Question Number 39655 by rahul 19 last updated on 09/Jul/18
Commented by rahul 19 last updated on 09/Jul/18
$$\mathrm{Given}\:\mathrm{Mass}\:\mathrm{of}\:\mathrm{Block}\:\mathrm{A}\:=\:\mathrm{m}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Mass}\:\mathrm{of}\:\mathrm{Block}\:\mathrm{B}\:=\:\mathrm{3m}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{acc}^{\mathrm{n}} \:\mathrm{of}\:\mathrm{blocks}\:\mathrm{at}\:\mathrm{the}\:\mathrm{instant} \\ $$$$\mathrm{when}\:\mathrm{string}\:\mathrm{between}\:\mathrm{block}\:\mathrm{A}\:\&\:\mathrm{ground} \\ $$$$\mathrm{is}\:\mathrm{cut}\:? \\ $$
Answered by MrW3 last updated on 09/Jul/18
$${Before}\:{the}\:{cut}: \\ $$$${T}_{\mathrm{1}} =\mathrm{3}{mg} \\ $$$$\mathrm{2}{T}_{\mathrm{1}} ={mg}+{T}_{\mathrm{2}} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\mathrm{5}{mg} \\ $$$$ \\ $$$${After}\:{the}\:{cut}: \\ $$$${T}_{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{3}{mg}−{T}_{\mathrm{1}} =\mathrm{3}{mg}×{a}_{{B}} \\ $$$$\mathrm{3}{mg}−\mathrm{3}{mg}=\mathrm{3}{mg}×{a}_{{B}} \\ $$$$\Rightarrow{a}_{{B}} =\mathrm{0} \\ $$$$\mathrm{2}{T}_{\mathrm{1}} −{mg}={ma}_{{A}} \\ $$$$\mathrm{6}{mg}−{mg}={ma}_{{A}} \\ $$$$\Rightarrow{a}_{{A}} =\mathrm{5}{g}\:\left(\uparrow\right) \\ $$
Commented by rahul 19 last updated on 09/Jul/18
Thank you sir !����