Question Number 55306 by Rio Mike last updated on 21/Feb/19
$${A}\:{mass}\:{of}\:\mathrm{6}{kg}\:{lies}\:{on}\:{an}\:{inclined}\:{plane} \\ $$$${which}\:{is}\:{smooth}\:{at}\:{angle}\:\theta\:{to}\:{the}\:{horizontal} \\ $$$${where}\:\:\mathrm{sin}\:\theta=\:\frac{\mathrm{1}}{\mathrm{3}}.{if}\:{it}\:{is}\:{connected}\:{to} \\ $$$${another}\:{mass}\:\mathrm{8}{kg}\:{by}\:{the}\:{same}\:{inelastic}\:{string} \\ $$$${passing}\:{over}\:{a}\:{smooth}\:{fixed}\:{pulley} \\ $$$${at}\:{the}\:{top}\:{of}\:{the}\:{plane}.{the}\:{partices}\:{are} \\ $$$${released}\:{from}\:{rest}\:,{Find}\: \\ $$$$\left.{a}\right)\:{the}\:{acceleration}\:{of}\:{each}\:{mass} \\ $$$$\left.{b}\right)\:{the}\:{tension}\:{in}\:{the}\:{string} \\ $$$$\left.{c}\right)\:{the}\:{speed}\:{of}\:{the}\:{masses}\:{after}\:\mathrm{4}{seconds} \\ $$$$\left.{d}\right)\:{the}\:{distanced}\:{covered}\:{with}\:{this}\:{time} \\ $$$${in}\:\left({c}\right)\:{above}\:{by}\:{the}\:{masses}. \\ $$$$ \\ $$$$\mathrm{2017}\:{CGCEB}\:{paper}\:\mathrm{3}\:{Mechanics}\:{a}/{v} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 21/Feb/19
$$\mathrm{8}×\mathrm{10}−{T}=\mathrm{8}{a}\:\left[\right. \\ $$$${T}−\mathrm{6}×\mathrm{10}×\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{6}{a} \\ $$$$\mathrm{14}{a}=\mathrm{60}\:\:\:{a}=\frac{\mathrm{30}}{\mathrm{7}}{m}/{sec}^{\mathrm{2}} \:\rightarrow{for}\:{both}\:{mass} \\ $$$${T}=\mathrm{80}−\mathrm{8}×\frac{\mathrm{30}}{\mathrm{7}}=\mathrm{80}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{7}}\right) \\ $$$${T}=\frac{\mathrm{320}}{\mathrm{7}}{N} \\ $$$${v}=\mathrm{0}+\frac{\mathrm{30}}{\mathrm{7}}×\mathrm{4}=\frac{\mathrm{120}}{\mathrm{7}}\rightarrow{for}\:{both}\:{mass} \\ $$$${s}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{30}}{\mathrm{7}}×\mathrm{4}^{\mathrm{2}} =\frac{\mathrm{240}}{\mathrm{7}}\boldsymbol{{meter}} \\ $$