Question

The system shown in the figure can move on a smooth surface. The spring is initially compressed by 6 cm and released. Find
(a) time period
(b) amplitude of 3 kg block
(c) maximum momentum of 6 kg block

Answer 1

$\begin{array}{c}\left(a\right)T=2\pi \sqrt{\frac{\mu }{k}}\mu =\frac{m_1{m}_2}{m_1+m_2}\\ =2\pi \sqrt{\frac{2}{800}}\mu =\frac{3\times 6}{3+6}=2\\ =\frac{2\pi }{20}=\frac{\pi }{10}\sec \end{array}$

$\begin{array}{c}\left(b\right)\frac{1}{2}.3.V_1^2+\frac{1}{2}.6.V_2^2=\frac{1}{2}.k.x^2\\ \frac{1}{2}.3.4V_1^2+\frac{1}{2}.6.V_2^2=\frac{1}{2}.k.x^2\\ 18.V_2^2=800\times \frac{6}{100}\times \frac{6}{100}\\ V_2=\frac{2}{5}m/s\\ V_1=\frac{4}{5}m/s\end{array}$

So, Max magn of momentum of $6kg=6\times \frac{2}{5}=\frac{12}{5}m/s$

$\left(c\right)$ Individual block can be assumed to perform $SHM$ about caom.

$\begin{array}{c}V=A.W\\ \frac{2}{5}=A\\ \frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2\\ \frac{1}{2}.500.x^2=\frac{1}{2}\times 3\times \frac{4}{25}\\ x^2=\frac{3}{25.200}\\ x=\frac{1}{50}\sqrt{\frac{3}{2}}m\end{array}$