A massless spring with spring constant k is placed between a block of mass m and a block of mass 3m. Initially the blocks are at rest on a frictionless surface and they are held together so that the spring between them is compressed by an amount D from its equilibrium length. The blocks are then released and the spring pushes them off in opposite directions. Find the speeds of the two blocks when they detach from the spring.

The spring constant is k.

The masses of the two blocks are m and 3m.

The compression in the spring is D.

The principle of conservation of linear momentum states that if two bodies collide with each other, the total linear momentum before and after the collision remains the same if no external force acts on the system.It is given as:

\(\begin{aligned}{c}{p_{{\rm{before}}}} = {p_{{\rm{after}}}}\\{\left( {mv} \right)_{{\rm{before}}}} = {\left( {mv} \right)_{{\rm{after}}}}\end{aligned}\)

According to the principal of conservation of energy, the total energy of the system before the collision is equal to the total energy of the system after the collision.

It is given as:

\({\left( {KE + PE} \right)_{{\rm{before}}}} = {\left( {KE + PE} \right)_{{\rm{after}}}}\)

There is no external force acting on the system; so the total momentum is conserved.

Let the final velocity of mass m be\({v_{\rm{m}}}\), and that of mass 3m be\({v_{{\rm{3m}}}}\). Initially, the blocks are at rest. Therefore, their initial linear momentum is zero.

\(\begin{aligned}{c}{p_{{\rm{before}}}} = {p_{{\rm{after}}}}\\0 = m{v_{\rm{m}}} + 3m\left( { - {v_{{\rm{3m}}}}} \right)\\{v_{\rm{m}}} = 3{v_{{\rm{3m}}}}\end{aligned}\) … (i)

In this interaction, energy is conserved, i.e., the initial potential energy of the mass-spring system is equal to the final kinetic energy of the moving blocks. Therefore, applying the law of conservation of energy, you get:

\(\begin{aligned}{c}P{E_{{\rm{spring initial}}}} = K{E_{{\rm{final}}}}\\\frac{1}{2}k{D^2} = \frac{1}{2}mv_{\rm{m}}^2 + \frac{1}{2}\left( {3m} \right)v_{{\rm{3m}}}^2\\k{D^2} = mv_{\rm{m}}^2 + \left( {3m} \right)v_{{\rm{3m}}}^2\end{aligned}\) … (ii)

Substitute equation (i) in (ii).

\(\begin{aligned}{c}k{D^2} = m{\left( {3{v_{{\rm{3m}}}}} \right)^2} + \left( {3m} \right)v_{{\rm{3m}}}^2\\k{D^2} = 12mv_{{\rm{3m}}}^2\\{v_{{\rm{3m}}}} = D\sqrt {\frac{k}{{12m}}} \end{aligned}\)

Substitute the value of\({v_{{\rm{3m}}}}\)in equation (i).

\( \Rightarrow {v_{\rm{m}}} = 3D\sqrt {\frac{k}{{12m}}} \)

Thus the velocity of the mass m is \(3D\sqrt {\frac{k}{{12m}}} \).