A block Q of mass M is placed on a horizontal | vedclass

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(D) Since there is no external horizontal force acting on the system consisting of block $Q$ and body $P$,the position of the centre of mass of the sy

Vedclass · Accepted Answer

$\frac{m L \cos 	heta}{(m+M)}$

Vedclass · Answer

(D) Since there is no external horizontal force acting on the system consisting of block $Q$ and body $P$,the position of the centre of mass of the system remains unchanged in the horizontal direction.Let the block $Q$ move a distance $x$ to the left. The body $P$ moves a distance $L \cos 	heta$ relative to the block $Q$ along the horizontal direction.Therefore,the displacement of $P$ with respect to the ground in the horizontal direction is $(L \cos 	heta - x)$ to the right.Using the principle of conservation of the centre of mass: $M(-x) + m(L \cos 	heta - x) = 0$.$-Mx + mL \cos 	heta - mx = 0$.$mL \cos 	heta = (M + m)x$.$x = \frac{mL \cos 	heta}{M + m}$.

$A$ block $Q$ of mass $M$ is placed on a horizontal frictionless surface $AB$ and a body $P$ of mass $m$ is released on its frictionless slope. As $P$ slides by a length $L$ on this slope of inclination $\theta$,the block $Q$ would slide by a distance:

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