Question

Law of Conservation of momentum.​

Answer 1

Definition of conservation of momentum

: a principle in physics: the total linear momentum of a system of particles not acted upon by external forces is constant in magnitude and direction irrespective of any reactions among the parts of the system.

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Answer 2

What is momentum ?

• The linear momentum of a body is defined as product of mass and velocity.

$Momentum = Mass \times velocity$

$\longrightarrow \sf \overrightarrow{p } = \overrightarrow{m } \times \overrightarrow{v }$

The conservation of momentum says that "For an isolated system (where there is no net external force), the total momentum of the system remains conserved."

This boils down to whenever one body gains momentum, then some other body must lose equal momentum.

Proving law of conservation of motion :-

• Suppose two balls A and B are moving in
• the same direction along a straight line
• with different velocities. And there are no other external
• unbalanced forces acting on them.

Let uA > uB and assuming that the two balls collide with each other which lasts for time t.

• During collision, the ball A exerts a force on ball B and the ball B exerts a force on ball A.

• For ball A, Mass = mA

• Before collision initial velocity = uA

• After collision final velocity = vA

• Initial momentum = mA uA

• Final momentum = mA vA

$\sf Rate \: of \: change \: of \: momentum = \dfrac{change \: in \: momentum}{time \: taken}$

$\implies \sf \: \dfrac{mA \: vA - mA \: uA}{?} = fAB$

For ball B

, Mass = mB

Before collision initial velocity = uB

After collision final velocity = vB

Initial momentum = mB uB

Final momentum = mB vB

$\sf Rate \: of \: change \: of \: momentum = \dfrac{Change \: in \: momentum }{Time \: taken}$

$\implies \sf \: \dfrac{mB \: vB– m \: uB}{t} = FBA$

• According to Newton’s 3rd law of motion
• FAB = - FBA

$\sf \implies \dfrac{mA \: \: vA \: - \: mA \: \: uA}{t} = \dfrac{mB \: \: vB \: - \: mV \: \: uB }{t}$

$\implies mA \: \: vA \: - \: mA \: \: uA = - (mB \: \: vB \: - \: mB \ \: uB )$

$\implies \: mA \: \: vA - mA \: \: uA = - mB \: \: vB + mB \: \: uB$

$\implies \: mA \: \: vA + mB \: \: vB = mA \: \: uA + mB \: \: uB$

$\implies \: mA \: \: uA + mB \: \: uB = mA \: \: vA + mB \: \: vB$

Hence, proved Total momentum of the two balls after collision = Total momentum of the two balls before collision