Animated Solution for Physics - System of Particles: Two masses $A$ and $B$, each of mass $M$ are fixed together by a massless spring. A force acts on the mass $B$ as shown in figure. If the mass $A$ starts moving away from mass $B$ with acceleration $a$, then the acceleration of mass $B$ wil be

A Spring, Two Masses, and a Pull

Imagine a perfectly smooth horizontal surface. On this surface, we have two identical blocks, A and B, each possessing a mass $M$. They are intimately connected by a massless spring, forming a single, unified system.

Suddenly, a constant external force $F$ begins to pull block B towards the left. As block B moves, the spring stretches, and this tension begins to drag block A along with it. We are told that block A accelerates with a value $a$. Our mission? To find the exact acceleration of block B, which we will call $a_B$.

When dealing with multiple interacting bodies, the Center of Mass (CM) is our greatest ally. Why? Because internal forces—like the tension in our spring—completely cancel out when we look at the system as a whole.

The only external force acting on our two-block system in the horizontal direction is $F$. According to Newton's Second Law for a system of particles, the acceleration of the center of mass $a_{CM}$ is dictated solely by this external force and the total mass of the system.

Since the total mass is $M+M=2M$, we can write:

We now know how the "invisible" center of mass is accelerating. But how does this relate to the actual physical blocks? Kinematics gives us a beautiful bridge. The acceleration of the center of mass is simply the mass-weighted average of the individual accelerations of the particles.

Let's plug in what we know. Both masses are $M$, the acceleration of A is $a$, and the acceleration of B is $a_B$.

Look at that equation! The mathematics is practically begging us to simplify it. Since both sides share a denominator of $2M$, we can cleanly cancel it out.

We are hunting for $a_B$. Let's isolate the term containing it by shifting $Ma$ to the other side.

Finally, dividing the entire equation by $M$ reveals our target:

And there we have it! The acceleration of block B is elegantly expressed in terms of the applied force, the mass, and the acceleration of block A.

Free Body Diagrams

While the Center of Mass method is conceptually profound, there is a faster, "ninja" way to solve this using Free Body Diagrams (FBDs).

Let the spring force be $kx$. For block A, the only horizontal force is the spring pulling it. So, $kx=Ma$. For block B, the external force $F$ pulls it forward, while the spring force $kx$ pulls it back. So, $F-kx=M{a}_B$.

Substitute the first equation into the second:

Divide by $M$, and boom! You arrive at the exact same answer in just two lines of algebra. Physics is beautiful when multiple paths lead to the same truth!