Derive the formula of conservation of …
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Meghna Thapar 4 years, 5 months ago
According to the law of conservation of momentum, the sum of momenta of two objects before collision is equal to the sum of momenta after collision, provided there is no external unbalanced force acting on the objects.
Suppose two balls A and B, of masses mA and mB are travelling in the same direction along a straight line at different velocities uA and uB. And there are no other external unbalanced forces acting on them. Let uA > uB and the two balls collide with each other. During collision which lasts for a time t, the ball A exerts a force FAB on ball B and the ball B exerts a force FBA on ball A.
Suppose vA and vB are the velocities of the two balls A and B after the collision, respectively.
The momenta of ball A before and after the collision are mA uA and mA vA. The rate of change of its momentum (or FAB, action) during the collision will be
mA (vA - uA)/t
Similarly, the rate of change of momentum of ball B (= FBA or reaction) during the collision will be
mB (vB - uB)/t.
According to the third law of motion, the force FAB exerted by ball A on ball B (action) and the force FBA exerted by the ball B on ball A (reaction) must be equal and opposite to each other. Therefore,
FAB = – FBA
or
mA (vA - uA)/t = -mB (vB - uB)/t. This gives,
mA uA + mB uB = mA uA + mB vB
where mA and mB are masses of the bodies, u1 and u2 are initial velocities of the bodies and v1 and v2 are final velocities of the bodies.
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