Conservation of Linear Momentum : If a force F is acting on particle of mass m, then according to Newton s second law of motion, we have F = dp /dt =

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2 Conservation of Linear Momentum : If a force F is acting on particle of mass m, then according to Newton s second law of motion, we have F = dp /dt = d (mv) /dt where p =mv is linear momentum of particle If external force acting on the particle, is zero,then dp/ dt = d ( mv) /dt = 0 p = mv = constant Thus in absence of external force, the linear momentum of particle is conserved. This is the conservation theorem for free particle.

3 Conservation of Angular momentum : The angular momentum of particle P of mass m about O is defined as J = r p ---- (a) T = r F if we differentiate (a) w.r.t t, then dj / dt = d (r p) = dr/ dt p + r dp/dt

4 SCATTERING OF PARTICLES: ELASTIC AND INELASTIC COLLISION. In elastic scattering total kinetic energy of all particles before scattering is equal to that after scattering, The internal structure of particle does not change and we can apply the law of conservation of energy without taking into consideration the internal energies at all. But most of the collision that occur in nature are inelastic. When two bodies collide, some amount of heat and sound energy is produced. Thus some part of the initial kinetic energy is lost. This loss must be taken into account while using the law of conservation of energy.

5 Laboratory and center of mass systems. Collision between two particles. Initial and final momenta k and p Initial and final energies k and T M = m 1 + m 2 = total mass V cm = velocity of C.M. u and v initial and final velocities. According to conservation law of linear momentum and kinetic energy we have, k 1 + k 2 = p 1 + p 2 k 1 + k 2 = T 1 + T 2 Equation (a) can be written as, (a) k 12 / 2m 1 + k 2 2 / 2m 2 = p 12 / 2m 1 + p 2 2 / 2m 2

6 Consider elastic collision in the lab and C.M. systems. In lab systems as in fig ( 3.2 ) linear moment p of the center of mass of the two bodies is not zero. Whereas in C.M. systems it is equal to zero. Final state of the scattered particles in the lab system is shown in fig. (3.3) p = M V cm = (m 1 +m 2 ) V cm In C.M. systems the position vectors of the particles are denoted by r 1 ` and r 2 ` respectively as in fig (3.3) then separation between particles is given by r = r 1 ` - r 2 ` = r 1 - r 2 r 1 - r 1 ` = r 2 - r 2 ` = R r 1 - r 1 ` = r 2 ` - r 2 ` = R

7 PHASE SPACE In Hamiltonian formulation of mechanics, we use momenta and position co-ordinates, i.e. we have a plot of p versus q. As a generalization if there are n generalized co-ordinate, we have n generalized momenta and we have 2n dimensional space. Thus we define 2n dimensional Cartesian space formed of co-ordinates q and p and known as phase space. The specification of point on the path in phase space provides six initial values for each particle. Such specification defines the state of the system completely. The equation of motion describe the motion of the system in this new formulation and involve the basis of (q,p,t). Specification of initial values of q and p at any instant on the path. There is only one possible path in phase space.

8 In our three dimensional space, a point is specified by three coordinates x, y, z or r, θ, Ψ. If we consider a gas consisting of N molecules, 3N co-ordinates will be needed to specify the position of all the molecules at any instant. If we are interested in energy of this system we require the momentum of each molecule which again have three independent components. Thus for complete specification of each molecule six co-ordinates viz. x, y, z, p x, p y, p z are required. For the entire system we shall need 6N co-ordinates.

9 To illustrate the concept of phase space, consider equation of one dimensional harmonic oscillator. E = P 2 / 2m + kx 2 / 2 where k= mω 2 1 = P 2 / 2mE + mω 2 x 2 / 2E 1 = P 2 / 2m E+ {mω 2 x 2 / [ 2E /mω 2 ] } This describe an ellipse in two dimensions one axis being x and other as p. Semi major axis = (2E /mω 2 )½ Semi minor axis = (2mE)½ The area of ellipse = 2E П / ω

10 ELASTIC SCATTERING substitute in (3.4) m 1 r 1 + m 2 r 1 - m 2 r = 0 r 1 = {m 2 / ( m 1 + m 2 )}. r µ = (m 1 * m 2 ) / ( m 1 + m 2 ) µ / m 1 = ( m 2 ) / ( m 1 + m 2 ) r 1 = (µ / m 1 ) r & r 2 = -(µ / m 2 ) r Differentiate above equations w.r.t. t we get velocities in C.M. system v 1 = (µ / m 1 ) v & v 2 = -(µ / m 2 ) v v = v 1 - v 2 = u u --- is relative velocity of first particle w.r.t second particle. m 1 v 1 = µ v = - m 2 v 2 Thus the linear momenta of particles are equal and opposite both in the initial as well as in the final state.

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12 Relations between different quantities in the Lab and C.M. systems. ( m 1 + m 2 ) V cm = (m 1 v 1 + m 2 v 2 ) V cm = (m 1 v 1 + m 2 v 2 ) / ( m 1 + m 2 ) In C.M. frame, this velocity is superposed on m 1 and m 2 u 2 = velocity of m 2 in C.M. = - V cm we have r 1 = r 1 - R v 1 = v 1 - R We have u 1 = u 1 - V cm u 1 = u 1 + V cm v 1 is the final velocity of the particle in Lab system and make scattering angle θ with the axis. Hence also, v 1 = v 1 + R V cm has same direction as that of incident velocity and it does not change during the process of collision.

13 we have to obtain the relation between the scattering angles in Lab and C.M. frame as tan θ = {(v 1 sin θ ) / (V cm + v 1 cos θ ) } tan θ = CD/OD = (v 1 sin θ ) / (OB + BD) tan θ = {(v 1 sin θ ) / (V cm + v 1 cos θ ) } tan θ = {( sin θ ) / (V cm / v 1 + cos θ ) } (b) since the system is conservative, the relative velocity after scattering i.e. when two particles are no longer in each others potentials must have the same magnitude as the initial velocity. r 1 = (µ / m 1 ) r & v 1 = (µ / m 1 ) r and ( m 1 + m 2 ) R = (m 1 u 1 )

14 V cm = (m 1 u 1 ) / ( m 1 + m 2 ) V cm = ( µ u 1 ) / ( m 2 ) V cm / v 1 = (m 1 ) / ( m 2 ) (c) from (b) & ( c ) tan θ = {( sin θ ) / ((m 1 ) / ( m 2 ) + cos θ ) } Thus scattering angle in Lab and C.M. Are related only through ratios of masses.

15 case 1 : If m 2 >> m 1 ( proton electron scattering ) tan θ = {( sin θ ) / ( cos θ ) } tan θ = tan θ θ = θ case 2: If m 2 = m 1 ( proton neutron scattering ) tan θ = {( sin θ ) / (1 + cos θ ) } tan θ = tan (θ /2) θ = (θ /2) Case 3 : If m 1 >> m 2 tan θ ~ 0 θ ~ 0 There is mostly scattering in forward direction.

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