Two bodies A and B of masses m and 2m respectively are placed on a smooth floor. They are connected

Q: Wait, why do we assume block C comes to a complete stop after hitting A?

This is a fundamental property of 1D perfectly elastic collisions between two objects of **equal mass**. When they collide, they completely exchange their velocities. Since block $A$ was initially at rest ($v=0$), block $C$ takes on that zero velocity, and block $A$ takes on block $C$'s initial velocity $v_0$.

Q: Why can't we just apply conservation of momentum to all three blocks (A, B, and C) from the very beginning to the very end?

You actually can! If you take the system as $A+B+C$, the initial momentum is $m v_0$. At maximum compression, $C$ is at rest, and $A$ and $B$ move with velocity $v$. So, $m v_0 = m(0) + (m+2m)v$, which gives $m v_0 = 3m v$. This yields the exact same common velocity $v = \frac{v_0}{3}$. However, breaking it into two distinct phases (the collision, then the spring compression) makes the energy analysis much clearer.

Q: Why are the velocities of block A and block B exactly the same at the moment of maximum compression?

Think about relative motion. If block $A$ was moving faster than block $B$, it would still be pushing into the spring, compressing it further. If block $A$ was moving slower than block $B$, the spring would be expanding. Therefore, the only instant the compression reaches its absolute maximum (and stops increasing) is when their relative velocity is zero—meaning they are moving at the exact same speed.

Q: Why is the total mass $3m$ used in the final kinetic energy term?

At the instant of maximum compression, both block $A$ (mass $m$) and block $B$ (mass $2m$) are moving together with the same common velocity $v$. Since kinetic energy is a scalar, we can simply add their kinetic energies together: $\frac{1}{2} m v^2 + \frac{1}{2} (2m) v^2 = \frac{1}{2} (3m) v^2$. It's mathematically equivalent to treating them as a single combined mass of $3m$ for that specific instant.

Q: Does the spring stay compressed at $x_0$ forever?

No! The state of maximum compression is just a fleeting moment. Immediately after $t_0$, the highly compressed spring will push block $A$ backward (relative to the center of mass) and push block $B$ forward. They will begin to oscillate back and forth while the entire system continues to drift to the right.

Question

Two bodies $A$ and $B$ of masses $m$ and $2m$ respectively are placed on a smooth floor. They are connected by a spring. A third body $C$ of mass $m$ moves with velocity $v_0$ along the line joining $A$ and $B$ and collides elastically with $A$ as shown in figure. At a certain instant of time $t_0$ after collision, it is found that the instantaneous velocities of $A$ and $B$ are the same. Further at this instant the compression of the spring is found to be $x_0$. Determine (a) the common velocity of $A$ and $B$ at time $t_0$ and (b) the spring constant.

Sigma Percentile · Accepted Answer

The beauty of physics often lies in how seemingly complex interactions can be elegantly unraveled using fundamental conservation laws. In this problem, we are presented with a fascinating sequence of events: a high-speed collision followed by the dynamic compression of a spring.

Imagine you are observing this setup on a perfectly smooth, frictionless floor. We have three blocks: $A$, $B$, and $C$. Blocks $A$ and $B$ are connected by a spring, patiently waiting. Block $C$, acting as the instigator, hurtles towards block $A$ with an initial velocity $v_0$.

### The Elastic Exchange

The first major event is the collision between block $C$ and block $A$. The problem states that this collision is perfectly elastic. But there is a crucial detail here: block $C$ and block $A$ have the exact same mass, $m$.

**When two identical masses undergo a perfectly elastic head-on collision, they completely exchange their velocities.**

This is a beautiful and highly useful shortcut in physics! Because block $A$ was initially at rest, block $C$ will transfer all of its momentum and kinetic energy to block $A$. Immediately after the impact, block $C$ comes to a dead stop, and block $A$ shoots forward with the velocity $v_0$. At this exact instant, the spring has not yet begun to compress, and block $B$ is still blissfully unaware of the chaos, remaining at rest.

### The Dance of Momentum

As block $A$ surges forward, it begins to compress the spring. The spring, resisting this compression, exerts a backward force on block $A$ (slowing it down) and a forward force on block $B$ (speeding it up).

