Question Number 20986 by Tinkutara last updated on 09/Sep/17
$$\mathrm{A}\:\mathrm{50}\:\mathrm{kg}\:\mathrm{log}\:\mathrm{rest}\:\mathrm{on}\:\mathrm{the}\:\mathrm{smooth}\:\mathrm{horizontal} \\ $$$$\mathrm{surface}.\:\mathrm{A}\:\mathrm{motor}\:\mathrm{deliver}\:\mathrm{a}\:\mathrm{towing}\:\mathrm{force} \\ $$$${T}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{below}.\:\mathrm{The}\:\mathrm{momentum}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{particle}\:\mathrm{at}\:{t}\:=\:\mathrm{5}\:\mathrm{s}\:\mathrm{is} \\ $$
Commented by Tinkutara last updated on 09/Sep/17
Answered by dioph last updated on 09/Sep/17
$${P}\left({t}\right)\:=\:\int{T}\left({t}\right){dt} \\ $$$${P}\left(\mathrm{5}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{4}} \mathrm{40}{t}^{\mathrm{2}} {dt}\:+\:\int_{\mathrm{4}} ^{\mathrm{5}} \mathrm{640}{dt} \\ $$$${P}\left(\mathrm{5}\right)\:=\:\frac{\mathrm{40}}{\mathrm{3}}{t}^{\mathrm{3}} \mid_{\mathrm{0}} ^{\mathrm{4}} \:+\:\mathrm{640}{t}\mid_{\mathrm{4}} ^{\mathrm{5}} \\ $$$${P}\left(\mathrm{5}\right)\:=\:\frac{\mathrm{40}×\mathrm{64}}{\mathrm{3}}\:+\:\mathrm{640}×\mathrm{1} \\ $$$${P}\left(\mathrm{5}\right)\:\cong\:\mathrm{1493}.\mathrm{33}\:\mathrm{kg}\:\mathrm{m}/\mathrm{s} \\ $$
Commented by Tinkutara last updated on 10/Sep/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$