Two particles with mass m_1 = 16 kg and m_2 | vedclass

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(C) The potential energy stored in the compressed spring is converted into the kinetic energy of the particles in the center of mass frame.Let the rel

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$1.88$

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(C) The potential energy stored in the compressed spring is converted into the kinetic energy of the particles in the center of mass frame.Let the relative velocity of separation be $v_{rel}$.The reduced mass of the system is $\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{16 	imes 2}{16 + 2} = \frac{32}{18} = \frac{16}{9} \ kg$.The energy conservation equation is $\frac{1}{2} kx^2 = \frac{1}{2} \mu v_{rel}^2$.Given $k = 100 \ Nm^{-1}$ and $x = 25 \ cm = 0.25 \ m = \frac{1}{4} \ m$.Substituting the values: $\frac{1}{2} 	imes 100 	imes (\frac{1}{4})^2 = \frac{1}{2} 	imes \frac{16}{9} 	imes v_{rel}^2$.$100 	imes \frac{1}{16} = \frac{16}{9} 	imes v_{rel}^2$.$v_{rel}^2 = \frac{100}{16} 	imes \frac{9}{16} = \frac{900}{256}$.$v_{rel} = \sqrt{\frac{900}{256}} = \frac{30}{16} = 1.875 \ ms^{-1}$.Rounding to two decimal places,we get $v_{rel} \approx 1.88 \ ms^{-1}$.

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