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76. [[underlined]] Theory of Spring Impulse-Meter [[/underlined]]. The theory of the spring impulse-meter is as follows: If we use the same rotation as in the preceding case, calling, in addition, the mass of the spring [[underlined]] m [[subscript]] s [[/subscript]] [[/underlined]], and the mass required for unit extension of the spring, [[underlined]] m [[subscript]] 1 [[/subscript]] [[/underlined]]; we have, by the same theory as that for the suspended gun suspended by a spring, ^[[overline]] V [[subscript]] c [[/subscript]] [[/overline]] = √m1g/√mc + 1/3ms S.]] Hence the momentum per unit area, communicated to the upper cap of the 12-inch pipe, when the chamber fired, is ^[[ [[underline]] I [[/underline]] / [[underline]] A [[/underline]] [[subscript]] c [[/subscript]] = (mc+1/3ms) [[overline]] V [[/overline]] c / [[underline]] A [[/underline]] c = √mc+1/3ms √m1g S/[[underline]] A [[/underline]] [[subscript]] c [[/subscript]] Hence the momentum that would be communicated to the [[underline]] suspended system [[/underline]] by the gaseous rebound, [[underline]] provided the system were at the top of the 12-inch pipe [[/underline]], would be ^[[underline]] A [[/underline]]g √mc+1/3ms √m1g S/[[underline]] A [[/underline]] [[subscript]] c [[/subscript]] and the percentage, [[underline]] Q [[/underline]], of the momentum communicated by the gaseous rebound to the observed momentum of the suspended system, would be ^[[Q = [[underline]] A [[/underline]] g √mc+1/3ms √m1g S/[[underline]] A [[/underline]] [[subscript]] c [[/subscript]] mov