A small ball B of mass m is suspended with light inelastic string of length L from a block A of same

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5.Work, Energy, Power and Collision

hard

$A$ small ball $B$ of mass $m$ is suspended with light inelastic string of length $L$ from $a$ block $A$ of same mass $m$ which can move on smooth horizontal surface as shown in the figure. The ball is displaced by angle $\theta$ from equilibrium position & then released. The displacement of block when ball reaches the equilibrium position is

$\frac{{L\sin \theta }}{2}$

$L\,sin\theta$

$L$

none of these

Solution

If ball and blocks are taken as system, then in horizontal direction no external force is present. Hence, centre of mass remains in rest in horizontal direction.

$m(L \sin \theta-x)-m_{x}=0$

Or $\mathrm{mL} \sin \theta-\mathrm{mx}-\mathrm{mx}=0$

$\mathrm{x}=\frac{\mathrm{L}}{2} \sin \theta$

Standard 11

Physics

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hard

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easy

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easy

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medium

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