Three blocks A, B and C are lying on a smooth horizontal surface as shown in the figure. A and B hav

Q: Why is kinetic energy not conserved in this collision?

In a perfectly inelastic collision, the bodies stick together. A significant amount of macroscopic kinetic energy is transformed into internal energy (like heat, sound, or deformation energy) to bind the bodies together. Therefore, kinetic energy is never conserved in inelastic collisions.

Q: Can we apply conservation of momentum if energy is lost?

Absolutely! The principle of conservation of linear momentum depends only on the absence of external forces. Since the forces during the collision are strictly internal to the system of blocks, momentum is perfectly conserved, regardless of what happens to the kinetic energy.

Q: Why did we calculate the final energy as $\frac{1}{6}$th of the initial energy instead of $\frac{5}{6}$th?

The problem states that $\frac{5}{6}$th of the initial kinetic energy is *lost*. The energy that remains in the system (the final kinetic energy) is the initial energy minus the lost energy: $1 - \frac{5}{6} = \frac{1}{6}$. Equating the final energy to $\frac{5}{6}$th is a very common silly mistake!

Q: Could we have treated this as a single collision where $A$ hits $B$ and $C$ simultaneously?

While the final momentum would be the same, physically, the collisions happen sequentially. $A$ hits $B$, they form a combined mass, and then this combined mass hits $C$. Treating them sequentially ensures we understand the physical process, even though the final velocity expression would mathematically look the same if we just equated initial momentum of $A$ to the final momentum of $A+B+C$.

Q: What would happen if the surface wasn't smooth?

If the surface had friction, there would be an external force acting on the system. We could no longer apply the conservation of linear momentum over the entire duration of motion. We would have to use the work-energy theorem to account for the work done by friction between the collisions.

Question

Three blocks $A, B$ and $C$ are lying on a smooth horizontal surface as shown in the figure. $A$ and $B$ have equal masses $m$ while $C$ has mass $M$. Block $A$ is given an initial speed $v$ towards $B$ due to which it collides with $B$ perfectly inelastically. The combined mass collides with $C$, also perfectly inelastically $\frac{5}{6}$th of the initial kinetic energy is lost in whole process. What is value of $\frac{M}{m}$?

Options:

(A) 4

(B) 2

(C) 3

(D) 5

Sigma Percentile · Accepted Answer

The problem of colliding blocks is a classic in physics, testing our understanding of two fundamental pillars: the conservation of momentum and the work-energy theorem. Let's embark on this journey to unravel the mystery of the missing energy!

### Analyzing the Setup

Imagine you are standing in a frictionless laboratory. On a perfectly smooth horizontal table, we have three blocks lined up: $A$, $B$, and $C$.

Block $A$ and block $B$ are identical twins, each possessing a mass $m$. Block $C$ is the larger sibling, with an unknown mass $M$. Initially, blocks $B$ and $C$ are completely at rest, just chilling on the surface. Suddenly, block $A$ is given a sharp push, sending it sliding towards block $B$ with a constant velocity $v$.

Because the surface is smooth, there are no external horizontal forces like friction to slow block $A$ down. This means the total linear momentum of our system is going to be our best friend throughout this problem.

### The First Collision: A Meets B

As block $A$ reaches block $B$, a **perfectly inelastic collision** occurs. What does this mean in the physical world? It means the two blocks don't bounce off each other; instead, they crash, deform slightly, and stick together, moving forward as a single combined entity.

Even though kinetic energy is lost in the deformation, the **conservation of linear momentum** holds absolutely true. Let's apply it to this first impact.

The initial momentum of the system just before the collision is entirely due to block $A$:
$$p_i = m v$$

After the collision, blocks $A$ and $B$ combine to form a new mass of $(m + m) = 2m$. Let's say this new combined block moves with a velocity $v'$. The final momentum is:
$$p_f = (2m) v'$$

Equating the initial and final momentum:
$$m v = 2m v'$$

Solving for the new velocity, we find:
$$v' = \frac{v}{2}$$

So, the combined mass of $2m$ is now lumbering forward at exactly half the original speed.

