Question

In the figure a block ${}^{\prime }{A}^{\prime }$ of mass ${}^{\prime }{m}^{\prime }$ is attached at one end of a light spring and the other end of the spring is connected to another block ${}^{\prime }{B}^{\prime }$ of mass $2m$ through a light string. ${}^{\prime }{A}^{\prime }$ is held and $B$ is in static equilibrium. Now $A$ is released. The acceleration of $A$ just after that instant is ${}^{\prime }{a}^{\prime }$. In the next case, $B$ is held and $A$ is in static equilibrium. Now when $B$ is released, its acceleration immediately after the release is ${}^{\prime }{b}^{\prime }$. The value of $a/b$ is: (Pulley, string and the spring are massless)

• A. $0$
• B. $\frac{1}{2}$
• C. $2$
• D. undefined

Answer 1

The correct option is A $0$

Acceleration do not change instantaneously in a spring and thus a = 0, So a/b = 0.