Question

Define conservation of momentum and drive its relation = (m1u1+ m2u2) = m1v1+ m2v2.​

Answer 1

Answer:

Explanation:

Conservation of momentum:

➝ This states that the total momentum of an isolated system which is free from all other external forces remains constant.

➝ In an isolated system the mutual force between the interacting particles may cause momentum change in the individual quantities, but the force between each pair is equal and opposite and they cancel each other. Hence the total momentum remains unchanged.

➝ Momentum before collission = Momentum after collission

Derivation:

Based on Newton's third law:

➟ Consider two bodies A and B moving with initial momentums $\sf P_A$ and $\sf P_B$.

➟  Let the final momentum after collission be $\sf{P_{A^{1} }}$ and $\sf{P_{B^{1} }}$

➟ Let their masses be m₁ and m₂ and initial velocities be u₁ and u₂ respectively.

➟ Let their final velocities be v₁ and v₂.

➟ Let us assume that they interact within a short interval of time. That is an impulse is produced.

➟ But we know that,

    Impulse = Change in momentum

    Impulse = Force × Time

➟ Hence,

    $\sf{I_A = F_{AB}\times t = P_{A^{1} }-P_A}$

    $\sf{I_B=F_{BA}\times t=P_{B^{1} }-P_B}$

➟ By Newton's third law we know that,

    $\sf F_{AB}=-F_{BA}$

➟ Hence,

    $\sf F_{AB}\times t = -F_{BA}\times t$

    $\sf {P_{A^{1} }-P_A=-(P_{B^{1} }-P_B)}$

    $\sf {P_{A^{1} }-P_A=-P_{B^{1} }+P_B}$

    $\sf {P_{A^{1} }+P_{B^{1} }=P_A+P_B}$

➟ But we know that,

    Momentum = Mass × Velocity

➟ Therefore,

    m₁v₁ + m₂v₂ = m₁u₁ + m₂v₂

➟ Hence proved.

Based on Newton's second law:

➟ By Newton's second law we know that rate of change of momentum is directly proportional to the force applied.

     $\sf \dfrac{\Delta P}{\Delta t} =F$

➟ But we know that the system is isolated, there are no external forces acting on it.

➟ Hence,

    F = 0

    $\sf \dfrac{\Delta P}{\Delta t} =0$

    Δ P = 0

➟ Momentum change = 0

➟ There is no change in momentum. That is P is a constant.