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

normal

$A$ 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. Tension in string when it is vertical, is

A

$mg$

B

$mg(2-cos\theta )$

C

$mg (3 - 2cos\theta )$

D

none of these

Solution

Loss in $P E=\operatorname{gain}$ in $K E$

$m g(L-L \cos \theta)=\frac{1}{2} m v_{1}^{2}+\frac{1}{2} m v_{2}^{2}$

Applying momentum conservation principle in horizontal direction. $P_{i x}=P_{f x}$

$\overrightarrow{0}=m v_{1}-m v_{2}$

$\therefore v_{1}=v_{2}=v$

$\therefore M g(L-L \cos \theta)=m v^{2}$

$\therefore v=\sqrt{g l(1-\cos \theta)}$

$=T-m g=\frac{m\left(2 v^{2}\right)}{l}$

$\therefore T=m g+\frac{4 m g l(1-\cos \theta)}{l}$

$\therefore T=m g+4 m g(1-\cos \theta)$

Standard 11

Physics

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