Question Number 137148 by nadovic last updated on 30/Mar/21
Answered by ajfour last updated on 30/Mar/21
$$\left(\mathrm{3}{kg}\right)\left(\mathrm{4}{m}/{s}\right)=\left(\mathrm{3}{kg}\right){v}+\left(\mathrm{5}{kg}\right){V} \\ $$$$\left.{V}−{v}=\left(\mathrm{0}.\mathrm{6}\right)\left(\mathrm{4}{m}/{s}\right)\:\:\:\right\}×\mathrm{3} \\ $$$$\mathrm{3}{v}+\mathrm{5}{V}=\mathrm{12}{m}/{s} \\ $$$${Adding} \\ $$$$\mathrm{8}{V}=\mathrm{19}.\mathrm{2}{m}/{s} \\ $$$${V}=\mathrm{2}.\mathrm{4}{m}/{s} \\ $$$${v}=\mathrm{0}\:{m}/{s} \\ $$$${Overall}\:{loss}\:{in}\:{Kinetic}\:{energy} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}{kg}\right)\left(\mathrm{4}{m}/{s}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}{kg}\right)\left(\mathrm{2}.\mathrm{4}{m}/{s}\right)^{\mathrm{2}} \\ $$$$\:\:=\mathrm{24}\left(\mathrm{1}−\mathrm{0}.\mathrm{6}\right){kgm}^{\mathrm{2}} /{s}^{\mathrm{2}} \\ $$$$\:\%\:{loss}\:{in}\:{K}_{{tot}} =\mathrm{9}.\mathrm{6}\:{J} \\ $$
Commented by nadovic last updated on 31/Mar/21
$${Thank}\:{you}\:{Sir} \\ $$