Newton's Third Law Of Motion

Law of conservation of linear momentum

Collision: If a number of bodies collide with one  another then total momentum of the bodies, just before collision is equal to the total momentum just after collision.
        i.e. m₁u₁+m₂u₂ = m₁v₁+m₂v₂
i)  In the absence of an external force, the linear momentum of a particle or a body remains constant i.e. if  \(\vec F=0\) then \(\vec P\) constant 
ii) In the absence of external force, the linear momenta of individual particles can change but the total linear momentum of the whole system remains constant. 
iii) The law of conservation of linear momentum is based on the Newton’s third law of motion.  This is the fundamental law of nature and there is no exception to it.
iv) This law is valid only for linear motion.
v)  Internal forces cannot change the total momentum of the system, however they may change the momentum of each particle of the  system.
vi) Motion of a rocket, firing of a bullet from a gun and explosion of a shell fired from a  cannon are some examples where we can apply the law of conservation of linear momentum.
 
Application Of The Principle Of Conservation Of Linear Momentum

1) Recoiling of a gun. When a bullet is fired from a gun, the gun recoils i.e. moves in a direction opposite to the direction of motion of the bullet. The recoil velocity of the gun can be calculated from the principle of conservation of linear momentum.
    Suppose m₁ = mass of bullet,
    m₂ = mass of gun, 
   \( \vec v_1 \) = velocity of  the bullet, 
   \(\vec v_2\) = velocity of recoil of the gun.
Before firing, the gun and the bullet both, are at rest. Therefore , total linear momentum before firing = 0. Therefore, total linear momentum before firing = 0. The vector sum of linear momenta on firing \( m_1 \vec v_1 + m_2 \vec v_2 = 0 \) . According to the principle of conservation of linear momentum, total linear momentum after firing should also be zero. 
           \( \therefore m_1 \vec v_1 + m_2 \vec v_2 = 0 \)  
    or  \( m_2 \vec v_2 = - m_1 \vec v_1 \) ............................. (25) 
    or \( \vec v_2 = - \frac{{m_1 \vec v_1 }} {{m_2 }} \) ............................... (26)
The negative sign shows that direction of \(\vec v_2\) is  opposite to the direction of  \(\vec v_1\) i.e. the gun recoils. Further, as m₂ > > m₁ therefore ,  \( \vec v_2 < < \vec v_1 \)i.e. velocity of recoil of the gun is much smaller than the velocity of the bullet . 
    From eq. (26) , \( v_2 \propto \frac{1} {{m_2 }} \)
It means that a heavier gun will recoil with a smaller velocity and vice-versa. 
Initial K.E of the system is zero, as both the gun and the bullet are at rest.
Final K.E. of the system = \( \left( {\frac{1} {2}m_1 v_1^2 + \frac{1} {2}m_2 v_2^2 } \right) > 0 \). Thus K.E of the system increases ( and is not constant). 
If P.E. is assumed to be constant, mechanical energy(=K.E = P.E) will also increase. 
        As M.E. is conserved, therefore, chemical energy of gun powder must have been converted into K.E.

While firing the gun must be held tightly to the shoulder. This would save hurting the shoulder. When the gun is held tightly, the body of the shooter and the gun behave as one body. Total mass becomes large and therefore, recoil velocity of the body and the gun becomes too small.

2) Flight of rockets and jet planes. In rockets and jet planes,the fuel is burnt in the presence of some oxidising agent in combustion chamber. The hot and highly compressed gases escape through the narrow opening (i.e., exhaust nozzle) with large velocity. As a result of it, the escaping gases acquire a large backward momentum. This in turn, imparts an equal forward momentum to the rocket in accordance with the law of conservation of linear momentum. 

3) When a man jumps out of a boat to the shore, the boat is pushed slightlyaway from the shore. The momentum of the boat is equal and opposite to that of the man in accordance with the law of conservation of linear momentum.
 
4) Explosion of bomb. When a bomb falls vertically downwards its horizontal velocity is zero and hence its horizontal momentum is zero. When bomb explodes, its pieces are scattered horizontally in different directions so that the vector sum of momenta of these pieces becomes zero in accordance with the law of conservation of linear momentum.