Question

As shown in figure, a particle of mass $2\text{kg}$ is attached to a bead. The bead can slide on a smooth straight wire. Length of the string which connect the particle and the bead is $l$. Initially, the particle is held in contact with the wire with the string taut as shown in figure, and then it is let to fall. If the bead has a mass $4\text{kg}$, then, when the string makes an angle $\theta ={60}^{\circ }$ with the wire, find the distance it slides upto this instant.

• A. $\frac{l}{6}$ towards left
• B. $\frac{l}{6}$ towards right
• C. $\frac{l}{2}$ towards left
• D. $\frac{l}{2}$ towards right

Answer 1

The correct option is B $\frac{l}{6}$ towards right

When string make an angle ${60}^{\circ }$ with the wire:


Let the bead travel $x$ distance as shown above.
Then, the distance covered by the $2\text{kg}$ particle in $x$ direction is $=[-(l-l\cos \theta )+x]$
and the distance covered by the bead $=x$ (towards right)
As we know, there is no external force acting on the system. So, $x_{com}$ will not change. It will be zero.
$x_{com}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}=0$
(Assuming motion towards right to be positive.)
$x_{com}=\frac{4\times \left(x\right)+2(l\cos {60}^{\circ }-l+x)}{4+2}$
$\Rightarrow 4x-2l+2l\cos {60}^{\circ }+2x=0$
$\Rightarrow 6x-l=0$
$\Rightarrow x=\frac{l}{6}$
+ve sign means the bead will move towards right.