Question

In system shown in figure $M_1>M_2$. System is held at rest by thread BC. Just after the thread BC is burnt.

• A. acceleration of $M_1$ will be equal to zero
• B. acceleration of $M_2$ will be downwards
• C. magnitudes of acceleration of two blocks will be nonzero and unequal
• D. magnitude of acceleration of both of blocks will be $\left(\frac{M_1-M_2}{M_1+M_2}\right)g$

Answer 1

The correct option is A acceleration of $M_1$ will be equal to zero

Initial values of tension in spring $T_1=M_{1}g$
Even after the thread is burnt, the tension in the spring remains the same
$\therefore$ block A is in equilibrium state, acceleration$=0$ and the spring pulls the string connected to block B with a force $M_{1}g$
$\therefore$ acceleration of block B$=\frac{M_1}{M_2}g$