A body 'A' with a momentum 'P' collides with another identical stationery body 'B' one dimensionally. During the collision, 'B' given an impulse 'J' to the body 'A'. Then the coefficient of restitution is

A body 'A' with a momentum 'P' collides with another identical station

1.	A body 'A' with a momentum 'P' collides with another identical stationery body 'B' one dimensionally. During the collision, 'B' given an impulse 'J' to the body 'A'. Then the coefficient of restitution is
Answer» Solution :ACCRODING to law of conservation of linear momentum, `m_(1)u_(1) + m_(2)u_(2)= m_(1)v_(1) + m_(2)v_(2)` i.e., `m u + m(0)= mv_(1) + mv_(2)` `rArr P- P_(1) = P_(2) " where " P_(2)=J` given `THEREFORE` Coefficient of restitution, `e= (v_(2)-v_(1))/(u)= (mv_(2)-mv_(1))/(m u)= (P_(2)-P_(1))/(P)` `=(P_(2) -(P- P_(2)))/(P)= (2p_(2)-P_(1))/(P) therefore e= (2J-P)/(P)= (2J)/(P)-1`