Two masses A and B connected with an inextensible string of length l lie on a smooth horizontal plane. A is…

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

1. $\frac{2m{v}^2}{3l}$
    
2. $\frac{3m{v}^2}{2l}$
3. $\frac{m{v}^2}{2l}$
4. $\frac{4m{v}^2}{3l}$

Solution

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

${\text{y}}_{\text{cm}}=\frac{0\times \text{2m}+\text{m}\times l}{\text{2m}}\Rightarrow {\text{y}}_{\text{cm}}=\frac{l}{3}$



blocks will perform circular motion in the center of mass frame

$\text{velocity of A in COM frame is}{\text{v}}_{\text{A}}^{\prime }=\text{v}-2\text{v/3}\Rightarrow {\text{v}}_{\text{A}}^{\prime }=\text{v/3}$

Tension will provide the required centripetal force

$\text{T}=\frac{\left(\text{2m}\right){\text{v}}_{\text{A}}^{\prime 2}}{\left(\ell /3\right)}\Rightarrow \text{T}=\frac{\left(\text{2m}\right){\left(\text{v/3}\right)}^2}{\left(\ell /3\right)}\Rightarrow \text{T}=\frac{{\text{2mv}}^2}{3\ell }$

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