Question

A small bar $\text{A}$ resting on a smooth horizontal plane is attached by threads to a point $\text{P}$ and by means of a weightless pulley, to a weight $\text{B}$ possessing the same mass as the bar itself. The bar is also attached to a point $\text{O}$ by means of a light non-deformed spring of length $l_0=\frac{32}{19}\text{cm}$ and stiffness $k=\frac{5mg}{l_0},$ where $m$ is the mass of the bar. The thread $\text{PA}$ having been burned, the bar starts moving to the right. Find its velocity at the moment when it breaks off the plane.(Take all the $g$ in ${\text{m/sec}}^2)$

• A. $\sqrt{2g}\text{m/s}$
• B. $2\sqrt{g}\text{m/sec}$
• C. $\sqrt{g}\text{m/sec}$
• D. $2\sqrt{2g}\text{m/sec}$

Answer 1

The correct option is C $\sqrt{g}\text{m/sec}$

Let the spring be elongated by $x$.


Let $\theta$ be angle between spring and vertical at the instant when block $\text{A}$ breaks off the plane.

From free body diagram,
$T=mg....\left(1\right)$

$kx\cos \theta =mg....\left(2\right)$

From figure, $\cos \theta =\frac{l_0}{l_0+x}$ and given $k=\frac{5mg}{l_0}$.

Substituting the values in $\left(2\right)$, we get

$\frac{5mg}{l_0}\times x\times \frac{l_0}{l_0+x}=mg$

$\Rightarrow x=\frac{l_0}{4}....\left(3\right)$

Distance covered by $\text{A}$ and $\text{B}$ , $d=\sqrt{(l_0+x{)}^2-l_0^2}=\frac{3l_0}{4}....\left(4\right)$

From energy conservation:
Gain in potential energy=Loss in kinetic energy

$\Rightarrow \Delta P.E_A+\Delta P.E_B+\Delta P.E_{Spring}=\Delta K.E_A+\Delta K.E_B+\Delta K.E_{Spring}$

$\Rightarrow (0-0)+(mgd-0)+(0-\frac{1}{2}k{x}^2)=(\frac{1}{2}m{v}^2-0)+(\frac{1}{2}m{v}^2-0)+(0-0)$

$\Rightarrow mgd=2\left(\frac{1}{2}m{v}^2\right)+\frac{1}{2}k{x}^2....\left(5\right)$

Velocity is same because they are connected.

From $\left(3\right),\left(4\right)$, $\left(5\right)$ and $k=\frac{5mg}{l_0}$ we get,
$mg\times \frac{3l_0}{4}=m{v}^2+\frac{1}{2}\times \frac{5mg}{l_0}\times \frac{l_0^2}{16}$

$\Rightarrow v^2=\frac{3g{l}_0}{4}-\frac{5g{l}_0}{32}$

$\Rightarrow v=\sqrt{\frac{19g{l}_0}{32}}$

$\Rightarrow v=\sqrt{\frac{19g}{32}\times \frac{32}{19}}=\sqrt{g}\text{m/s}$

Hence, option (c) is the correct answer.