. Say at t = 0 x = x0 and v = v 0 T2 + U 2 = T 1 + U 1 (the total energy is conserved) 1 2 mv 2 (ma)x = 1 2 mv 2 0 (ma)x 0 = constant v 2 = v a(x x 0) Gravitational potential energy U( y) = y2 y 1 F(y)dy U( y) = y2 y 1 ( mg)dy = mg(y 2 y 1 ) 51

52 Example 2. Simple harmonic oscillator Equation of motion: F = m d 2 x dt 2 = kx x Potential energy: U(x) = 0 Fdx = x kx 2 0 ( kx)dx = 2 Total energy: E = T (x) + U(x) = 1 2 mẋ 2 + kx 2 2 Check conservation of energy: EOM : mẍ + kx = 0 [multiply by ẋ] mẍẋ + kxẋ = m d dt (ẋ 2 ) k d dt (x 2 ) = 0 Integrate wrt t: 1 2 mẋ kx 2 = constant i.e. energy conserved. 52

53 SHM potential energy curve E = U(x) mv 2 The particle can only reach locations x that satisfy U < E 53

54 Example 3. Minimum approach of a charge A particle of mass m and charge +Q 1 starts from x = + with velocity v 0. It approaches a fixed charge +Q. Calculate its minimum distance of approach x min. Force on charge +Q1 : F(x) = + QQ 1 4πɛ 0 x 2 (+ve direction) Potential energy at point x: U(x) = x F(x)dx = + QQ 1 4πɛ 0 x (where PE = 0 at x = ) Conservation of energy : 1 2 mv = 1 2 mv 2 + U(x) Min. dist. when v = 0: 1 2 mv 2 0 = QQ 1 4πɛ 0 x min x min = QQ 1 2πmɛ 0 v

55 5.2 Potential with turning points U is a maximum: unstable equilibrium U is a minimum: stable equilibrium 55

56 5.2.1 Oscillation about stable equilibrium 56 For SHM : U(x) = 1 2 k(x x 0) 2 Taylor expansion [ about ] x0 : [ du d U(x) = U(x 0 )+ (x x 0 )+ 1 2 ] U dx 2! x=x }{{} 0 dx 2 (x x 0 ) x=x }{{} 0 =0 =k

57 Example: The Lennard-Jones potential The Lennard-Jones potential describes the potential energy between two atoms in a molecule: U(x) = ɛ[(x 0 /x) 12 2(x 0 /x) 6 ] (ɛ and x 0 are constants and x is the distance between the atoms). Show that the motion for small displacements about the minimum is simple harmonic and find its frequency. U(x) = U(x0 ) + [ ] [ du dx (x x x=x 0 ) + 1 d 2 U 0 2! (x x dx ]x=x 2 0 ) U(x 0 ) = ɛ[(x 0 /x) 12 2(x 0 /x) 6 ] x=x0 = ɛ du(x) dx x=x0 = 12ɛ[ 1 x 0 (x 0 /x) x 0 (x 0 /x) 7 ] x=x0 = 0 as expected. d 2 U(x) dx 2 x=x0 = 12ɛ [13(x 0 /x) 14 7(x 0 /x) 8 ] x=x0 x 2 0 Hence U(x) ɛ + 72ɛ (x x 2!x0 2 0 ) 2 F(x) = du dx 1 2 = 72ɛ x ɛ 2( )(x x x0 2 0 ) = k(x x 0 ) SHM about x 0 Angular frequency of small oscillations : ω 2 = k m = 72ɛ mx 2 0

58 5.2.2 Bounded and unbounded potentials Bounded motion : E = E1 : x constrained x 1 < x < x 2 Unbounded motion : E = E2 : x unconstrained at high x x 0 < x < 58

59 6.1 Lab & CM frames of reference From hereon we will deal with 2 inertial frames: The Laboratory frame: this is the frame where measurements are actually made The centre of mass frame: this is the frame where the centre of mass of the system is at rest and where the total momentum of the system is zero 59

