Question

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

• A. Acceleration $M_1$ and $M_2$ will be upward
• B. Magnitude of acceleration of both masses will be $\frac{M_1-M_2}{M_1+M_2}g$
• C. Acceleration of $M_1$ and $M_2$ will be equal to zero
• D. Acceleration of $M_1$ will be equal to zero, which that of $M_2$ will be $\frac{M_1-M_2}{M_2}g$ upward

Answer 1

Answer: (D)

As $M_1>M_2$, so tension in thread connecting the block $B$ and support $C$ is $\left(M_1-M_2\right)g$. When this thread is burnt, this tension disappears. In this case tension in spring is $M_{1}g$ and it is elongated.
So, tension in stringconnecting $A$ and $B$ is $M_{1}g$. Hence, resultant force on $A$ remains zero because tension in string is balanced by tension in spring. The block $B$ has a net upward force $\left(M_1-M_2\right)g$.
so initial acceleration of block $B$ is $\frac{\left(M_1-M_2\right)g}{M_2}$
Thus, initial acceleration of $M_1$ is zero and that of $M_2$ is $\frac{\left(M_1-M_2\right)g}{M_2}$ upward.