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Two particles with mass $m_1$ = $16\ kg$ and $m_2$ = $2\ kg$ slide as unit with a common velocity of $12\ ms^{-1}$ on a level frictionless surface. Between them is a compressed massless spring with spring constant $k$ = $100\ Nm^{-1}$ . The spring, originally compressed by $25\ cm$ , is suddenly released, sending the two masses, which are connected to the spring, flying apart from each other. The orientation of the spring w.r.t. the initial velocity is shown in diagram. What is the relative velocity of separation in $ms^{-1}$ , after the particles lose contact? ................$m/s$
$3.88$
$22$
$1.88$
$0$
Solution
By using reduced mass concept Here $\mathrm{u}_{\mathrm{rel}}=0$
$v_{\text {rel }}=?, \mu=\frac{m_{1} m_{2}}{m_{1}+m_{2}}=\frac{16 \times 2}{16+2}=\frac{16}{9} \mathrm{kg}$
$\Rightarrow \frac{1}{2} \mathrm{kx}^{2}=\frac{1}{2} \mu \mathrm{v}_{\mathrm{rel}}^{2} \Rightarrow \mathrm{v}_{\mathrm{rel}}=\sqrt{\frac{\mathrm{kx}^{2}}{\mu}}$
$=\sqrt{\frac{100 \times(1 / 4)^{2}}{16 / 9}}=\frac{30}{16} \mathrm{ms}^{-1}=1.875 \mathrm{ms}^{-1}$
$=1.88 \mathrm{ms}^{-1}$
