Question

By what fraction does the mass of a spring change when it is compressed by 1 cm? The mass of the spring is 200 g at its natural length and the spring constant is 500 N m${}^{-1}$

Answer 1

Given : $x=1$ cm $=0.01$ m $m_o=0.2$ kg

Using energy conservation : $P.E_i+K.E_i=P.E_f+K.E_f$

$\therefore$ $\frac{1}{2}k{x}^2+0=0+\frac{1}{2}{m}_o{v}^2$ $⟹v=\sqrt{\frac{k}{m_o}}x$

$\therefore$ $v=\sqrt{\frac{500}{0.2}}\times 0.01=0.5$ m/s

Mass of the boy while running $m=\frac{m_o}{\sqrt{1-(\frac{v^2}{c^2}{)}^2}}$

$\therefore$ Fractional increase in the mass $\frac{\Delta m}{m_o}=\frac{\frac{m_o}{\sqrt{1-(v/c{)}^2}}-m_o}{m_o}=[1-(v/c{)}^2{]}^{-1/2}-1$

$⟹$ $\frac{\Delta m}{m_o}=1+\frac{v^2}{2c^2}-1=\frac{v^2}{2c^2}$ (for $v<<c$)

$\therefore$ $\frac{\Delta m}{m_o}=\frac{(0.5{)}^2}{2\times (3\times {10}^8{)}^2}=1.4\times {10}^{-18}$ kg