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Question Number 118944 by mathdave last updated on 20/Oct/20
Answered by Dwaipayan Shikari last updated on 21/Oct/20
ForstationaryboatsΣPboat,man=0(Astheboatandthemansystemwasstationary)MV+M′Ψ=0(UsingconservationofMomentum)Ψ=−MVM′(Ψ=velocityboatV=velocitymanM,M′(massofmanandboat)Ψ=−7134.4.6msΨ=−9.60588msVelocityoftheboat9.60588msinbackwarddirection.11.3)v2−Ψ2=−2aSΨ2=2aS9.62=2.a.19a=9.6238.ms2=92.1638=2.425ms2SoForceneeded=M′×2.425=34×2.425N=82.45N
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