
Applying linear momentum conservation,
$0.5 \times 3=(1+0.5) v \text { or } v =1 m / s$
By conversation of energy,
After collision
$\frac{1}{2}(1+0.5) v ^2=\frac{1}{2} kA ^2$
$\Rightarrow A =\sqrt{\frac{1.5}{ k }} \times v$
$\Rightarrow A =\sqrt{\frac{1.5}{600}} \times 1=\frac{1}{20} m =0.05 m$
$A =5 cm$
Time period of oscillation,
$T =2 \pi \sqrt{\frac{ m _1+ m _2}{ k }}=2 \pi \sqrt{\frac{1.5}{600}}=\frac{2 \pi}{20}=\frac{\pi}{10} s$

$x = 3\,sin\, 20\pi t + 4\, cos\, 20\pi t$ ,
where $x$ is in $cms$ and $t$ is in $seconds$ . The amplitude is ..... $cm$

