LINEAR MOMENTUM AND ITS CONSERVATION PRINCIPLE:
Linear momentum is defined as 
For a system of n particles =
If m is constant,
Suppose 
=zero = 
= constant. This gives the principle of momentum conservation.
“The linear momentum of a system remains constant (both magnitude and direction) if external forces acting on the system adds up to zero”.
Illustration:
A man of mass 5kg climbs a rope of length 10m suspended below a balloon of mass 10kg. The balloon is stationary with respect to ground. If the man begins to climb up the rope at a speed 2m/s (relative to rope) in upward direction, what is the speed with which the balloon moves w.r.t ground?
Solution:
Given 
= 2 m/s (upward). Now take the balloon and man as one system.
If center of mass does not move = 
(Downwards 2/3 m/s).
Dumb Question:
<!--[if !supportLists]--> 1) Why center of mass does not move here?
Ans: Since initially the balloon + Man system was stationary. So even if both move as no extra external force is acting (except gravity) over the center of mass. It continues to remain in its position. Hence center of mass does not move.
Illustration:
A gun of mass m1 fires a bullet of mass hm1 with speed V0 relative to the barrel of gun. Which is inclined at an angle of a0 with horizontal. The gun is placed on smooth horizontal surface. Find racial speed of gun?
Solution:
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So velocity components for bullet are:
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| VrsinA |
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For bullet 
(-V)
As no external force is acting over the system barrel + bullet. Hence we can apply principle of momentum conservation [in horizontal direction].
Dumb Question:
<!--[if !supportLists]--> 1) What is the speed of bullet w.r.t ground?
Ans: 
Hence the speed as seen from ground will be addition of both speeds (barrel speed + its speed w.r.t barrel).
<!--[if !supportLists]--> 2) What happens to vertical direction momentum?
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Ans: The vertical direction momentum = hmVrsina is transferred to ground. The normal reaction N gives an impulse in normal direction to take care of this momentum.
Vrsin a
