Blocks of masses m, 2m, 4m and 8m are arranged in a line on a frictionless floor. Another block of m

Q: Why is kinetic energy not conserved here?

Because the collisions are perfectly inelastic. In such collisions, the bodies stick together, and a maximum amount of kinetic energy is permanently lost to the environment in the form of heat, sound, and internal deformation.

Q: Do I need to calculate the velocity after each individual collision?

No, you don't! Because momentum is conserved for the system as a whole from the very beginning to the very end, you can skip the intermediate steps and just equate the initial state to the final combined state.

Q: What exactly does 'perfectly inelastic' mean?

It means the coefficient of restitution ($e$) is zero. Practically, it means the colliding bodies stick together and move with a common, shared velocity after the impact.

Q: Why did we add all the masses together for the final momentum?

Because after all the perfectly inelastic collisions have occurred, every single block is stuck to the one in front of it, forming one giant block of mass $16m$ moving together.

Q: How do we find the percentage loss?

It's the ratio of the lost energy to the original energy, expressed as a percentage. You calculate it using the formula: $(\Delta K / K_i) \times 100$.

Question

Blocks of masses $m, 2m, 4m$ and $8m$ are arranged in a line on a frictionless floor. Another block of mass $m$, moving with speed $v$ along the same line (see figure) collides with mass $m$ in perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time, the last block of mass $8m$ starts moving, the total energy loss is $p\%$ of the original energy. Value of $p$ is close to

Options:

(A) 77

(B) 87

(C) 94

(D) 37

Sigma Percentile · Accepted Answer

Imagine you are standing on a perfectly frictionless floor, watching a fascinating chain reaction unfold. A single block of mass $m$ is sliding smoothly with a velocity $v$. Lined up ahead of it are four stationary blocks, progressively getting heavier: $m$, $2m$, $4m$, and a massive $8m$.

I know this might look like a terrifying chain reaction of collisions, but let's take a breath. The problem states that every single collision is **perfectly inelastic**. This is a crucial keyword! It means that whenever two blocks collide, they don't bounce off each other; instead, they stick together like wet clay and move as one combined mass.

### The Master Equation: Conservation of Momentum

Here is the beautiful part about physics: no matter how messy or sticky the collisions get, as long as there are no external horizontal forces (like friction), the **total linear momentum of the system is strictly conserved**.

We don't need to calculate the velocity after the first collision, then the second, and so on. We can just look at the very beginning and the very end!

Initially, only the first block is moving. So, the initial momentum is simply:
$$P_{	ext{initial}} = mv$$

Finally, after all the dominoes have fallen, every single block is stuck together into one giant super-block. The total mass of this super-block is the sum of all the individual masses:
$$M_{	ext{total}} = m + m + 2m + 4m + 8m = 16m$$

Let's call the final velocity of this giant block $v'$. The final momentum is:
$$P_{	ext{final}} = 16m \cdot v'$$

Equating the initial and final momentum:
$$mv = 16m \cdot v'$$
$$v' = \frac{v}{16}$$

Notice how the momentum is perfectly conserved, but the velocity has dropped drastically because the mass increased so much!

### Calculating the Energy Loss

Now, let's talk about energy. In perfectly inelastic collisions, kinetic energy is ruthlessly sacrificed to the gods of heat, sound, and deformation. Let's find out exactly how much was lost.

The initial kinetic energy ($K_i$) was entirely in the first block:
$$K_i = \frac{1}{2}mv^2$$

The final kinetic energy ($K_f$) is the energy of the giant $16m$ block moving at $v/16$:
$$K_f = \frac{1}{2}(16m)\left(\frac{v}{16}ight)^2$$
$$K_f = \frac{1}{2}(16m)\left(\frac{v^2}{256}ight) = \frac{1}{32}mv^2$$

The energy lost ($\Delta K$) is the difference between what we started with and what we ended up with:
$$\Delta K = K_i - K_f = \frac{1}{2}mv^2 - \frac{1}{32}mv^2$$
$$\Delta K = \frac{16 - 1}{32}mv^2 = \frac{15}{32}mv^2$$

### The Final Percentage

The question asks for the percentage of the original energy that was lost, denoted as $p\%$. We find this by dividing the lost energy by the initial energy and multiplying by 100:
$$p\% = \frac{\Delta K}{K_i} 	imes 100\%$$
$$p\% = \frac{\frac{15}{32}mv^2}{\frac{1}{2}mv^2} 	imes 100\%$$
$$p\% = \frac{15}{16} 	imes 100\% = 93.75\%$$

The value of $p$ is $93.75$, which is closest to the integer **94**.

This tells us a profound physical truth: when a small moving mass collides and sticks to a much larger stationary mass, almost all of its initial kinetic energy is dissipated! Only a tiny fraction ($6.25\%$) survives to keep the giant mass moving.

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Visualized Solution

$Visualizing the Setup$

$Conservation of Linear Momentum$

$Momentum Equation$

$Calculating Final Velocity$

$Initial Kinetic Energy$

$Final Kinetic Energy$

$Loss in Kinetic Energy$

$Percentage Loss$

$Conclusion$

The Sigma Insight: Head-on Collision

The Master Equation

Calculating the Energy Loss

The Final Percentage

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Linear Momentum

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

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

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

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

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

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

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