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100 PHYSICS problems without explicitly outlining and where F is the force on the particle. Let us justifying the procedure. We now realise that in consider a system of n particles with masses m 1 earlier studies we assumed, without saying so, m ,...m respectively and velocitiesv ,v ,.......v that rotational motion and/or internal motion 2 n 1 2 n respectively. The particles may be interacting of the particles were either absent or negligible. and have external forces acting on them. The We no longer need to do this. We have not only found the justification of the procedure we linear momentum of the first particle is m 1 1 , followed earlier; but we also have found how to of the second particle ism 2 2 and so on. describe and separate the translational motion For the system of n particles, the linear of (1) a rigid body which may be rotating as well, or (2) a system of particles with all kinds momentum of the system is defined to be the of internal motion. vector sum of all individual particles of the system, P = p1+ p 2...+ p n =m v1+1 v +2.2+m v n n (6.14) Comparing this with Eq. (6.8) P = M V (6.15) Thus, the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass. Differentiating Eq. (6.15) with respect to time, d P d V Fig. 6.12 The centre of mass of the fragments = M = MA (6.16) of the projectile continues along the dt dt same parabolic path which it would Comparing Eq.(6.16) and Eq. (6.11), have followed if there were no d P = F explosion. dt ext (6.17) Figure 6.12 is a good illustration of Eq. (6.11). Thisisthestatementof Newton’s second law A projectile, following the usual parabolic of motion extended to a system of particles. trajectory, explodes into fragments midway in Suppose now, that the sum of external air. The forces leading to the explosion are forces acting on a system of particles is zero. internal forces. They contribute nothing to the Then from Eq.(6.17) motion of the centre of mass. The total external force, namely, the force of gravity acting on the dP = 0 or P = Constant (6.18a) body, is the same before and after the explosion. dt Thus, when the total external force acting The centre of mass under the influence of the external force continues, therefore, along the on a system of particles is zero, the total linear same parabolic trajectory as it would have momentum of the system is constant. This is the law of conservation of the total linear followed if there were no explosion. momentum of a system of particles. Because of 6.4 LINEAR MOMENTUM OF A SYSTEM OF Eq. (6.15), this also means that when the PARTICLES total external force on the system is zero the velocity of the centre of mass remains Let us recall that the linear momentum of a constant. (We assume throughout the particle is defined as p = m v (6.12) discussion on systems of particles in this chapter that the total mass of the system Let us also recall that Newton’s second law remains constant.) written in symbolic form for a single particle is Note that on account of the internal forces, dp i.e. the forces exerted by the particles on one F = dt (6.13) another, the individual particles may have Reprint 2026-27