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19 Sep 2007 15:44:12 IST

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velocity of seperation...and velocity of approach

None

what is the velocity of seperation...and velocity of approach discribe it in methematical form....i dont want their definations ..(u may consider any example )

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PARTHESH BULBULE

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Joined: 25 Jul 2007

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19 Sep 2007 18:58:56 IST

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Suppose m & M are two masses which are to be collided, and u₁ & u₂ are their initial velocities while v₁ & v₂ are their final velocities respectively.

Then according to the law of conservation of momentum,

mu₁ + Mu₂ = mv₁ + Mv₂

Newton's experimental law says that, If two bodies with initial speeds u₁ & u₂ collide, so they have final velocities v₁ & v₂, then,

-e = v₂ - v₁/ u₂ - u₁

where e is the coefficient of restitution v₂- v₁ is the velocity of seperation and u₂ - u₁is the velocity of approach.

For example if two bodies with initial velocities 3m/s & 4m/s suffer a head-on collision then, and their final velocities are 1m/s and 5m/s then,

velocity of approach is 4 - 3 = 1m/s & velocity of seperation is 5 - 1 = 4m/s.

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shami pratap

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Joined: 5 Aug 2007

Posts: 38

19 Sep 2007 21:55:00 IST

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there is nothing great in its math. just understand basic cocept and U will never face any difficlty. Suppose U R running after a thief, Your velocity is Ui and thief's T. Then you are appraching thief with a net vlocity of Ui-T. With this resultant velocity you are approaching. Thus Ui-T is velocty of approach. But, just on catching him he slips and runs away at speed Tf and you are following him with velocity Uf, the thief is seperating from you with a resultant velocity of Tf-uf. (normally we assume, Tf is higher than Uf) . It is found in elastic collisions that ratio Tf-uf/Ui-T is 1. But if this ratio is less tha 1 , You can confidently declare it is not elastic collision. It will be inelastic collision.

ayesha arora

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Joined: 8 Aug 2007

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19 Sep 2007 22:42:15 IST

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thanks 4 d ans....

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