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In a head on collision between two ident...

In a head on collision between two identical particles A and B moving on a horizontal surface, B is stationary and A has momentum `P_(0)` before impact. Due to the impact, A recrieves an impulse I. Then coefficient of restitution between the two is :

A

`(2I)/(P_(0))-1`

B

`(2I)/(P_(0))+1`

C

`(I)/(P_(0))+1`

D

`(I)/(P_(0))-1`

Text Solution

Verified by Experts

The correct Answer is:
A

Impulse =`DeltaP`
So, For B `I=mv_(2)` ....(i)
and For `A-I=mv_(1)-P_(0)`....(ii)
`rArre=(v_(2)-v_(1))/(-(0-v_(0)))=(mv_(2)-mv_(1))/(mv_(0))=(I-(P_(0)-I))/(P_(0))=(2I-P_(0))/(P_(0))`
`e=(2I)/(P_(0))-1`
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