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.