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 $t_1$

$mg-kx=ma$
$v\frac{dv}{dx}=\frac{mg-kx}{m}$
$\underset{0}{\overset{0}{\int }}vdv=\underset{0}{\overset{t}{\int }}\left(g-\frac{k}{m}x\right)dx$
where $x_0$ is maximum elongation in spring
$0=g{x}_0-\frac{k{x}_0^2}{2m}$
$x=\frac{2mg}{k}$
At the time of break off of block $B$
$Mg=k{x}_0$
$Mg=2mg$
$m=\frac{M}{2}$