60 6.2 Internal forces and reduced mass Internal forces only: F 12 = m 1 r 1 ; F 21 = m 2 r 2 Then r 2 r 1 = F 21 m 2 F 12 m 1 NIII : F 21 = F 12 = F int 60 Define r = r2 r 1 ṙ = ṙ 2 ṙ 1 Fint ( 1 m m 2 ) = r Define 1 µ = 1 m m 2 F int = µ r µ = m 1m 2 m 1 +m 2 is the reduced mass of the system This defines the equation of motion of the relative motion of the particles under internal forces, with position vector r & mass µ

61 6.3 The Centre of Mass The centre of mass (CM) is the point where the mass-weighted position vectors (moments) relative to the point sum to zero ; the CM is the mean location of a distribution of mass in space. Take a system of n particles, each with mass m i located at positions r i, the position vector of the CM is defined by: n i=1 m i(r i r cm ) = 0 Solve for rcm : r cm = 1 M n i=1 m i r i where M = n i=1 m i 61

62 Example : SHM of two connected masses in 1D SHM between two masses m 1 and m 2 connected by a spring x = x2 x 1 ; Natural length L Fint = k (x L) = µ ẍ (µ = m 1m 2 m 1 +m 2 = reduced mass) ẍ + k µ (x L) = 0 Solution: x = x 0 cos(ωt + φ) + L where ω = k µ With respect to the CM: xcm = m 1x 1 +m 2 x 2 M where M = m 1 + m 2 x 1 = x 1 x cm = Mx 1 m 1 x 1 m 2 x 2 M x 2 = x 2 x cm = Mx 2 m 1 x 1 m 2 x 2 Eg. take m 1 = m 2 = m ω = 62 M = m 2x M = m 1x M 2k m ; x 1 = 1 2 x, x 2 = 1 2 x

63 6.3.1 CM of a continuous volume If the mass distribution is continuous with density ρ(r) inside a volume V, then: n i=1 m i(r i r cm ) = 0 becomes V ρ(r)(r r cm)dv = 0 where dm = ρ(r)dv Solve for rcm r cm = 1 M V ρ(r) r dv where M is the total mass in the volume 63

64 Example: the CM of Mount Ranier Mount Ranier has approximately the shape of a cone (assume uniform density) and its height is 4400 m. At what height is the centre of mass? We have cylindrical symmetry - just need to consider the y direction. Integrate from top (y = 0) to bottom (y = h) h ycm 0 = ydm h 0 M = ydm h 0 dm dm = ρ(πr 2 )dy = ρ(πy 2 tan 2 θ)dy ycm = h 0 yρ(πy 2 tan 2 θ)dy h 0 ρ(πy 2 tan 2 θ)dy h 0 y cm = y 3 dy = 3h4 h = 3h 0 y 2 dy 4h 3 4 (measured from the top) 64 h = 4400m ycm = 1100m above the base

65 6.3.2 Velocity in the Centre of Mass frame The position of the centre of mass is given by: r cm = 1 n M i=1 m i r i where M = n i=1 m i The velocity of the CM: v cm = ṙ cm = 1 M n i=1 m i ṙ i In the Lab frame, total momentum ptot : p tot = n i=1 m i ṙ i = Mv cm Hence the total momentum of a system in the Lab frame is equivalent to that of a single particle having a mass M and moving at a velocity v cm 65

66 6.3.3 Momentum in the CM frame Velocity of the CM: vcm = ṙ cm = n i=1 m i ṙ i i m i = n i=1 m i v i i m i Velocity of a body in the CM relative to Lab v i = v i v cm The total momentum of the system of particles in the CM: i p = i i m iv i = i m i(v i v cm ) = i m iv i i m i j m j v j j m j = i m iv i j m jv j = 0 Hence the total momentum of a system of particles in the CM frame is equal to zero In addition, the total energy of the system is a minimum compared to all other inertial reference frames. 66

67 i Motion of CM under external forces Force on particle i: mi r i = F ext int i + F i n n n ext int m i r i = F i + F i = n ext i F i }{{} all masses i }{{} external forces i }{{} internal forces = zero rcm = i m i r i M where M = i m i m i r i M rcm = i M r CM = i F i ext The motion of the system is equivalent to that of a single particle having a mass M acted on by the sum of external forces 67 (The CM moves at constant velocity if no external forces)

