Question

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

• A. 2 M
• B. M
• C. $\frac{M}{2}$
• D. a function of $M$ and the force constant of the spring

Answer 1

Answer: (C)

For minimum mass of $m$, mass $M$ breaks off contact when elongation in spring is maximum. At the time of break off, block $A$ is at lowest position and its speed is zero. At an instant $t_{1}$

$ m g-k x =m a $
$ v \frac{d v}{d x} =\frac{m g-k x}{m} $
$ \int\limits_{0}^{0} v \,d v =\int\limits_{0}^{t}\left(g-\frac{k}{m} x\right) d x $
where $x_{0}$ is maximum elongation in spring
$ 0 =g x_{0}-\frac{k x_{0}^{2}}{2 m} $
$x =\frac{2 m g}{k} $
At the time of break off of block $B$
$M g =k x_{0}$
$M g =2 m g$
$m =\frac{M}{2}$