Answer on Question #47714-Physics-Mechanics-Kinematics-Dynamics
By reference to Newton's laws of motion, deduce that when two particles collide, momentum is conserved.
Answer
From Newton's Third law – when objects are in contact, the forces exerted by the objects on each other are equal and opposite:
F A B → = − F B A → . \overrightarrow{F_{AB}} = - \overrightarrow{F_{BA}}. F A B = − F B A .
From Newton's Second law (collision time is the same) - impulses are equal and opposite:
F A B → Δ t = m A Δ v A → ; F B A → Δ t = m B Δ v B → \overrightarrow{F_{AB}}\Delta t = m_A \overrightarrow{\Delta v_A}; \quad \overrightarrow{F_{BA}}\Delta t = m_B \overrightarrow{\Delta v_B} F A B Δ t = m A Δ v A ; F B A Δ t = m B Δ v B F A B → Δ t = − F B A → Δ t → m A Δ v A → = Δ p A → = − m B Δ v B → = − Δ p B → . \overrightarrow{F_{AB}}\Delta t = - \overrightarrow{F_{BA}}\Delta t \rightarrow m_A \overrightarrow{\Delta v_A} = \overrightarrow{\Delta p_A} = - m_B \overrightarrow{\Delta v_B} = - \overrightarrow{\Delta p_B}. F A B Δ t = − F B A Δ t → m A Δ v A = Δ p A = − m B Δ v B = − Δ p B .
Therefore changes in momentum are equal and opposite. Total change in momentum is zero:
Δ p → = Δ p A → + Δ p B → = − Δ p B → + Δ p B → = 0. \overrightarrow{\Delta p} = \overrightarrow{\Delta p_A} + \overrightarrow{\Delta p_B} = - \overrightarrow{\Delta p_B} + \overrightarrow{\Delta p_B} = 0. Δ p = Δ p A + Δ p B = − Δ p B + Δ p B = 0.
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