1.

A body .A. with a momentum .P. collides with another identical stationery body .B. gives an impulse .J. to the body .A.. Then the coefficient of restitution is

Answer»

Solution :According 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 = (2J-P)/(P)=(2J)/(P)=1`


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