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4-1.Newton's Laws of Motion
hard
For the system shown in figure, $m_1>m_2>m_3 >m_4$ . Initially, the system is at rest in equilibrium condition. If the string joining $m_4$ and ground is cut, then just after the string is cut :
Statement $I$ $m_1$ ,$m_2$ ,$m_3$ remain stationary.
Statement $II$ The value of acceleration of all the $4$ blocks can be determined.
Statement $III$ Only $m_4$ remains stationary.
Statement $IV$ Only $m_4$ accelerates.
Statement $V$ All the four blocks remain stationary.
Now, choose the correct option.
AAll the statements are correct
BOnly $I$ , $II$ and $IV$ are correct
COnly $II$ and $V$ are correct
DOnly $II$ and $IV$ are correct.
Solution
Before cutting the string, the tension in string joining $m_{4}$ and ground is $T=\left(m_{1}+m_{2}-m_{3}-m_{4}\right) g$
And the spring force in the spring joing $m_{3}$ and $m_{4}$ is $T-m_{4} g .$
As the string is cut, the spring force do not change instantly, so just after cutting the string the equilibrium of $m_{1}, m_{2}$ and $m_{3}$would be maintained but $m_{4}$ accelerates in upward direction with acceleration given by
$a=\frac{T+m_{4} g-m_{4} g}{m_{4}}$
And the spring force in the spring joing $m_{3}$ and $m_{4}$ is $T-m_{4} g .$
As the string is cut, the spring force do not change instantly, so just after cutting the string the equilibrium of $m_{1}, m_{2}$ and $m_{3}$would be maintained but $m_{4}$ accelerates in upward direction with acceleration given by
$a=\frac{T+m_{4} g-m_{4} g}{m_{4}}$
Standard 11
Physics
