Question

A spring is held compressed so that its stored energy is $2.4 \,J$. its ends are in contact with masses 1 $g$ and $48\, g$ placed on a functionless table. When the spring is released, the heavier mass wilt acquire a speed of

• A. $ \frac{2.4}{49}m{{s}^{-1}} $
• B. $ \frac{2.4\times 48}{49}m{{s}^{-1}} $
• C. $ \frac{{{10}^{3}}}{7}cm{{s}^{-1}} $
• D. $ \frac{{{10}^{6}}}{7}cm{{s}^{-1}} $

Answer 1

Answer: (C)

$\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}=2.4$
or $m_{1} v_{1}^{2}+m_{2} v_{2}^{2}=4.8 ?( i )$
Now, $m_{1} v_{1}=m_{2} v_{2}$
or $v_{1}=48 v_{2}$
Using Eq. (i) $\frac{1}{1000}\left(48 v_{2}\right)^{2}+\frac{48}{1000} v_{2}^{2}=4.8$
or $\frac{48}{1000}\left(49 v_{2}^{2}\right)=4.8$
$\therefore v_{2}=\frac{10}{7} m / s=\frac{10^{3}}{7} cm / s$