### The Second Collision: The Giant Mass Meets C

But the journey isn't over! This new combined mass ($2m$) is now on a collision course with the stationary block $C$ (mass $M$).

Once again, the problem states this is a perfectly inelastic collision. The combined mass $2m$ crashes into $M$, and they all stick together to form an even larger super-block of mass $(2m + M)$. Let's call their final, ultimate velocity $v''$.

We apply the conservation of momentum one more time. The momentum before this second collision is the momentum of the $2m$ block:
$$p_i = (2m) v' = (2m) \left(\frac{v}{2}\right) = m v$$

Notice how the total momentum of the system has remained exactly $m v$ from the very beginning!

The momentum after they all stick together is:
$$p_f = (2m + M) v''$$

Equating them gives us the final velocity of the entire system:
$$m v = (2m + M) v''$$
$$v'' = \frac{m v}{2m + M}$$

### The Energy Constraint: The Crucial Clue

Now we arrive at the heart of the problem. We are told that $\frac{5}{6}$th of the initial kinetic energy is lost during this entire chaotic process.

This is where many students make a critical error. If $\frac{5}{6}$th is *lost*, we must focus on what *remains*. The final kinetic energy $K_f$ must be exactly $\frac{1}{6}$th of the initial kinetic energy $K_i$.

Let's write down our energies. The initial kinetic energy was just block $A$ moving:
$$K_i = \frac{1}{2} m v^2$$

The final kinetic energy is the entire super-block moving together:
$$K_f = \frac{1}{2} (2m + M) (v'')^2$$

According to our constraint:
$$K_f = \frac{1}{6} K_i$$

### Final Calculation

Now, it's time for the grand finale. Let's substitute our expressions into the energy constraint equation:
$$\frac{1}{2} (2m + M) (v'')^2 = \frac{1}{6} \left( \frac{1}{2} m v^2 \right)$$

We already found our final velocity $v'' = \frac{m v}{2m + M}$. Let's plug that in carefully:
$$\frac{1}{2} (2m + M) \left( \frac{m v}{2m + M} \right)^2 = \frac{1}{12} m v^2$$

Squaring the velocity term:
$$\frac{1}{2} (2m + M) \frac{m^2 v^2}{(2m + M)^2} = \frac{1}{12} m v^2$$

Notice how beautifully the physics simplifies the math! One factor of $(2m + M)$ cancels out in the numerator and denominator on the left side:
$$\frac{1}{2} \frac{m^2 v^2}{(2m + M)} = \frac{1}{12} m v^2$$

Now, we can divide both sides by $m v^2$ (since neither mass nor velocity is zero):
$$\frac{m}{2(2m + M)} = \frac{1}{12}$$

Multiply both sides by 2:
$$\frac{m}{2m + M} = \frac{1}{6}$$

Cross-multiplying yields a simple linear equation:
$$6m = 2m + M$$

Subtracting $2m$ from both sides reveals the mass of block $C$:
$$M = 4m$$

The question asks for the ratio of $M$ to $m$. Dividing by $m$, we get our final, elegant answer:
$$\frac{M}{m} = 4$$

Through the rigorous application of momentum conservation and careful accounting of energy, we've successfully decoded the system!

Select Answer:

Visualized Solution

$Initial Setup$

$First Collision (A and B)$

$Velocity after 1st Collision$

$Second Collision (A+B and C)$

$Velocity after 2nd Collision$

$Kinetic Energy Loss$

$Equating Final Kinetic Energy$

$Substituting v^{''}$

$Solving for M$

$Final Ratio$

$Food for Thought$

The Sigma Insight: Head-on Collision

Analyzing the Setup

The First Collision

The Second Collision

The Energy Constraint

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Collision (A and B)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity after 1st Collision

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Second Collision (A+B and C)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity after 2nd Collision

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kinetic Energy Loss

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Final Kinetic Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting v′′

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for M

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Food for Thought