Page 163 - Class-11-Physics-Part-1_Neat

SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 149

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 velocities v v 2 ,.......v n
,
1
that rotational motion and/or internal motion 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 v ,
1 1
followed earlier; but we also have found how to
of the second particle is m v and so on.
2
2
describe and separate the translational motion
For the system of n particles, the linear
of (1) a rigid body which may be rotating as
momentum of the system is defined to be the
well, or (2) a system of particles with all kinds
vector sum of all individual particles of the
of internal motion.
system,
P = p + p + ... + p
1 2 n
+
= m v + m v + ... m v (7.14)
1 1 2 2 n n
Comparing this with Eq. (7.8)
P = M V (7.15)
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. (7.15)
with respect to time,
dP dV
Fig. 7.12 The centre of mass of the fragments dt = M dt = MA (7.16)
of the projectile continues along the
same parabolic path which it would Comparing Eq.(7.16) and Eq. (7.11),
have followed if there were no dP
explosion. = F ext (7.17)
dt
Figure 7.12 is a good illustration of Eq. This is the statement of Newton’s second law
(7.11). 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.(7.17)
motion of the centre of mass. The total external
force, namely, the force of gravity acting on the dP = 0 or P = Constant (7.18a)
body, is the same before and after the explosion. dt
The centre of mass under the influence of the Thus, when the total external force acting
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
followed if there were no explosion. the law of conservation of the total linear
momentum of a system of particles. Because of
7.4 LINEAR MOMENTUM OF A SYSTEM OF Eq. (7.15), 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
particle is defined as constant. (We assume throughout the
discussion on systems of particles in this
p = m v (7.12)
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 = (7.13) another, the individual particles may have
dt

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