Question

state law of conservation of momentum give its formula ​

Answer 1

Answer:

For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2

Answer 2

Law of Conservation of Momentum

It 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 affected by any action, reaction only in case is no external unbalanced force is applied on the bodies.

Formula

$\huge\boxed{\bf m_A v_A+m_B v_B = m_A u_A + m_B u_B}$

Derivation of Formula

Let,

• $\sf m_A=mass\:of\:ball\:A$

• $\sf m_B =mass\:of\:ball\:B$

• $\sf u_A=initial\:velocity\:of\:ball\:A$

• $\sf u_B=initial\:velocity\:of\:ball\:B$

• $\sf v_A=velocity\:after\:the\:collision\:of\:ball\:A$

• $\sf v_B=velocity\:after\:the\:collision\:of\:ball\:B$

• $\sf F_{AB}=Force \:exerted \:by \:A \:on \:B$

• $\sf F_{BA}=Force \:exerted \:by \:B \:on \:A$

Now,

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

$\boxed{\sf Change \:in \:the \:momentum \:of \:A= m_A v_A -m_A u_A}$

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

$\boxed{\sf Rate \:of \:change \:of \:momentum \:A=\dfrac{m_A v_A-m_A u_A}{t}}$

• Force exerted by B on A = Change in momentum of A / time taken

$\boxed{\sf F_{BA}=\dfrac{m_A v_A-m_A u_A}{t}}$     -----(1)

• Rate of change of momentum B = Change in momentum of B / time taken

$\boxed{\sf Rate \:of \:change \:of \:momentum \:B=\dfrac{m_B v_B-m_B u_B}{t}}$

• Force exerted by A on B = Change in momentum of B / time taken

$\boxed{\sf F_{AB}=\dfrac{m_B v_B-m_B u_B}{t}}$     -----(2)

We know,

• By Newton's third law of motion states that every action has an equal and opposite reaction. Then, $\bf F_{AB}=-F_{BA}$

• The negative(-) sign is used to indicate that 1 object is moving in opposite direction after collision.

Using (1) and (2) ,

$\implies\sf \dfrac{m_B v_B-m_B u_B}{t} = -\dfrac{m_A v_A-m_A u_A}{t}$

$\implies\sf m_B v_B-m_B u_B=m_A v_A-m_A u_A$

$\implies\sf m_B v_B+m_A v_A=m_B u_B+m_A u_A$

And we got the final equation,

$\huge\boxed{\bf m_A v_A+m_B v_B = m_A u_A + m_B u_B}$