Question

In the system shown $m_1>m_2.$ System is held at rest by thread BC. Just after lower thread is Burnt,

• A. Acceleration of $m_2$ is upward
• B. Magnitude of acceleration of both blocks will be $\left(\frac{m_1-m_2}{m_1+m_2}\right)g$
• C. Acceleration of $m_1$ will be equal to zero
• D. Magnitudes of acceleration of two blocks will be non-zero and unequal

Answer 1

Answer: (A), (C)

Acceleration of $m_1$ will be equal to zero

Before burn,

$T=m_{1}g$ & $T=T_1+m_{2}g$
After burning the wire,
Tension in the spring does not change instantaneously.

so, for $m_1$
Fnet= $T-m_{1}g$ = 0
$\therefore a_{m1}=0$
For $m_2$
$T_1$ becomes zero and $m_2$ will have a non-zero acceleration
$\therefore T^{{}^{\prime }}-m_{2}g=m_{2}a$
$\Rightarrow a=\frac{T^{{}^{\prime }}}{m_2}-g$