Question Number 190902 by Spillover last updated on 13/Apr/23
Answered by mr W last updated on 13/Apr/23
$$\left({i}\right) \\ $$$$\left({m}_{{A}} +{m}_{{B}} \right){a}=\left({m}_{{B}} −{m}_{{A}} \right){g} \\ $$$$\Rightarrow{a}=\frac{\left({m}_{{B}} −{m}_{{A}} \right){g}}{{m}_{{B}} +{m}_{{A}} }=\frac{\left(\mathrm{0}.\mathrm{7}−\mathrm{0}.\mathrm{3}\right)×\mathrm{10}}{\mathrm{0}.\mathrm{7}+\mathrm{0}.\mathrm{3}}=\mathrm{4}\:{m}/{s}^{\mathrm{2}} \\ $$$${T}={m}_{{B}} \left({g}−{a}\right)={m}_{{B}} \left(\mathrm{1}−\frac{{m}_{{B}} −{m}_{{A}} }{{m}_{{B}} +{m}_{{A}} }\right){g}=\frac{\mathrm{2}{m}_{{A}} {m}_{{B}} {g}}{{m}_{{A}} +{m}_{{B}} } \\ $$$$\:\:\:=\frac{\mathrm{2}×\mathrm{0}.\mathrm{3}×\mathrm{0}.\mathrm{7}×\mathrm{10}}{\mathrm{0}.\mathrm{3}+\mathrm{0}.\mathrm{7}}=\mathrm{4}.\mathrm{2}\:{N}\:\checkmark \\ $$$$\left({ii}\right) \\ $$$${v}={at}=\mathrm{1}.\mathrm{6} \\ $$$$\Rightarrow{t}=\frac{\mathrm{1}.\mathrm{6}}{\mathrm{4}}=\mathrm{0}.\mathrm{4}\:{s} \\ $$$${h}=\frac{{at}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{4}×\mathrm{0}.\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0}.\mathrm{32}\:{m} \\ $$$$\mathrm{0}.\mathrm{32}+\mathrm{0}.\mathrm{52}=−\mathrm{1}.\mathrm{6}{t}+\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{5}{t}^{\mathrm{2}} −\mathrm{1}.\mathrm{6}{t}−\mathrm{0}.\mathrm{84}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\mathrm{1}.\mathrm{6}+\sqrt{\mathrm{1}.\mathrm{6}^{\mathrm{2}} +\mathrm{4}×\mathrm{5}×\mathrm{0}.\mathrm{84}}}{\mathrm{2}×\mathrm{5}}=\mathrm{0}.\mathrm{6}\:{s}\:\checkmark \\ $$
Commented by Spillover last updated on 14/Apr/23
$${nice}\:{solution}.{thanks} \\ $$