68 6.4 Kinetic energy and the CM Lab kinetic energy : TLab = 1 2 mi v 2 i ; v i = v i v CM where v i is velocity of particle i in the CM TLab = 1 2 mi v 2 i + m i v i v CM mi v 2 CM But mi v mi v i i v CM = M v } M CM = 0 {{} = 0 T Lab = T CM Mv 2 cm The kinetic energy in the Lab frame is equal to the kinetic energy in CM + the kinetic energy of the CM 68

69 7.1 Two-body collisions - general concepts No external forces. Collision via massless springs or other force type. 69 ti : collision starts. All energy is kinetic. T i = 1 2 m 1u m 2u2 2 t : collision peaks. Some kinetic is converted into potential (of the spring). E = 1 2 m 1x m 2x 2 2 +E int tf : collision ends. All energy is kinetic again. T f = 1 2 m 1v m 2v 2 2 ; T i = T f + E ( inelastic)

70 7.1.1 Momentum exchange and impulse During collision: internal force causes change of momentum F = dp dt At ti : total momentum p = p 1 + p 2 = m 1 u 1 + m 2 u 2 At t : m1 dp 1 = F 12 dt m 2 dp 2 = F 21 dt At tf : m 1 v 1 and m 2 v 2 Impulse p1 = I 1 = t f t i p 2 = I 2 = t f t i Since F12 + F 21 = 0 p 1 + p 2 = 0 Momentum conserved. F 12 dt F 21 dt 70

71 7.1.2 An off-axis collision in 2D 71 Impulse is along line of centres p 1 = t f t i p 2 = t f t i F 12 dt F 21 dt v1 = 1 m 1 tf t i F 12 dt + u 1 v2 = 1 m 2 tf t i F 12 dt

72 7.2 Elastic collisions in the Lab frame Before After Conservation of momentum: m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 Conservation of energy: 1 2 m 1u m 2u 2 2 = 1 2 m 1v m 2v 2 2 [Note that the motion is in a plane, and the 2D representation can be trivially extended into 3D by rotation of the plane]. 72

73 7.2.1 Elastic collisions in 1D in the Lab frame Momentum : m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 (1) m 1 (v 1 u 1 ) = m 2 (u 2 v 2 ) (2) Energy : 1 2 m 1u m 2u 2 2 = 1 2 m 1v m 2v 2 2 m 1 (v 1 u 1 )(v 1 + u 1 ) = m 2 (u 2 v 2 )(u 2 + v 2 ) (3) Divide (2) & (3) : (v 1 + u 1 ) = (u 2 + v 2 ) (u 1 u 2 ) = (v 2 v 1 ) (4) Relative speed before collision = Relative speed after 73

74 7.2.2 Special case in 1D where target particle is at rest u2 = 0 ; From (1) & (4) : m 1 u 1 = m 1 v 1 + m 2 (u 1 + v 1 ) v1 = (m 1 m 2 )u 1 m 1 +m 2 Similarly : m 1 u 1 = m 1 (v 2 u 1 ) + m 2 v 2 v2 = 2m 1u 1 m 1 +m 2 Special cases: 74 v 1 = m 1 m 2 m 1 +m 2 u 1 and v 2 = 2m 1 m 1 +m 2 u 1 m1 = m 2 : v 1 = 0, v 2 = u 1 (complete transfer of momentum) m 1 >> m 2 : Gives the limits v 1 u 1, v 2 2u 1 (m 2 has double u 1 velocity) m1 << m 2 : Gives the limits v 1 u 1, v 2 0 ( brick wall collision)

75 Example: Newton s cradle Consider here just 3 balls If the balls are touching, the most general case is: 75 Momentum after collision : mu = mv 1 + mv 2 + mv 3 1 Energy after collision : 2 mu2 = 1 2 mv mv mv equations, 3 unknowns The obvious solution: v1 = v 2 = 0, v 3 = u But other solution(s) possible: Momentum : mu = 1 3 mu mu mu 1 Energy : 2 mu2 = 1 18 mu mu mu2 So why does the simple solution always prevail?

