Gujarati
4-1.Newton's Laws of Motion
normal

A block of mass $2 M$ is attached to a massless spring with spring-constant $k$. This block is connected to two other blocks of masses $M$ and $2 M$ using two massless pulleys and strings. The accelerations of the blocks are $a_1, a_2$ and $a_3$ as shown in figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is $x _0$. Which of the following option($s$) is/are correct? [g is the acceleration due to gravity. Neglect friction]

A$x_0=\frac{4 Mg }{ k }$
BWhen spring achieves an extension of $\frac{x_0}{2}$ for the first time, the speed of the block connected to the spring is $3 g \sqrt{\frac{M}{5 k }}$
C$a_2-a_1=a_1-a_3$
DAt an extension of $\frac{x_0}{4}$ of the spring, the magnitude of acceleration of the block connected to the spring is $\frac{3 g }{10}$
(IIT-2019)

Solution

$T=\frac{2(2 m)(m)}{3 m}\left(g-a_1\right)$
$=\frac{4 m}{3}\left(g-a_1\right)$
$\frac{8 m}{3}\left(g-a_1\right)-k x=2 m a_1$
$\frac{8 Mg }{3}-\frac{8 ms }{3}- kg =2 ma _1$
$\frac{8 Mg }{3}-\operatorname{lax}=\frac{14 ma _1}{3}$
$\frac{8 Mg -3 kx }{14 m }=3$
$a _1=\frac{8 Mg -3 lg }{14 m }$
$frac{\pi d r}{d x}-\left(\frac{8 Mg }{14 m }-\frac{3 kx }{14 m }\right)$
$\int r d x=\frac{1}{14 m} \int(3 M g-3 lcx ) d x$
$\text { for may elongation }$
$0=\frac{1}{14 m} \int_0^1(3 Mz -3 h x) d x$
$=\frac{1}{14 m }\left(8 Mg _1-\frac{3 lx _1^2}{2}\right)$
$8 MgX _{ a }=\frac{3 kx _2^2}{2}$
$x_1=\frac{16 Mg }{3 g }$
$\text { at } x=\frac{x_3}{2}$
$\int_0^V dd =\frac{1}{14 m } \int_1^{4 / 2}(3 Mg -3 kx ) d dx$
$\frac{\tau^2}{2}=\frac{1}{1+ m }\left(\frac{8 MgL }{2}-\frac{3 kxx _0^2}{2 \times 4}\right)$
$v ^2=\frac{1}{7 m }\left(\frac{8 Mg }{2} \times \frac{16 Mg }{3 m }-\frac{3 \pi}{8} \times \frac{16 M ^2 g ^2}{3 \pi \times 3 \pi }\right)$
$=\frac{1}{7 m }\left(\frac{64 M ^2 g ^2}{3 x }-\frac{2 M ^2 g ^2}{3 x }\right)$
$v ^2=\frac{62 Mg ^2}{21 k }$
For acc. $2 z_1-a_2+a_3$ therefore
$a_2-a_1=a_1-a_3$
$a _1=\frac{8 Mg -3 kx _0 / 4}{14 m }$
$=\frac{8 g }{14}-\frac{3 kx _0}{14 m \times 4}$
$=\frac{8 g }{14}-\frac{3 x }{14 m \times 4} \times \frac{16 Mg }{3 x }$
$=\frac{8 g }{14}-\frac{4 g }{14}$
$=\frac{4 g }{14}=\frac{2 g }{7}$
Standard 11
Physics

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