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Derivation of conservation of momentum

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Derivation of conservation of momentum
  • 1 answers

Abhinav Asthana 7 years ago

Let us take masses of two objects mA and mB velocity=uA and uB Let them collide eachother object A exerts FAB on object B object B exerts FBA on object A Final velocity=vA and vB Momentum of object A is mA uA and mA v Rate of change of momentum =mA vA-uA/t similarly mB=uB-vB/t According to third law of motion FAB=FBA (OR) mAuA+mBuB=mAvA+mBvB Hence we derived the formula Hope you understood
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