76 7.2.3 Collision in 2D : equal masses, target at rest m 1 = m 2 = m, u 2 = 0 Momentum: mu1 = mv 1 + mv 2 u 1 = v 1 + v 2 Squaring u 2 1 = v2 1 + v v 1.v 2 Energy: 1 2 mu2 1 = 1 2 mv mv2 2 u2 1 = v2 1 + v2 2 Hence 2v1.v 2 = 0 EITHER v 1 = 0 & v 2 = u 1 OR (θ 1 + θ 2 ) = π 2 Either a head-on collision or opening angle is 90 76

77 Relationship between speeds and angles m 1 = m 2 = m, u 2 = 0 From cons. mom. (v2 ) 2 = (u 1 v 1 ) 2 v 2 2 = v u2 1 2u 1v 1 cos θ 1 Energy: u 2 1 = v v 2 2 v 2 2 = u2 1 v 2 1 Equate : 2v 2 1 = 2u 1 v 1 cos θ 1 cos θ 1 = v 1 u 1 and by symmetry cos θ 2 = v 2 u 1 Note we can also do this via components of momentum : u 1 = v 1 cos θ 1 + v 2 cos θ 2 and v 1 sin θ 1 = v 2 sin θ 2 (u 1 v 1 cos θ 1 ) 2 = v2 2 cos2 θ 2 and v1 2 sin2 θ 1 = v2 2 sin2 θ 2 Add : v 2 2 = v 2 1 sin2 θ 1 + u 2 1 2u 1v 1 cos θ 1 + v 2 1 cos2 θ 1 77 Gives : v 2 2 = v u2 1 2u 1v 1 cos θ 1

78 8.1 Elastic collisions in the CM frame Lab frame: Centre of mass frame (zero momentum frame) 78 Conservation of momentum in CM: m 1 u 1 + m 2 u 2 = 0 ; m 1 v 1 + m 2 v 2 = 0 Conservation of energy in CM: 1 2 m 1u m 2u 2 2 = 1 2 m 1v m 2v 2 2

79 8.2 Lab to CM : 2-body 1D elastic collision vcm = (m 1u 1 +m 2 u 2 ) (m 1 +m 2 ) Before in CM : m 1 u 1 + m 2u 2 = 0 After in CM : m 1 v 1 + m 2v 2 = 0 From last lecture u 1 u 2 = v 2 v 1 Sub for u 2, v 2 : u 1 (1 + m 1/m 2 ) = v 1 (1 + m 1/m 2 ) v 1 = u 1 v 2 = u 2 79

80 8.2.1 Collision in 1D : numerical example 1) Find CM velocity relative to laboratory frame : v cm = m 1u 1 +m 2 u 2 m 1 +m 2 = = 2 ms 1 2) Transform initial velocities into CM : u = u v cm u 1 = 3 2 = 1 ms 1 ; u 2 = = 1.5 ms 1 3) Conservation of energy : v 1 = u 1 : v 2 = u 2 v 1 = 1 ms 1 ; v 2 = 1.5 ms 1 after collision 4) Transform final velocities back to Laboratory frame : v = v + v cm v 1 = = 1 ms 1 ; v 2 = = 3.5 ms 1 80

81 8.3 Relationship between speeds in CM in 2D 81 Momentum : m1 u 1 + m 2u 2 = 0 ; m 1v 1 + m 2v 2 = 0 (1) Dot products : m1 u 1 2 2u 1 u 2 ; m 2u 2 2 1u 1 u 2 m 1 v 1 2 2v 1 v 2 ; m 2v 2 2 1v 1 v 2 Energy : 1 2 m 1u m 2u 2 2 = 1 2 m 1v m 2v 2 2 Hence (m 1 + m 2 )u 1 u 2 = (m 1 + m 2 )v 1 v 2 u 1 u 2 = v 1 v 2 : magnitudes u 1 u 2 = v 1 v 2 back-to-back From (1), magnitudes u 2 = m 1 m 2 u 1 ; v 2 = m 1 m 2 v 1 Hence u 1 ( m 1 m 2 u 1 ) = v 1 ( m 1 m 2 v 1 ) u 2 1 = v 1 2 ; u 2 2 = v 2 2 v 1 = u 1 ; v 2 = u 2 Back-to-back in direction as Speeds before = speeds after shown in diagrams.

