A block moving horizontally on a smooth surface with a speed of 40 ms^-1 splits into two equal parts

Q: Why is momentum conserved but not kinetic energy?

Momentum is conserved because there are no external forces acting on the system. Kinetic energy is not conserved because the internal forces that cause the block to split do positive work, converting internal potential energy into kinetic energy.

Q: The question asks for 'fractional change', why do we calculate the ratio?

Strictly speaking, fractional change is $\frac{\Delta KE}{KE_i}$. However, in competitive exams like JEE, terminology can sometimes be loose. The official answer of 5 corresponds to the ratio $\frac{KE_f}{KE_i}$. It's important to be flexible and let the answer format guide your interpretation.

Q: Does the direction of the split matter?

Absolutely. Momentum is a vector quantity. The problem states the first part moves in the 'same direction', so we use a positive sign. If it moved backwards, we would use a negative sign in our momentum equation.

Q: What if the second part's velocity came out negative?

A negative velocity would simply mean that the second fragment is moving in the opposite direction to the original motion. The math naturally takes care of the directions if you set up your sign convention correctly.

Q: Where did the extra kinetic energy come from?

The extra kinetic energy comes from the internal energy of the block. For a block to split and accelerate its fragments, some internal mechanism (like a compressed spring or a small explosive charge) must release energy, doing work on the pieces.

Question

A block moving horizontally on a smooth surface with a speed of $40 	ext{ ms}^{-1}$ splits into two equal parts. If one of the parts moves at $60 	ext{ ms}^{-1}$ in the same direction, then the fractional change in the kinetic energy will be $x : 4$, where $x$ is

Sigma Percentile · Accepted Answer

The phenomenon of an object splitting or exploding into multiple fragments is a classic scenario in physics that beautifully illustrates the interplay between momentum and energy. In this problem, we are presented with a block moving at a constant velocity that suddenly splits into two equal halves. Let's embark on a detailed journey to uncover the mechanics behind this event.

### Analyzing the Setup

Imagine a block of mass $m$ gliding effortlessly across a perfectly smooth, frictionless horizontal surface at a brisk speed of $40 	ext{ m/s}$. Suddenly, an internal mechanism causes it to split into two identical pieces, each with a mass of $m/2$. We are told that one of these fragments continues to move in the original direction but at an increased speed of $60 	ext{ m/s}$. Our goal is to determine the state of the second fragment and subsequently analyze the change in the system's kinetic energy.

The key to unlocking this problem lies in understanding the nature of the forces involved. The split is caused entirely by internal forces within the block. Because the surface is smooth, there are no external horizontal forces, such as friction, acting on the system.

### The Master Equation: Conservation of Momentum

According to Newton's Second Law of Motion, the rate of change of momentum of a system is directly proportional to the net external force applied to it. Since the net external horizontal force is zero ($F_{	ext{ext}} = 0$), the total linear momentum of the system must remain constant. This is the Principle of Conservation of Linear Momentum.

Let's set up our momentum equation. The initial momentum of the system is simply the mass of the original block multiplied by its velocity:
$$p_i = m 	imes 40$$

After the split, the total momentum is the vector sum of the momenta of the two individual fragments. Let the velocity of the second fragment be $v_2$. The final momentum is:
$$p_f = \left(\frac{m}{2}ight) 	imes 60 + \left(\frac{m}{2}ight) 	imes v_2$$

Equating the initial and final momenta, we get:
$$m(40) = \frac{m}{2}(60) + \frac{m}{2} v_2$$

### Calculating the Missing Velocity

Now, let's perform the algebraic manipulation to find $v_2$. We can divide the entire equation by the common mass term $m$:
$$40 = \frac{60}{2} + \frac{v_2}{2}$$
$$40 = 30 + \frac{v_2}{2}$$

Subtracting $30$ from both sides yields:
$$10 = \frac{v_2}{2}$$

Multiplying by $2$, we find the velocity of the second fragment:
$$v_2 = 20 	ext{ m/s}$$

The positive sign indicates that the second fragment also continues to move in the same original direction, albeit at a slower speed than the first fragment.

### The Kinetic Energy Shift

With the velocities of all parts known, we can now turn our attention to the kinetic energy of the system. Unlike momentum, kinetic energy is not necessarily conserved during an explosion or a split. In fact, the internal forces that cause the split do positive work on the fragments, converting some form of internal potential energy (like chemical or elastic energy) into additional kinetic energy.

Let's calculate the initial kinetic energy ($KE_i$) of the solid block:
$$KE_i = \frac{1}{2} m u^2 = \frac{1}{2} m (40)^2$$
$$KE_i = \frac{1}{2} m (1600) = 800 m$$

Next, we calculate the final kinetic energy ($KE_f$), which is the sum of the kinetic energies of the two fragments:
$$KE_f = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$$
$$KE_f = \frac{1}{2} \left(\frac{m}{2}ight) (60)^2 + \frac{1}{2} \left(\frac{m}{2}ight) (20)^2$$
$$KE_f = \frac{m}{4} (3600) + \frac{m}{4} (400)$$
$$KE_f = 900 m + 100 m = 1000 m$$

As anticipated, the final kinetic energy ($1000 m$) is greater than the initial kinetic energy ($800 m$). The system has gained $200 m$ Joules of kinetic energy due to the internal work done during the split.

### The Final Calculation and a Note on Terminology

The problem asks us to find the value of $x$ given that the "fractional change in the kinetic energy will be $x : 4$".

Strictly speaking in physics, the term "fractional change" refers to the change in a quantity divided by its initial value:
$$	ext{Fractional Change} = \frac{\Delta KE}{KE_i} = \frac{KE_f - KE_i}{KE_i} = \frac{1000 m - 800 m}{800 m} = \frac{200}{800} = \frac{1}{4}$$

If we equate this to $x/4$, we would get $x = 1$. However, the official answer key for this JEE problem states that the answer is $5$. This implies that the question author intended for the phrase "fractional change" to actually mean the **ratio of the final kinetic energy to the initial kinetic energy**.

Let's calculate this ratio:
$$	ext{Ratio} = \frac{KE_f}{KE_i} = \frac{1000 m}{800 m} = \frac{5}{4}$$

Equating this ratio to $x/4$:
$$\frac{5}{4} = \frac{x}{4}$$

This gives us the final answer:
**$$x = 5$$**

**A crucial takeaway for competitive exams:** While precise terminology is important, you must also be adaptable. If a strict interpretation leads to an answer that doesn't align with the provided options or the expected format, consider alternative interpretations of the phrasing. In this case, recognizing that the author meant "ratio" instead of "fractional change" is key to arriving at the intended solution.

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Split}

\text{Conservation of Linear Momentum}

\text{Momentum Equation Setup}

\text{Solving for } v_2

\text{Kinetic Energy Formula}

\text{Initial Kinetic Energy}

\text{Final Kinetic Energy}

\text{Ratio of Kinetic Energies}

The Sigma Insight: Conservation of Linear Momentum

Analyzing the Setup

The Master Equation

Calculating the Missing Velocity

The Kinetic Energy Shift

The Final Calculation and a Note on Terminology

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