The problem asks us to analyze the system at a specific time $t_0$, when the instantaneous velocities of $A$ and $B$ are identical. Let's call this common velocity $v$. This state of identical velocities corresponds to the moment of **maximum compression** of the spring, denoted as $x_0$.

To find this common velocity, we look at the system comprising blocks $A$ and $B$ (and the massless spring). Because the floor is smooth, there are absolutely no external horizontal forces acting on this system.

**Therefore, the linear momentum of the system must be conserved.**

Let's set up our momentum equation. The initial momentum of our system (just after the collision) is entirely due to block $A$:
$$P_{	ext{initial}} = m_A v_0 = m v_0$$

At time $t_0$, both blocks are moving together with velocity $v$. The final momentum is:
$$P_{	ext{final}} = (m_A + m_B) v = (m + 2m) v = 3m v$$

Equating the initial and final momentum:
$$m v_0 = 3m v$$

Solving for $v$, we find the common velocity:
$$v = \frac{v_0}{3}$$

This elegantly answers the first part of our problem!

### The Energy Ledger

Now, we must determine the spring constant, $k$. While momentum tells us about the velocities, energy will tell us about the forces and compressions.

Since there is no friction, the total mechanical energy of the system is conserved. The energy we start with must equal the energy we end up with at time $t_0$.

Just after the collision, the only energy in the system is the kinetic energy of block $A$:
$$E_{	ext{initial}} = \frac{1}{2} m_A v_0^2 = \frac{1}{2} m v_0^2$$

At time $t_0$, the energy is distributed into two forms: the kinetic energy of the moving blocks and the elastic potential energy stored in the compressed spring. 
$$E_{	ext{final}} = \frac{1}{2} (m_A + m_B) v^2 + \frac{1}{2} k x_0^2$$
$$E_{	ext{final}} = \frac{1}{2} (3m) v^2 + \frac{1}{2} k x_0^2$$

Now, we substitute the common velocity $v = \frac{v_0}{3}$ that we found earlier into our energy equation:
$$\frac{1}{2} m v_0^2 = \frac{1}{2} (3m) \left(\frac{v_0}{3}ight)^2 + \frac{1}{2} k x_0^2$$

Let's carefully simplify the kinetic energy term on the right side:
$$\frac{1}{2} (3m) \left(\frac{v_0^2}{9}ight) = \frac{3m v_0^2}{18} = \frac{1}{6} m v_0^2$$

Substituting this back into our main equation:
$$\frac{1}{2} m v_0^2 = \frac{1}{6} m v_0^2 + \frac{1}{2} k x_0^2$$

To isolate the spring energy, we subtract the final kinetic energy from the initial kinetic energy:
$$\frac{1}{2} k x_0^2 = \frac{1}{2} m v_0^2 - \frac{1}{6} m v_0^2$$

Finding a common denominator (which is 6):
$$\frac{1}{2} k x_0^2 = \frac{3}{6} m v_0^2 - \frac{1}{6} m v_0^2 = \frac{2}{6} m v_0^2 = \frac{1}{3} m v_0^2$$

Finally, we solve for the spring constant $k$ by multiplying both sides by 2 and dividing by $x_0^2$:
$$k x_0^2 = \frac{2}{3} m v_0^2$$
$$k = \frac{2 m v_0^2}{3 x_0^2}$$

### Conclusion

By sequentially applying the conservation of momentum and the conservation of energy, we have successfully decoded the dynamics of this system. The problem beautifully illustrates how energy is transferred from a free-moving mass into the internal potential energy of a bound system.

Visualized Solution

$I ni t ia l S e t u p$

$E l a s t i c C o l l i s i o n$

$P os t - C o l l i s i o n S t a t e$

$M a x im u m C o m p r ess i o n$

$C o n ser v a t i o n o f M o m e n t u m$

$M o m e n t u m E q u a t i o n$

$C o mm o nV e l oc i t y$

$C o n ser v a t i o n o f E n er g y$

$E n er g y E q u a t i o n$

$S u b s t i t u t in g V e l oc i t y$

$S im pl i f y in g K in e t i c E n er g y$

$I so l a t in g S p r in g E n er g y$

$S p r in g C o n s t an t$

The Sigma Insight: Head-on Collision

The Elastic Exchange

The Dance of Momentum

The Energy Ledger

Conclusion

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Visualized Solution

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