Question

Two masses $A$ and $B$ connected with an inextensible string of length $l$ lie on a smooth horizontal plane. $A$ is given a velocity of $vm{s}^{-1}$ along the ground perpendicular to line $AB$ as shown in the figure. Find the tension in a string during their subsequence motion.

• A. $\frac{2m{v}^2}{3l}$
• B. $\frac{3m{v}^2}{2l}$
• C. $\frac{m{v}^2}{2l}$
• D. $\frac{4m{v}^2}{3l}$

Answer 1

Answer: (A)

From conservation of linear momentum
​ ${\text{v}}_{\text{cm}}=\frac{\text{m}\times 0+\text{2mv}}{\text{m}+\text{2m}}\Rightarrow {\text{v}}_{\text{cm}}=\frac{\text{2v}}{3}$
position of the centre of mass
${\text{y}}_{\text{cm}}=\frac{0\times \text{2m}+\text{m}\times l}{\text{2m}}\Rightarrow {\text{y}}_{\text{cm}}=\frac{l}{3}$

velocity of $A$ in COM frame is $v_A^{\prime }=v-2v/3\Rightarrow v_A^{\prime }$
$=v/3$
Tension will provide the required centripetal force
$T=\frac{(2m)v_A^2}{(\ell /3)}\Rightarrow T=\frac{(2m)(v/3{)}^2}{(\ell /3)}\Rightarrow T=\frac{2m{v}^2}{3\ell }$