Question

CONSERVATION OF MOMENTUM

I WANT DERIVATION WITH AN EXAMPLE !!!!

PLS BE FAST

Answer 1

This law states that the sum of the momenta before collison is equal to the sum of momenta after collision,provided no external unbalanced force acts on it.
Let two bodies have mass m1 and m2 moving with a initial velocity u1 and u2 respectively.After collision the first body attains a final vecity v1 and the second body attains the final velocity as v2.
The momentum of the first body and second body before collision is:-
P1=m1u1 P2=m2u2
The momentum of the first body and second body after collision is:-
P1 '=m1v1 P2 '=m2v2
force exerted by the first body on the second body is
F12=P1 '-P1/t
F12=m1v1-m1u1/t
similarly
force exerted by the second body on the first body is
F21=P2 '-P2/t
F21=m2v2-m2u2/t
According to newtons third law of motion, the force exerted by the first body on the second body should be equal and opposite.
Thus,
=>F12= -F21
=>m1v1-m1u1/t= -(m2v2-m2u2)/t
=>m1v1-m1u1= -m2v2+m2u2 (by cancelling "t"on both sides)
=>m1v1+m2v2=m1u1+m2u2
=>m1u1+m2u2=m1v1+m2v2
thus,
the total momentum of the two bodies before collision is equal to the total momentum of the two bodies after the collision.
Hence,the momentum of two bodies are conserved,provided no external force acts on it...

Answer 2

Conservation of momentum :
it means that the momentum of the body remains conserved if no ext. Force applied .

lets solve this by using second law the real law of motion .

Consider an isolated system of two bodies A and B . 

Suppose the mutually interact with each other .

Let F₂₁ be the force exerted by A on B . 


Let d pB / dt  be the resulting change of momentum of B .

Let F₁₂ be the force  exerted by B on A .

Let d pA /dt  be the resulting change of momentum of A .

According to newton's second law ..

F₂₁ = $\frac{d pB}{ dt }$           and    F₁₂ =$\frac{d pA}{dt}$ 

∴  F₂₁ + F₁₂ = $\frac{d pB}{dt} + \frac{d pA}{dt}$

In the absence of the external force rate of change of momentum must be  0 .

  here now  = d( pB + pA ) /dt = 0 


∴ F₁₂ + F₂₁ = 0 


here the momentum of the body A

initially = p
final = p' 
dp ( change in momentum ) = p'1 - p1

same with body B 
initially = p2 
final = p '2 
dp = p'2 - p2

Now 
F₁₂ = -F₂₁

(p'1 - p1 ) / dt = ( p'2 - p2 )  /dt 


p'1-p1 = p'2 - p2 

m1v1 - m1u1 = m2v2 - m2u2 

m1v1+ m2v2 = m1u1 + m2u2 

here is the desired result 

the final momentum of the two bodies equal to initial momentum of the bodies ..............so the momentum is conserved  .
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