82 8.4 Lab to CM : 2-body 2D elastic collision 1) Find centre of mass velocity v CM (u1 v CM )m 1 + (u 2 v CM )m 2 = 0 vcm = m 1u 1 +m 2 u 2 m 1 +m 2 2) Transform initial Lab velocities to CM u 1 = u 1 v CM, u 2 = u 2 v CM 3) Get final CM velocities 82 v 1 = u 1 ; v 2 = u 2

83 4) Transform vectors back to the Lab frame v1 = v 1 + v CM ; v 2 = v 2 + v CM 5) Can then use trigonometry to solve 83

84 9.1.1 Example 1: Equal masses, target at rest Before Magnitude of velocities: vcm = m 1u 1 +m 2 u 2 m 1 +m 2 = u 0 2 u 1 = u 0 v CM = u 0 2 u 2 = v CM = u 0 2 v 1 = u 1 = u 0 2 v 2 = u 2 = u 0 2 After 84

85 Angles: Relationships between angles and speeds Cosine rule: ( u 0 2 ) 2 = ( u 0 2 ) 2 + v 2 1 2v 1 u 0 2 cos θ 1 v1 u 0 cos θ 1 = v1 2 cos θ1 = v 1 u 0 as before cos θ2 = v 2 u 0 Opening angle: Cosine rule: u 2 0 = v v 2 2 2v 1v 2 cos(θ 1 + θ 2 ) But u 2 0 = v1 2 + v 2 2 (conservation of energy) cos(θ1 + θ 2 ) = 0 θ 1 + θ 2 = π 2 NB: Lines joining opposite corners of rhombus cross at 90 85

86 9.1.2 Example 2: Elastic collision, m 2 = 2m 1, θ 1 = 30 Find the velocities v 1 and v 2 and the angle θ 2 Magnitude of velocities: vcm = m 1u 1 +m 2 u 2 m 1 +m 2 = u 0 3 u 1 = u 0 v CM = 2u 0 3 u 2 = v CM = u 0 3 v 1 = u 1 = 2u 0 3 v 2 = u 2 = u

87 Relationships between angles and speeds Sine rule: (sin 30/ 2u 0 3 ) = (sin α/ u 0 3 ) sin α = 1 4 α = 14.5 β = 30 + α = 44.5 sin 30/ 2u 0 3 = sin( )/v 1 v 1 = 0.93u 0 87 Cosine rule: v 2 2 = ( u 0 3 ) 2 + ( u 0 3 ) 2 2( u 0 3 ) 2 cos β v 2 = 0.25u 0 Sine rule: (sin 44.5/v 2 ) = (sin θ 2 / u 0 3 ) θ 2 = 68.0

88 9.2 Inelastic collisions in the Lab frame in 1D (u 2 = 0) An inelastic collision is where energy is lost (or there is internal excitation). Take m 2 at rest & in 1D. Momentum : m 1 u 1 = m 1 v 1 + m 2 v 2 (1) 1 Energy : 2 m 1u1 2 = 1 2 m 1v m 2v2 2 + E (2) Square Equ.(1) and subtract 2m1 Equ.(2) m 2 (m 2 m 1 )v m 1m 2 v 1 v 2 2m 1 E = 0 Substitute for m 1 v 1 from Equ.1 to get quadratic in v 2 m 2 (m 2 + m 1 )v 2 2 2m 1m 2 u 1 v 2 + 2m 1 E = 0 Solve, taking consistent solutions with elastic case ( E = 0) 88 v 2 = 2m 1m 2 u 1 + 4m 2 1 m2 2 u2 1 8m 1m 2 (m 1 +m 2 ) E 2m 2 (m 1 +m 2 ) (3) v 1 = 2m2 1 u 1 4m 2 1 m2 2 u2 1 8m 1m 2 (m 1 +m 2 ) E 2m 1 (m 1 +m 2 ) (4)

89 1D inelastic collisions viewed in the Lab frame (u 2 = 0) We see from Equ. (3) & (4) there is a limiting case: 4m 2 1 m2 2 u2 1 8m 1m 2 (m 1 + m 2 ) E 0 i.e. E m 1 m 2 u 2 1 2(m 1 +m 2 ) This corresponds to the two bodies sticking together in a single object of mass (m 1 + m 2 ) v 1 = v 2 From momentum cons. m1 u 1 = m 1 v 1 + m 2 v 2 if v 1 = v 2 = v, then v = m 1u 1 (m 1 +m 2 ) (the CM velocity) ] For equal mass m 1 = m 2 v 2, v 1 = u 1 2 [1 ± 1 4 E mu1 2 89

