Question

Two identical beads, each of m = 100g, are connected by an inextensible massless string, which can slide along the two arms AC and BC of a rigid smooth wire frame in v vertical plane. If the system is released from rest, the kinetic energy of the first bead when the second bead has moved down by a distance of 0.1 m is $x\times {10}^{-3}$ J. Find the value of x (g = 10 $m{s}^{-2}$) (shown situation is after movement of 0.1m).

Answer 1

let, the string make angle $\theta$ with the horizontal bar.

let, the first bead has velocity $=V_1$

Second bead velocity $=V_2$

now, equaliting the componets of velocity of

the beads along the string,

$V_1\cos \theta =V_2\sin \theta$

$V_1\times \frac{0.9}{0.5}=V_2\times \frac{0.3}{0.5}$

$V_2=\frac{9}{3}{V}_1$

now, the second bead descend by $0.1m$

so, from conservation of energy

$mgh=\frac{1}{2}m{V}_1^2+\frac{1}{2}m{V}_2^2$

or $10\times 0.1=\frac{1}{2}(V_1^2+\frac{16}{9}{V}_1^2)$

or $2=\frac{25}{9}{V}_1^2$

or $V_1^2=\frac{18}{25}$

kinetic energy of the first bead

$\begin{array}{cc}{K}_1& =\frac{1}{2}\times \frac{100}{1000}\times \frac{18}{25},\because m=\frac{100}{1000}kq\\ & =36\times {10}^{-3}\because V_1^2=\frac{18}{25}\end{array}$

The correct answer is $36$