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Standard XII

Physics

Introduction

Two bars of m...

Question

Two bars of masses $m_{1}$ and $m_{2}$ , connected by a weightless spring of stiffness $k$ , rest on a smooth horizontal plane. Bar 2 is shifted by a small distance $x_{0}$ to the left and released. The velocity of the centre of mass of the system when bar 1 breaks off the wall is :

x_{0} \sqrt{\frac{k m_{2}}{m_{1} + m_{2}}}

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\frac{x_{0}}{m_{1} + m_{2}} \sqrt{k m_{2}}

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x_{0} k \frac{m_{1} + m_{2}}{m_{2}}

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x_{0} \frac{\sqrt{k m_{1}}}{(m_{1} + m_{2})}

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Solution

The correct option is B $\frac{x_{0}}{m_{1} + m_{2}} \sqrt{k m_{2}}$
$\frac{1}{2} k x_{0}^{2} = \frac{1}{2} m_{2} v_{2}^{2}$
$∴ v_{2} = \sqrt{\frac{k}{m_{2}}} x_{0}$
$v_{C M} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$= \frac{\sqrt{k m_{2}}}{(m_{1} + m_{2})} x_{0} ($ as $v_{1} = 0$ )

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Two bars of masses m1 and m2, connected by a weightless spring of stiffness k, rest on a smooth horizontal plane. Bar 2 is shifted by a small distance x0 to the left and released. The velocity of the centre of mass of the system when bar 1 breaks off the wall is :

The correct option is B x0m1+m2√km212kx20=12m2v22∴v2=√km2x0vCM=m1v1+m2v2m1+m2=√km2(m1+m2)x0( as v1=0)

Two bars of masses $m_{1}$ and $m_{2}$ , connected by a weightless spring of stiffness $k$ , rest on a smooth horizontal plane. Bar 2 is shifted by a small distance $x_{0}$ to the left and released. The velocity of the centre of mass of the system when bar 1 breaks off the wall is :

The correct option is B $\frac{x_{0}}{m_{1} + m_{2}} \sqrt{k m_{2}}$
$\frac{1}{2} k x_{0}^{2} = \frac{1}{2} m_{2} v_{2}^{2}$
$∴ v_{2} = \sqrt{\frac{k}{m_{2}}} x_{0}$
$v_{C M} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$= \frac{\sqrt{k m_{2}}}{(m_{1} + m_{2})} x_{0} ($ as $v_{1} = 0$ )