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Prove that in inelastic collision in one-dimension, always energexpression for the loss of energ |
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Answer» Let Two bodies m₁ and m₂ moves with speed v₁ and v₂ , collide after some interval of time and new bodies of mass (m₁ + m₂) form and it moves with v velocity as shown in figure. now, There is no external force act on Bodies so, linear momentum of bodies are conserved . so, initial momentum = final momentum m₁v₁ + m₂v₂ = (m₁ + m₂)v ----(1) Now, change in kinetic energy = final kinetic energy - intial kinetic energy = 1/2(m₁ + m₂)v² - 1/2m₁v₁² - 1/2m₂v₂² Putting equation (1)= 1/2(m₁ + m₂)[(m₁v₁ + m₂v₂)/(m₁ + m₂)]² - 1/2m₁v₁² - 1/2m₂v₂² = 1/2[m₁v₁ + m₂v₂]²/(m₁ + m₂) - 1/2m₁v₁² - 1/2m₂v₂² = 1/2[ (m₁²v₁² + m₂²v₂² + 2m₁m₂v₁v₂ - m₁²v₁² - m₁m₂v₁² - m₂²v₂² - m₁m₂v₂²)]/(m₁+m₂)= 1/2[2m₁m₂v₁v₂ - m₁m₂v₁² - m₁m₂v₂² ]/(m₁ + m₂) = -1/2{m₁m₂/(m₁+m₂)}(v₁²+v₂²-2v₁v₂) = -1/2μ (v₁ - v₂)² where μ = m₁m₂/(m₁ + m₂) here negative sign shows that kinetic energy is lost . Ty😘 |
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