90 9.2.1 Coefficient of restitution General definition : e = v 2 v 1 Speed of relative separation = u 1 u 2 Speed of relative approach 90 From Equ.(3) & (4) previously v 2 v 1 = 2m 1m 2 u 1 + 4m 2 1 m2 2 u2 1 8m 1m 2 (m 1 +m 2 ) E 2m 2 (m 1 +m 2 ) 2m2 1 u 1 4m1 2m2 2 u2 1 8m 1m 2 (m 1 +m 2 ) E 2m 1 (m 1 +m 2 ) Factorizing, then simplifying, then dividing by u1 gives e = 1 2(m 1+m 2 ) E m 1 m 2 u 2 1 = 1 E T where T = 1 2 µu2 1 with µ = m 1m 2 m 1 +m 2 (the reduced mass) We see later that T is the initial energy in the CM frame, hence e is related to the fractional energy loss in this frame e = 1 completely elastic; e = 0 perfectly inelastic, in general 0 < e < 1

91 9.3 Inelastic collisions viewed in the CM frame Case of perfectly inelastic collision (e = 0) After collision, total mass (m 1 + m 2 ) is at rest in CM: KE in CM: TCM = T LAB 1 2 (m 1 + m 2 )vcm 2 Differentiate: Loss in KE TCM = T LAB (obvious) Max. energy that can be lost = TCM = 91 = 1 2 m 1u m 2u (m 1 + m 2 )v 2 CM

92 9.3.1 Kinetic energy in the CM : alternative treatment Revisit kinetic energy in the CM frame: T Lab = T CM Mv 2 CM TCM = 1 2 m 1u m 2u 2 2 x 1 = m 2 m 1 +m 2 x = m 2 M x, x 2 = m 1 M x u 1 = m 2 M ẋ, u 2 = m 1 TCM = 1 ( 2 m1 ( m 2 M )2 + m 2 ( m 1 M )2) ẋ 2 M ẋ TCM = 1 m 1 m 2 2 (m M m 1 ) ẋ 2 = 1 m 1 m 2 2 M ẋ 2 Also ẋ = ẋ2 ẋ 1 = u 2 u 1 T CM = 1 m 1 m 2 2 M ẋ 2 = 1 2 µẋ 2 = 1 2 µ(u 1 u 2 )2 = 1 2 µ(u 1 u 2 ) 2 These expressions give the CM kinetic energy in terms of the relative velocities in the CM & Lab and the reduced mass µ 92

93 9.3.2 Coefficient of restitution in the CM Initial KE in the CM : T i CM = 1 2 µ(u 1 u 2 )2 Final KE in the CM : T f CM = 1 2 µ(v 1 v 2 )2 Conservation of energy : T i CM = TCM f + E 1 2 µ(u 1 u 2 )2 = 1 2 µ(v 1 v 2 )2 + E ) 2 = 1 E ( v 1 v 2 u 1 u 2 ( v 1 v 2 u 1 u 2 ) T i CM = ± 1 E T i CM Same expression as before with T = T i CM Coefficient of restitution e = v 2 v 1 = v u 1 u 2 2 v 1 = 1 E CM u 1 u 2 LAB TCM i ONLY in CM frame can ALL the KE be used to create E For e = 0 the two particles coalesce and are at rest in CM 93

94 9.3.3 Example of inelastic process A calcium nucleus (A=20), mass m, travels with velocity u 0 in the Lab. It decays into a sulphur nucleus (A=16), mass 4 5m, and an α-particle (A=4), mass 1 5m. Energy T is released as KE in the calcium rest frame (CM). A counter in the Lab detects the sulphur nucleus at 90 to the line of travel. What is the speed and angle of the α-particle in the Lab? 94

95 Energy T is released as KE in the CM. v CM = u 0 Momentum in CM: 4 5 mv mv 2 = 0 v 2 = 4v 1 Energy: T = 1 2 ( 4 5 m)v ( 1 5 m)16v 12 = 2mv 1 2 v 1 = [ T 2m ] 1 2 v 2 = [ 8 T m ] 1 2 Transform to Lab by boosting by v CM (= u 0 ) 2 + cos α = u 0 v 1 = [ 2mu2 0 T ] 1 2 Cosine rule: v 2 2 = v 22 + u v 2 u 0 cos α Sine rule: sin θ 2 95 v 2 = sin α v 2 Solve for v 2, θ 2

