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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$1.88$

Vedclass · Answer

(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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