Question

derive anexpression for the law of conservation of momentum​

Answer 1

Law of conservation of momentum :

Let us consider object A of mass m₁ with a velocity of u₁ and object B of mass m₂ with a velocity of u₂. After collision u₁ changes into v₁ and u₂ changes into v₂.

➞ Momentum of object A before collision = m₁ × u₁

➞ Momentum of object A before collision = m₁u₁

➞ Momentum of object B before collision = m₂ × u₂

➞ Momentum of object B before collision = m₂u₂

➞ Total momentum before collision = m₁u₁ + m₂u₂ .......i)

➞ Momentum of object A after collision = m₁ × v₁

➞ Momentum of object A after collision = m₁v₁

➞ Momentum of object B after collision = m₂ × v₂

➞ Momentum of object B after collision = m₂v₂

➞ Total momentum before collision = m₁v₁ + m₂v₂ .......ii)

According to Newtons third law of motion,

★ To every action there is an equal and opposite reaction.

➞ F₁ = - F₂

➞ m₁(v₁ - u₁)/t = - m₂(v₂ - u₂)/t

➞ m₁(v₁ - u₁) = - m₂(v₂ - u₂)

➞ m₁v₁ - m₁u₁ = - m₂v₂ + m₂u₂

➞ m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂

➞ m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

∴ Total momentum before collision = Total momentum after collision.

°.° Hence, Proved.

Note : Refer to the diagram in the attachment.