Question

Two blocks of masses 20 kg and 0.50 kg are attached at the two ends of a
compressed spring. The elastic potential energy stored in the spring is 10 J. Find the
velocities of the blocks if the spring delivers its energy to the blocks when released. ​

Answer 1

Mass of first block, $\displaystyle\sf {m_1=20\ kg}$

Mass of second block, $\displaystyle\sf {m_2=0.5\ kg}$

At equilibrium, let the velocities of the blocks be $\displaystyle\sf {v_1}$ and $\displaystyle\sf {v_2}$ respectively.

Here the center of mass of the system has no displacement, no velocity, since it has no acceleration.

$\displaystyle\longrightarrow\sf{\bar v=0}$

$\displaystyle\longrightarrow\sf{\dfrac {m_1v_1+m_2v_2}{m_1+m_2}=0}$

$\displaystyle\longrightarrow\sf{m_1v_1+m_2v_2=0}$

$\displaystyle\longrightarrow\sf{m_1v_1=-m_2v_2}$

$\displaystyle\longrightarrow\sf{20v_1=-0.5v_2}$

$\displaystyle\longrightarrow\sf{v_2=-40v_1\quad\quad\dots (1)}$

When spring is compressed, the blocks have no velocity, hence their kinetic energies are zero. But potential energy due to compression of spring (10 J) exists.

At equilibrium, blocks have velocities, hence their kinetic energies exist. But spring has zero potential energy because it is in normal state.

Thus by conservation of total mechanical energy,

$\displaystyle\longrightarrow\sf{\dfrac {1}{2}m_1(v_1)^2+\dfrac {1}{2}m_2(v_2)^2=10}$

$\displaystyle\longrightarrow\sf{20(v_1)^2+0.5(v_2)^2=20}$

From (1),

$\displaystyle\longrightarrow\sf{20(v_1)^2+0.5(-40v_1)^2=20}$

$\displaystyle\longrightarrow\sf{20(v_1)^2+800(v_1)^2=20}$

$\displaystyle\longrightarrow\sf{820(v_1)^2=20}$

$\displaystyle\longrightarrow\sf{(v_1)^2=\dfrac {1}{41}}$

$\displaystyle\longrightarrow\underline {\underline {\sf{v_1=\sqrt {\dfrac {1}{41}}\ m\,s^{-1}}}}$

And from (1),

$\displaystyle\longrightarrow\underline {\underline {\sf{v_2=-40\sqrt {\dfrac {1}{41}}\ m\,s^{-1}}}}$

Negative sign indicates the blocks mone in opposite directions.