Derive the equation for the conservation of momentum

Law of conservation of momentum is more basic than third law of motion. However here we derive the law of conservation of momentum from the third law of motion.  

Let us suppose a body of mass m1 travelling with velocity u1 collides with another body of mass m2 and travelling with velocity u2. After collision let the first body changes its velocity to v1 and the second body to v2.

Since there is change in velocity of the either bodies there must be change in momentum and force experienced by both of them due to each other.

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science mam ne jo question diya meritnation mein paste kar diya........!! wah wah deenanath chauhan!...............vase yeh answer muje bhi nahi aata :P :P

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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 attainthe 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.

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its very complicated plz explain in some easy way it dosent mention if it is long

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