Question

Derive law of conservation of momentum

Answer 1

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.

Answer 2

Law of conservation of momentum

It states that

If there is no external force applied to a system then momentum remains conserved after collision .

We knowbthat
F=ma
a=v/t

F=mv/t

P=mv
F=P/t
According to law of conservation of momentum force is conserved or net foece = 0
0 = P/t
0×t =P
P=0
P is final momentum _ initial momentum
Final momentum _ initial momentum =0
Final momentum = initial momentum