Question

A spring is held compressed so that its stored energy is $2.4J$. its ends are in contact with masses 1 $g$ and $48g$ 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 $v_1=48v_2$
Using Eq. (i) $\frac{1}{1000}{\left(48v_2\right)}^2+\frac{48}{1000}{v}_2^2=4.8$
or $\frac{48}{1000}\left(49v_2^2\right)=4.8$
$\therefore v_2=\frac{10}{7}m/s=\frac{10^3}{7}cm/s$