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(C) Let $m$ be the mass of ball $A$. When ball $A$ is released from rest,it moves downwards,stretching the spring. Let the spring stretch by a distanc

Vedclass · Accepted Answer

$M/2$

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(C) Let $m$ be the mass of ball $A$. When ball $A$ is released from rest,it moves downwards,stretching the spring. Let the spring stretch by a distance $x$ at the lowest point of the ball's motion.By the law of conservation of energy,the loss in gravitational potential energy of ball $A$ is equal to the gain in elastic potential energy of the spring:$mgx = \frac{1}{2}kx^2$$mg = \frac{1}{2}kx$$kx = 2mg$ ... $(1)$For block $B$ of mass $M$ to leave contact with the ground,the upward spring force must be at least equal to the weight of block $B$:$kx \ge Mg$ ... $(2)$Substituting the value of $kx$ from equation $(1)$ into equation $(2)$:$2mg \ge Mg$$m \ge M/2$Thus,the minimum mass of $A$ required is $M/2$.

In the figure,the ball $A$ is released from rest when the spring is at its natural (unstretched) length. For the block $B$ of mass $M$ to leave contact with the ground at some stage,the minimum mass of $A$ must be

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