Two blocks 1 and 2 of equal mass m are connected by an ideal string (see figure below) over a fricti

Gujarati

Hindi

4-1.Newton's Laws of Motion

hard

Two blocks $1$ and $2$ of equal mass $m$ are connected by an ideal string (see figure below) over a frictionless pulley. The blocks are attached to the ground by springs having spring constants $k_1$ and $k_2$ such that $k_1 > k_2$. Initially, both springs are unstretched. Block $1$ is slowly pulled down a distance $x$ and released. Just after the release the possible values of the magnitudes of the accelerations of the blocks $a_1$ and $a_2$ can be

Aeither $\left(a_1=a_2=\frac{\left(k_1+k_2\right) x}{2 m}\right)$ or $\left(a_1=\frac{k_1 x}{m}-g\right.$ and $\left.a_2=\frac{k_2 x}{m}+g\right)$

B$\left(a_1=a_2=\frac{\left(k_1+k_2\right) x}{2 m}\right)$ only

C$\left(a_1=a_2=\frac{\left(k_1-k_2\right) x}{2 m}\right)$ only

Deither $\left(a_1=a_2=\frac{\left(k_1-k_2\right) x}{2 m}\right)$ $\text { or }\left( \alpha _1=a_2=\frac{\left(k_1 k_2\right) x}{\left(k_1+k_2\right) m}-g\right)$

(KVPY-2012)

Solution

(b)
Free body diagram for block $1$ and $2$ are as shown below.
So, equations of force balance are
$T+k_1 x-m g =m a_1$
$\text { and }$ $k_2 x+m g-T =m a_2$
As string is in extensible,
$a_1=a_2=a \text { (let) }$
Then adding both equations, we have
$\left(k_1+k_2\right) x=2 m a$
$\Rightarrow \quad a=\left(\frac{k_1+k_2}{2 m}\right) \cdot x$

Standard 11

Physics