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Tension in a String

The blocks m ...

Question

The blocks $(m_{2} > m_{1})$ are held in such a manner that spring is unstretched initially. Consider the situation at the time of release of the block $(t = 0)$ and short time after releasing the block $(t > 0)$ . Accelerations of the blocks of masses $m_{1}$ and $m_{2}$ are $a_{1}$ and $a_{2}$ respectively

t = 0, a_{1} = a_{2}

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t = 0, a_{1} < a_{2}

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t > 0, a_{1} = a_{2}

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t > 0, a_{1} < a_{2}

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Solution

The correct options are
A $t = 0, a_{1} = a_{2}$
D $t > 0, a_{1} < a_{2}$
Free body diagram of blocks
For $m_{1}$ :
$a_{1} = \frac{m_{1} g - T}{m_{1}}$
$= g - \frac{T}{m_{1}}$
For $m_{2}$ :
$a_{2} = \frac{m_{2} g - T}{m_{2}}$
$= g - \frac{T}{m_{2}}$
At $t = 0 : T = 0 \Rightarrow a_{1} = a_{2}$
At $t > 0 : T > 0 \Rightarrow a_{1} < a_{2}$

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The blocks (m2>m1) are held in such a manner that spring is unstretched initially. Consider the situation at the time of release of the block (t=0) and short time after releasing the block (t>0). Accelerations of the blocks of masses m1 and m2 are a1 and a2 respectively

The correct options are A t=0,a1=a2 D t>0,a1<a2Free body diagram of blocks For m1: a1=m1g−Tm1 =g−Tm1 For m2: a2=m2g−Tm2 =g−Tm2 At t=0:T=0⇒a1=a2 At t>0:T>0⇒a1<a2