Use the definition of linear pometum form the previous qestion. Can we state the principle of conservation of linear pomentum for a single particle?

Write the law of conservation of total linear momentum for the system of particle.

Law of conservation of linear momentum is derived by using Newton’s which laws ?

Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass : (a) Show p=p_(i).+m_(i)V where p_(i) is the momentum of the ith particle (of mass m_(i) ) and p_(i).=m_(i)v_(i). . Note v_(i). is the velocity of the i^(th) particle relative to the centre of mass Also, prove using the definition of the centre of mass Sigmap_(i).=0 (b) Show K=K.+(1)/(2)MV^(2) where K is the total kinetic energy of the system of particles, K. is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and (1)/(2)MV^(2) is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14). (c ) Show vecL=vecL.+vecRxxvec(MV) where vecL.=Sigmavec(r_(i)).xxvec(p_(i)). is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember vec(r._(i))=vec(r_(i))-vecR , rest of the notation is the velcities taken relative to the centre of mass. Remember vec(r._(i))=vec(r_(i))-vecR rest of the notation is the standard notation used in the chapter. Note vecL , and vec(MR)xxvecV can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles. (d) Show vec(dL.)/(dt)=sumvec(r_(i).)xxvec(dp.)/(dt) Further, show that vec(dL.)/(dt)=tau._(ext) where tau._(ext) is the sum of all external torques acting on the system about the centre of mass. (Hint : Use the definition of centre of mass and Newton.s Thrid Law. Assume the internal forces between any two particles act along the line joining the particles.)

When a particle is restricted to move aong x axis between x =0 and x = a , where a is of nanometer dimension. Its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a . The wavelength of this standing wave is realated to the linear momentum p of the particle according to the de Breogile relation. The energy of the particl e of mass m is reelated to its linear momentum as E = (p^(2))/(2m) . Thus, the energy of the particle can be denoted by a quantum number 'n' taking values 1,2,3,"......." ( n=1 , called the ground state) corresponding to the number of loop in the standing wave. Use the model decribed above to answer the following three questions for a particle moving in the line x = 0 to x =a . Take h = 6.6 xx 10^(-34) J s and e = 1.6 xx 10^(-19) C . If the mass of the particle is m = 1.0 xx 10^(-30) kg and a = 6.6 nm , the energy of the particle in its ground state is closet to

When a particle is restricted to move aong x axis between x =0 and x = a , where a is of nanometer dimension. Its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a . The wavelength of this standing wave is realated to the linear momentum p of the particle according to the de Breogile relation. The energy of the particl e of mass m is reelated to its linear momentum as E = (p^(2))/(2m) . Thus, the energy of the particle can be denoted by a quantum number 'n' taking values 1,2,3,"......." ( n=1 , called the ground state) corresponding to the number of loop in the standing wave. Use the model decribed above to answer the following three questions for a particle moving in the line x = 0 to x =a . Take h = 6.6 xx 10^(-34) J s and e = 1.6 xx 10^(-19) C . The speed of the particle, that can take disrete values, is proportional to

When a particle is restricted to move aong x axis between x =0 and x = a , where a is of nanometer dimension. Its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a . The wavelength of this standing wave is realated to the linear momentum p of the particle according to the de Breogile relation. The energy of the particl e of mass m is reelated to its linear momentum as E = (p^(2))/(2m) . Thus, the energy of the particle can be denoted by a quantum number 'n' taking values 1,2,3,"......." ( n=1 , called the ground state) corresponding to the number of loop in the standing wave. Use the model decribed above to answer the following three questions for a particle moving in the line x = 0 to x =a . Take h = 6.6 xx 10^(-34) J s and e = 1.6 xx 10^(-19) C . The allowed energy for the particle for a particular value of n is proportional to

An accelration produces a narrow beam of protons, each having an initial speed of v_(0) . The beam is directed towards an initially uncharges distant metal sphere of radius R and centered at point O. The initial path of the beam is parallel to the axis of the sphere at a distance of (R//2) from the axis, as indicated in the diagram. The protons in the beam that collide with the sphere will cause it to becomes charged. The subsequentpotential field at the accelerator due to the sphere can be neglected. The angular momentum of a particle is defined in a similar way to the moment of a force. It is defined as the moment of its linear momentum, linear replacing the force. We may assume the angular momentum of a proton about point O to be conserved. Assume the mass of the proton as m_(P) and the charge on it as e. Given that the potential of the sphere increases with time and eventually reaches a constant velue. The limiting electric potential of the sphere is

An accelration produces a narrow beam of protons, each having an initial speed of v_(0) . The beam is directed towards an initially uncharges distant metal sphere of radius R and centered at point O. The initial path of the beam is parallel to the axis of the sphere at a distance of (R//2) from the axis, as indicated in the diagram. The protons in the beam that collide with the sphere will cause it to becomes charged. The subsequentpotential field at the accelerator due to the sphere can be neglected. The angular momentum of a particle is defined in a similar way to the moment of a force. It is defined as the moment of its linear momentum, linear replacing the force. We may assume the angular momentum of a proton about point O to be conserved. Assume the mass of the proton as m_(P) and the charge on it as e. Given that the potential of the sphere increases with time and eventually reaches a constant velue. If the initial kinetic energy of a proton is 2.56 ke V , then the final potential of the sphere is

An accelration produces a narrow beam of protons, each having an initial speed of v_(0) . The beam is directed towards an initially uncharges distant metal sphere of radius R and centered at point O. The initial path of the beam is parallel to the axis of the sphere at a distance of (R//2) from the axis, as indicated in the diagram. The protons in the beam that collide with the sphere will cause it to becomes charged. The subsequentpotential field at the accelerator due to the sphere can be neglected. The angular momentum of a particle is defined in a similar way to the moment of a force. It is defined as the moment of its linear momentum, linear replacing the force. We may assume the angular momentum of a proton about point O to be conserved. Assume the mass of the proton as m_(P) and the charge on it as e. Given that the potential of the sphere increases with time and eventually reaches a constant velue. After a long time, when the potential of the sphere reaches a constant value, the trajectory of proton is correctly sketched as

An accelration produces a narrow beam of protons, each having an initial speed of v_(0) . The beam is directed towards an initially uncharges distant metal sphere of radius R and centered at point O. The initial path of the beam is parallel to the axis of the sphere at a distance of (R//2) from the axis, as indicated in the diagram. The protons in the beam that collide with the sphere will cause it to becomes charged. The subsequentpotential field at the accelerator due to the sphere can be neglected. The angular momentum of a particle is defined in a similar way to the moment of a force. It is defined as the moment of its linear momentum, linear replacing the force. We may assume the angular momentum of a proton about point O to be conserved. Assume the mass of the proton as m_(P) and the charge on it as e. Given that the potential of the sphere increases with time and eventually reaches a constant velue. One the potential of the sphere has reached its final, constant value, the minimum speed v of a proton along its trajectory path is given by