When a nucleus at rest emits a beta particle, it is found that the velocities of the recoiling nucleus and th beta particle ae not along the same straight line. How can this be possible in view of the principle of conservation of momemtum?

Radioactive nuclei emit beta^- particles. Electrons exist inside the nucleus.

which a U^(238) nucleus original at rest , decay by emitting an alpha particle having a speed u , the recoil speed of the residual nucleus is

The figure shows the velocity time graph of the particle which moves along a straight line starting with velocity at 5 m//sec and coming to rest at t=30 s . Then :-

Assertion: When a particle moves with constant velocity its average velocity, its insttantaneous velocity and its speed are all equal in magnitude. Reason: If the average velocity of a particle moving on a straight line is zero in a time interval, then it is possible that the instantaneous velocity is never zero in the internal.

A continuous stream of particles, of mass m and velocity r , is emitted from a source at a rate of n per second. The particles travel along a straight line, collide with a body of mass M and get embedded in the body. If the mass M was originally at rest, its velocity when it has received N particles will be

A body starts from rest and travels a distance x with uniform acceleration, then it travels a distance 2x with uniform speed, finally it travels a distance 3x with uniform retardation and comes to rest. If the complete motion of the particle is along a straight line, then the ratio of its average velocity to maximum velocity is

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.)

Two particle are oscillating along two close parallel straight lines side by side, with the same frequency and amplitude. They pass each other, moving on opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is.........

Two particles each of mass m are connected by a light inextensible string and a particle of mass M is attached to the midpoint of the string. The system is at rest on a smooth horizontal table with string just taut and in a straight line. The particle M is given a velocity v along the table perpendicular to the string. Prove that when the two particles are about to collide : (i) The velocity of M is (Mv)/((M + 2m)) (ii) The speed of each of the other particles is ((sqrt2M (M + m))/((M + 2m)))v .