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(B) Let the horizontal surface be the reference line for the $x$-axis. Since there is no external horizontal force acting on the system,the centre of

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$\frac{L}{2}(1 - \cos 	heta )$

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(B) Let the horizontal surface be the reference line for the $x$-axis. Since there is no external horizontal force acting on the system,the centre of mass of the system does not shift in the horizontal direction. Let $x_A$ and $x_B$ be the horizontal positions of block $A$ and ball $B$. The horizontal position of the centre of mass is $X_{cm} = \frac{m x_A + m x_B}{m + m} = \frac{x_A + x_B}{2}$. Initially,let $x_A = 0$ and $x_B = L \sin 	heta$. So,$X_{cm, initial} = \frac{0 + L \sin 	heta}{2} = \frac{L \sin 	heta}{2}$. When the string becomes vertical,the block $A$ and ball $B$ will be at the same horizontal position $x'$. Thus,$X_{cm, final} = \frac{x' + x'}{2} = x'$. Since the external horizontal force is zero,$X_{cm, initial} = X_{cm, final}$,so $x' = \frac{L \sin 	heta}{2}$. The displacement of the centre of mass in the horizontal direction is zero. Now,consider the vertical direction. The vertical position of the centre of mass is $Y_{cm} = \frac{m y_A + m y_B}{m + m} = \frac{y_A + y_B}{2}$. Initially,$y_A = 0$ and $y_B = -L \cos 	heta$. So,$Y_{cm, initial} = \frac{0 - L \cos 	heta}{2} = -\frac{L \cos 	heta}{2}$. Finally,$y_A = 0$ and $y_B = -L$. So,$Y_{cm, final} = \frac{0 - L}{2} = -\frac{L}{2}$. The vertical displacement of the centre of mass is $\Delta Y_{cm} = Y_{cm, final} - Y_{cm, initial} = -\frac{L}{2} - (-\frac{L \cos 	heta}{2}) = -\frac{L}{2}(1 - \cos 	heta)$. The magnitude of the vertical displacement is $\frac{L}{2}(1 - \cos 	heta)$ in the downward direction.

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