A spring is compressed between two blocks of masses $m_1$ and $m_2$ placed on a horizontal frictionless…
A spring is compressed between two blocks of masses $m_1$ and $m_2$ placed on a horizontal frictionless surface as shown in the figure. When the blocks are released, they have initial velocity of $v_1$ and $v_2$ as shown. The blocks travel distances $x_1$ and $x_2$ respectively before coming to rest. The ratio $\left(\frac{x_1}{x_2}\right)$ is
$\frac{m_2}{m_1}$
$\frac{m_1}{m_2}$
$\sqrt{\frac{m_2}{m_1}}$
$\sqrt{\frac{m_1}{m_2}}$
Solution
Since the blocks are released from rest, the initial kinetic energy of the system is zero. When the blocks are released, they gain kinetic energy due to the potential energy stored in the compressed spring.
The final kinetic energy of block 1 is $\frac{1}{2}m_1v_1^2$ and that of block 2 is $\frac{1}{2}m_2v_2^2$. According to the conservation of energy, the total kinetic energy of the system remains constant. Therefore, we can equate the initial and final kinetic energy of the system.
The initial kinetic energy of the system is zero, so:
$
\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_2v_2^2
$
Solving this equation gives us the relationship between $v_1$ and $v_2$:
$
v_1 = \sqrt{\frac{m_2}{m_1}}v_2
$
Now, we know that the work done by the force of friction (which is zero) is equal to the change in kinetic energy. Therefore, the work done on each block by the spring force is equal to the final kinetic energy of the block. This gives us:
$
\frac{1}{2}m_1v_1^2 = m_1gx_1
$
$
\frac{1}{2}m_2v_2^2 = m_2gx_2
$
Solving these equations gives us:
$
x_1 = \frac{v_1^2}{2g}
$
$
x_2 = \frac{v_2^2}{2g}
$
Substituting the value of $v_1$ from equation (2) into equation (5) gives us the ratio $\frac{x_1}{x_2}$:
$
\frac{x_1}{x_2} = \frac{\frac{m_2}{m_1}v_2^2}{2g} \div \frac{v_2^2}{2g} = \frac{m_2}{m_1}
$
Therefore, the correct answer is Option A) $\frac{m_2}{m_1}$.
