Page 113 - Class-11-Physics-Part-1_Neat

LAWS OF MOTION 99

The total momentum of an isolated system
of interacting particles is conserved.
An important example of the application of the
law of conservation of momentum is the collision
of two bodies. Consider two bodies A and B, with
initial momenta p and p . The bodies collide,
A B
get apart, with final momenta p′ and p′
A B
respectively. By the Second Law
Fig. 5.7 Equilibrium under concurrent forces.
F t ∆ = p′ − p and In other words, the resultant of any two forces
AB A A

say F and F , obtained by the parallelogram
1
2
F t ∆ = p′ − p
BA B B law of forces must be equal and opposite to the
(where we have taken a common interval of time third force, F . As seen in Fig. 5.7, the three
3
forces in equilibrium can be represented by the
for both forces i.e. the time for which the two
sides of a triangle with the vector arrows taken
bodies are in contact.)
in the same sense. The result can be
Since F = − F by the third law, generalised to any number of forces. A particle
AB BA
is in equilibrium under the action of forces F ,
1
p ′ − p A = ( − p ′ − p B ) F ,... F if they can be represented by the sides
A
B
n
2
of a closed n-sided polygon with arrows directed
i.e. p ′ + ′ =p B p A + p B (5.9) in the same sense.
A
which shows that the total final momentum of Equation (5.11) implies that
the isolated system equals its initial momentum.
F + F + F = 0
Notice that this is true whether the collision is 1x 2x 3x
F + F + F = 0
elastic or inelastic. In elastic collisions, there is 1y 2y 3y
a second condition that the total initial kinetic F + F + F = 0 (5.12)
2z
3z
1z
energy of the system equals the total final kinetic where F , F and F are the components of F
1x 1y 1z 1
energy (See Chapter 6). along x, y and z directions respectively.
Example 5.6 See Fig. 5.8. A mass of 6 kg
t
5.8 EQUILIBRIUM OF A PARTICLE is suspended by a rope of length 2 m
from the ceiling. A force of 50 N in the
Equilibrium of a particle in mechanics refers to horizontal direction is applied at the mid-
the situation when the net external force on the point P of the rope, as shown. What is the
particle is zero.* According to the first law, this angle the rope makes with the vertical in
-2
means that, the particle is either at rest or in equilibrium ? (Take g = 10 m s ). Neglect
uniform motion. the mass of the rope.
If two forces F and F , act on a particle,
1 2
equilibrium requires
F = − F (5.10)
1 2
i.e. the two forces on the particle must be equal
and opposite. Equilibrium under three
concurrent forces F , F and F requires that
1 2 3
the vector sum of the three forces is zero.
F + F + F = 0 (5.11)
1 2 3 (a) (b) (c)
Fig. 5.8
* Equilibrium of a body requires not only translational equilibrium (zero net external force) but also rotational
equilibrium (zero net external torque), as we shall see in Chapter 7.

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