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# A light spring of spring constant $k$ is kept compressed between two blocks of masses $m$ and $M$ on a smooth horizontal surface. When released, the blocks acquire velocities in opposite directions. The spring loses contact with the blocks when it acquires natural length. If the spring was initially compressed through a distance $x$, find the final speed of the block of mass $M$:$(A){\left[ {\dfrac{{kM}}{{M(M + m)}}} \right]^{\dfrac{1}{2}}}x$ $(B){\left[ {\dfrac{{km}}{{M(M + m)}}} \right]^{\dfrac{1}{2}}}x$$(C){\left[ {\dfrac{M}{{M(M + m)}}} \right]^{\dfrac{1}{2}}}x$$(D){\left[ {\dfrac{m}{{M(M + m)}}} \right]^{\dfrac{1}{2}}}x$

Last updated date: 18th Jun 2024
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Hint: Use the formula of conservation of linear momentum by taking external force is zero. State an equation of conservation of energy where initial energy is due to the compression of the spring and final energy due to the motion of the blocks. Calculate the velocity of the big block from these two conservation formulas.

Formula used:
From the momentum conservation law,
$m{v_1} = M{v_2}$ where, $m$ and $M$ are the masses of block-1 and block-2 respectively.
${v_1}$ and ${v_2}$ are the velocities of block-1 and block-2 respectively after losing the contact of the spring.
From the energy conservation law,
$\dfrac{1}{2}k{x^2} = \dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}M{v_2}^2$
$k$ Is the spring constant and $x$ is the expansion due to the compression of the spring.

The two blocks of masses $m$ and $M$ are attached through a light spring of spring constant $k$ and expanded by the length $x$.
After losing the contact of the spring, block-1 is moving with a velocity ${v_1}$ and block-2 is moving with a velocity ${v_2}$.
The diagram is shown below,

According to the momentum for the conservation and we can write it as
$m{v_1} = M{v_2}$ [ $m$ and $M$ are the masses of block-1 and block-2 respectively.]
$\Rightarrow {v_1} = \dfrac{{M{v_2}}}{m}...................(1)$
The initial energy is due the compression of the spring, hence ${E_1} = \dfrac{1}{2}k{x^2}$
$k$ is the spring constant and $x$ is the expansion due to the compression of the spring.
And, the final energy will be due to the motion of the blocks, hence ${E_2} = \dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}M{v_2}^2$
From the energy conservation law,
${E_1} = {E_2}$
$\Rightarrow \dfrac{1}{2}k{x^2} = \dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}M{v_2}^2.................(2)$
By putting the value of ${v_1}$ from eq $(1)$ in the eq. $(2)$ we get,
$\Rightarrow \dfrac{1}{2}k{x^2} = \dfrac{1}{2}m{\left( {\dfrac{{M{v_2}}}{m}} \right)^2} + \dfrac{1}{2}M{v_2}^2$
$\Rightarrow k{x^2} = \dfrac{{{M^2}{v_2}^2}}{m} + M{v_2}^2$
$\Rightarrow M{v_2}^2\left( {\dfrac{M}{m} + 1} \right) = k{x^2}$
$\Rightarrow M{v_2}^2\left( {\dfrac{{M + m}}{m}} \right) = k{x^2}$
$\Rightarrow {v_2}^2M(M + m) = km{x^2}$
$\Rightarrow {v_2}^2 = \dfrac{{km}}{{M(M + m)}}{x^2}$
$\Rightarrow {v_2} = {\left[ {\dfrac{{km}}{{M(M + m)}}} \right]^{\dfrac{1}{2}}}x$
So, the velocity of the block-2 will be, ${v_2} = {\left[ {\dfrac{{km}}{{M(M + m)}}} \right]^{\dfrac{1}{2}}}x$

Hence the correct answer is in option $(A)$.

Note: We know that the force is the change of the linear momentum of the objects in a system. Since, there is o external force acting horizontally on the system consisting of two blocks and a spring, the linear momentum will be zero. Hence the equation be like, $m{v_1} - M{v_2} = 0$. This concept leads to the concept of conservation of linear momentum.