The system shown in the figure can move on a smooth surface. The spring is initially compressed by 6 cm and then released .

The correct options are
A The particles perform SHM with time period $\frac{\pi }{10}\text{sec}$
B The block of mass 3 kg perform SHM with amplitude $4\text{cm}$
C The block of mass 6 kg will have maximum momentum $2.40\text{kg m/s}$
Time period of the oscillation : $T=2\pi \sqrt{\frac{\mu }{K}}$
Where, $\mu =\frac{m_1{m}_2}{m_1+m_2}=\frac{3\times 6}{6+3}=\frac{18}{9}=2\text{kg},$
$K=800\text{N/m}$
$K=$ spring constant
Put these values in formula
$\therefore T=2\pi \sqrt{\frac{2}{800}}=2\pi \times \frac{1}{20}$
$\because T=\frac{\pi }{10}\text{sec.}$
(B) Here, $x_0=6\text{cm}$
$x_1+x_2=6...\left(1\right)$
$3x_1=6x_2$
$x_1=2x_2$
from eqn. (1)
$2x_2+x_2=6$
$3x_2=6$
$x_2=2\text{cm}$
put the value of $x_2$ in equation
and $x_1+2=6$
$x_1=4\text{cm}$
So, the block of mass $3\text{kg}$ perform SHM with amplitude $4\text{cm}$
(C) applying conservation of momentum -
$3v_1-6v_2=0$
$3v_1=6v_2$
$v_1=2v_2$
Applying energy conservation -
$\therefore \frac{1}{2}\times 3v_1^2+\frac{1}{2}\times 6\times v_2^2=\frac{1}{2}k{x}_0^2$
$\frac{3}{2}{v}_1^2+3v_2^2=\frac{1}{2}\times 800\times (0.06{)}^2$
put the value of $v_1$ in this -
$\frac{3}{2}(2v_2{)}^2+3v_2^2=\frac{1}{2}\times 800\times 0.0036$
$6v_2^2+3v_2^2=400\times 36\times {10}^{-4}$
$9v_2^2=1.44$
$v_2=\sqrt{\frac{1.44}{9}}=\frac{1.2}{3}$
$v_2=0.4\text{m /sec}$
So, the block of mass $6\text{kg}$ will have maximum momentum -
$P_{\text{max}}=6\times 0.4$
$P_{\text{max}}=2.40\text{kg~m/sec}.$