Conservation of Linear Momentum

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The conservation law for linear momentum is slight modification or change in Newtons First Law of momentum. In an isolated system it says, “the sum of the linear momentum of all the bodies does not change. It remains conserved for the system.” The isolated system is one that is unaffected from its surroundings.
 
For example, two cars approaching, a canon-bullet system, Passengers in a vehicle. Vehicle may be anything; it may be a bus, a train, a bicycle or anything.
 
Momentum tells “how much motion the body posses at any instant of time”.
 
P = M V    (P is momentum, M is mass and V is velocity)  
 
If I take the derivative of above equation then I get,
 
dp/dt = M dv/dt  and  
dv/dt =   acceleration produced in the body.
 
The term dp/dt tell the variation in momentum w.r.t. time and this is called the derivative of momentum wrt time.
 
But for the isolated system the total change in momentum is zero.
 
dp/dt =0  so   a  = 0
 
This would mean that either the body is at rest or the body would be moving with uniform velocity.
 
Example 1:  A golf ball of mass 50 g at rest is hit by striker if the ball stops at a distance 50m from the origin with uniform retardation of 4 m/s2. Calculate impulse??
 
Solution: By applying the relation  v2  -  u2  = 2as 
 
Let just after the hitting the velocity of the bullet is u and the final velocity is  v which is zero.     
a  =  u2/2s  ,      u  =  20 m/s  
Impulse   =  change in momentum  =  mv  -  mu  =  - 1.2 Ns
 
 
Example 2:  A machine gun has a mass of 20 g. It fires 25 g bullets at the rate of 600 bullets per minute with a speed of 200m/s. Calculate:
 
1. Recoil velocity of gun   
2. Force required to keep the gun in position.
 
Solution 1: Before firing the total momentum is zero for the system, so
           
0   =   m1v1 + m2v2    (where m1,  v1 and m2, v2 are the mass and velocity of bullet and cannon respectively.)
 
So,   V2   = -m1v1/m2     (here m1  =    25 g   and m2   =    20 kg  =  200 m/s)
 
On substituting values,                     
 
V2  =   2.5 m/s
 
Solution 2:  Force required to keep the gun in position   =    total change in momentum of bullets
 
                                                                                =d/dt x nmv
 
(Where ‘ nmv’ represents the total momentum of the bullets)
 
  =      600 x 0.025 x 200/ 60
 
   =     50 N   

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