Question

Derive the mathematical formula of conservation of momentum

Answer 1

Heya!

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Law of conservation of momentum states that when two objects collide with each other , the sum of their linear momentum always remains same or we can say conserved and is not effected by any action, reaction only in case is no external unbalanced force is applied on the bodies.

DERIVING THE LAW OF CONSERVATION OF MOMENTUM-

Let,

mA = Mass of ball A
mB= Mass of ball Ba

uA= initial velocity of ball A
uB= initial velocity of ball B

vA= Velocity after collision of ball A
vB= Velocity after collision of ball B

Fab= Force exerted by A on B
Fba= Force exerted by B on A

Now,

Change in momentum of A= momentum of A after collision - momentum of A before collision

= mA vA - mA uA

Rate of change of momentum A= Change in momentum of A/ time taken

= mA vA - mA uA/t

Force exerted by B on A (Fba)=

Fba= mA vA - mA uA/t. [i]

In the same way,

Rate of change of momentum of B=

mV vB - mB uB/t

Force exerted by A on B (Fab)=

Fab= mB vB - mB uB/t. [ii]

Newton's third law of motion states that every action has an equal and opposite reaction, then,

Fab= -Fba [ ' -- ' sign is used to indicate that 1 object is moving in opposite direction after collision]

Using [i] and [ii] , we have

mB vB - mB uB/t = - (mA vA - mA uA/t)

mB vB - mB uB= - mA vA + mA uA

Finally we get,

mB vB + mA vA = mB uB + mA uA

This is the derivation of conservation of linear momentum.
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Hope it helps...!!!

Answer 2

Let us take masses of two objects mA and mB
velocity=uA and uB
Let them collide eachother
object A exerts FAB on object B
object B exerts FBA on object A
Final velocity=vA and vB
Momentum of object A is mA uA and mA v

Rate of change of momentum =mA vA-uA/t
similarly mB=uB-vB/t
According to third law of motion
FAB=FBA
(OR)
mAuA+mBuB=mAvA+mBvB

Hence we derived the formula
Hope you understood

AKSHAYA SHRUTI