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Unformatted text preview: 5.68) Maximum P and E of a Rocket The rocket begins at rest with mass M , ejecting exhaust at speed u . We can use equation 5.54 to find the rockets current speed as a function of its remaining mass m ( t ) v ( t ) = u log M m ( t ) the rockets forward momentum is mv . If we subsitute in for velocity as a function of m , we see p = mu log M m Find when this is a maximum by taking the derivative with respect to m dp dm = u log M m- mu m M M m 2 = 0 log M m = 1 So momentum is maximal when m = 1 e M Kinetic energy is given by 1 2 mv 2 = 1 2 mu 2 parenleftbigg log M m parenrightbigg 2 = 1 2 mu 2 parenleftBig log m M parenrightBig 2 Find the maximum by differentiating with respect to m ... 1 2 u 2 parenleftBig log m M parenrightBig 2 + mu 2 log parenleftBig m M parenrightBig 1 m = 0 log parenleftBig m M parenrightBig =- 2 Which means m = 1 e 2 M 5.70) Snow on a Sled, Quantitative Sled begins with mass M and speed V , with snow accumulating on it at a rate . The essential fact in all three cases is that the momentum of the sled and the sled, along the direction of the track, is conserved. If the snow is pushed off the sled, perpendicular to the tracks in the frame of the ground, then it will have no forward momentum, and the sled will have the same mass and momentum it had originally. Mv ( t ) = MV v ( t ) = V If nothing is done, we can determine from conservation of momentum that ( M + t ) v ( t ) = MV v ( t ) = M M + t V 1 If the snow is pushed off perpendicular to the sleds tracks in the sleds frame, its forward velocity will be the same as the sleds (at the time the snow was pushed off) This means the sled will lose forward momentum at a rate proportional to its velocity times the rate snow is accumulating M dv dt =- v v ( t ) dv dt =- M v which has solution v ( t ) =...
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