Question

Derivation of law of conservation of momentium

Answer 1

See the answer below Mate:-

#Law of conservation of momentum:-

$*$law of conservation of Linear momentum is a extremely important consequence of Newton's third law of motion combination with the second law of motion.

$*$According to conservation of momentum:-When two or more bodies acts upon each other their total momentum remains constant provided no external forces are acting.

$*$So a momentum can't be created or destroyed.

$*$When this law is applied for a collision between two bodies, the total momentum of the colliding bodies before collision is equal to the Total Momentum After collision.

$*$We can apply this law for the collision between 2 vehicles. This type of law is Applicable for all types of collision.

$*$Consider two particles say A and B of mass $M_1$ and $M_2$collide with each other and forces acting on these particles are only the ones they exert on each other.

$*$Let $U_1$ and $V_1$ be the initial and final velocities of particle A and similarly, $U_2$and $V_2$ for particle B. Let the two particles be in contact for a time t. So, Change in momentum of A=m1 (v1-u1) Change in momentum of B=m2 (v2-u2)

$*$During the collision, let A impart an average force equal to $F_B_A$ on B and let B exert an average $F_A_B$ on A. We know that from third law of motion $F_B_A$=$-F_A_B$         ............(4)

Here,

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