Question Number 205685 by MathedUp last updated on 27/Mar/24
Answered by TonyCWX08 last updated on 27/Mar/24
$$\frac{\mathrm{1}}{\mathrm{10}}{g} \\ $$
Answered by mr W last updated on 27/Mar/24
Commented by mr W last updated on 27/Mar/24
$${m}_{{B}} ={mass}\:{of}\:{object}\:{B} \\ $$$${decrease}\:{of}\:{potential}\:{energy}\:{of}\:{B}\:{is} \\ $$$${m}_{{B}} {gh} \\ $$$${increase}\:{of}\:{kinetic}\:{energy}\:{of}\:{B}\:{is} \\ $$$$\frac{{m}_{{B}} \left({v}_{\mathrm{2}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} \right)}{\mathrm{2}} \\ $$$${it}\:{is}\:{given}: \\ $$$${m}_{{B}} {gh}=\mathrm{10}×\frac{{m}_{{B}} \left({v}_{\mathrm{2}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} \right)}{\mathrm{2}} \\ $$$$\Rightarrow{gh}=\mathrm{5}\left({v}_{\mathrm{2}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} \right) \\ $$$${on}\:{the}\:{other}\:{side}\:{v}_{\mathrm{2}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{2}{ah} \\ $$$$\Rightarrow{gh}=\mathrm{5}×\mathrm{2}{ah}\: \\ $$$$\Rightarrow{a}=\frac{{g}}{\mathrm{10}}={B}'{s}\:{acceleration} \\ $$$${A}\:{and}\:{B}\:{have}\:{the}\:{same}\:{acceleration}, \\ $$$$\Rightarrow{A}'{s}\:{acceleration}\:{is}\:{a}=\frac{{g}}{\mathrm{10}} \\ $$
Commented by TonyCWX08 last updated on 28/Mar/24
$${MathedUp} \\ $$$${I}\:{believe}\:{you}\:{will}\:{say}\:{so}\:{if}\:{you}\:{look}\:{carefully}\:{on}\:{his}\:{profile}\:{picture}. \\ $$
Commented by TonyCWX08 last updated on 28/Mar/24
$${By}\:{the}\:{way},\:{please}\:{change}\:{your}\:{profile}. \\ $$$${Thanks}. \\ $$