96 10.1 Resisted motion and limiting speed Newton II: m dv dt = F ext + F R where F ext is the external force and F R is a resistive force If Fext = 0 and F R velocity, then v exp( αt) (see Lecture 4) If F ext 0 and e.g. F R v n then there exists a limiting speed corresponding to dv dt F R = F ext = 0 that satisfies 96

97 F R = 10.2 Air resistance av }{{} Laminar flow Laminer flow : Stoke s Law F = 6πηrv r is the radius of the sphere v is the velocity of the sphere η is the viscosity of the fluid Turbulent flow : S F = 1 2 πρc dr 2 v 2 ρ is the density of the fluid C d is the drag coefficient (e.g. for a smooth sphere C d bv 2 }{{} Turbulent flow 97

98 10.3 Example 1 : Resistive force, F R v Body fired vertically upwards under gravity air resistance velocity v = v 0 & x = 0 at t = 0 Equation of motion: m dv v dt = mg βv dv = t dt where α = β v 0 g+αv 0 m [ 1 log α e(g + αv) ] v = [ t ] t v 0 0 ( ) ( ) log (g+αv) e (g+αv 0 = αt 1+ αv = 1 + αv 0 exp( αt) ) g g v = g α [ (1 + αv 0 g ) exp( αt) 1 ] Terminal (limiting) velocity: t, vt g α Can show by expansion, as α 0, v v0 gt 98

99 Maximum height and distance travelled for F R v v = g α [ (1 + αv 0 g ) exp( αt) 1 ] At maximum height v = 0, t = tmax exp( αt max ) = (1 + αv 0 t max = 1 α log e ( Can expand log to show : t max v 0 g when α 0 g ) αv 0 g ) Distance travelled : x = t 0 = g α g α x = g α [ (1 + αv 0 g ) exp( αt) 1 ] dt [ ] t 1 α (1 + αv 0 g ) exp( αt) t 0 [( ) ] 1 α (1 + αv 0 g )(1 exp( αt) t Can show by expansion x v 0 t 1 2 gt2 when α 0 99

100 10.4 Example 2: Resistive force, F R v 2 Body falls vertically downwards under gravity with air resistance [velocity] 2, v = 0, x = 0 at t = 0 Equation of motion: m dv dt = mg βv 2 Terminal velocity when dv dt = 0 : v T = Equation of motion becomes dv dt v Integrate dv 0 g(1 v 2 /vt) = t 2 0 dt Standard integral : mg β 1 1 z 2 dz = 1 2 log e = g ( ) 1 v 2 /vt 2 ( 1+z 1 z ) [ ( )] v vt 2g log 1+v/vT e 1 v/v T = t 1+v/v T 1 v/v 0 T = exp(t/τ), where τ = v T 2g (1 v v T ) = (1 + v v T ) exp( t τ ) Velocity as a function of time: [ ] v = v 1 exp( t/τ) T 1+exp( t/τ) 100

101 Velocity as a function of distance for F R v 2 Equation of motion: dv dt = g ( 1 v 2 /vt 2 ) Write dv dt = dv dx dx dt = v dv dx v 0 v dv g(1 v 2 /vt) = x 2 0 dx [ v T 2 2g log ( e 1 v 2 /vt 2 ) ] v = x 0 ( 1 v 2 /v 2 T ) = exp ( x/xt ), where x T = v 2 T 2g v 2 = v 2 T [1 exp ( x/x T )] To get x vs. t integrate again : t 0 dt = x dx v

102 Work done on the body by the force for F R v 2 Equation of motion: m dv dt = mg βv 2 Work done: Fdx = x mg dx o }{{} Conservative x βv 2 dx 0 }{{} Dissipative Conservative term : Work done = mgx Dissipative term : Work done = x 0 βv 2 dx = x 0 βv T 2 [1 exp( x/x T )] dx = βvt 2[x + x T (exp( x/x T ) 1) ] }{{} = x T v 2 /vt 2 v 2 T = mg β x T = v 2 T 2g = βv 2 T (x v 2 /2g) = mg[x v 2 /2g] Energy dissipated = 1 2 mv 2 mgx As expected. 102