This question has statement I and statement II. Of the four choices given after the statements, choo

Q: Wait, why does maximum energy loss happen when they stick together?

In any collision, momentum is strictly conserved. However, kinetic energy can be converted into heat, sound, or deformation energy. When the objects stick together (a perfectly inelastic collision), they undergo the maximum possible permanent deformation, which means the maximum amount of kinetic energy is dissipated.

Q: Why can't the energy loss be 100%?

Energy loss can only be 100% if the total final momentum is zero. In this scenario, the initial momentum is $mv$, so the final combined mass must keep moving to conserve momentum. Because it must keep moving, it must retain some kinetic energy, meaning the loss cannot be 100%.

Q: What is reduced mass and why do we use it here?

Reduced mass, $\mu = \frac{mM}{m+M}$, is a mathematical tool that simplifies two-body collision problems. It allows us to treat the two interacting bodies as a single body of mass $\mu$ moving with the relative velocity $v_{rel}$, making the energy loss formula much more elegant.

Q: Why is Statement I false? It looks so similar to the correct answer.

Statement I claims the fraction $f$ is $\frac{m}{M+m}$. However, our derivation shows $f = \frac{M}{M+m}$. The numerator must be the mass of the *target* particle ($M$), not the projectile ($m$). It's a classic trap to test if you memorized the formula or actually derived it.

Q: What if the second mass $M$ was also moving initially?

If $M$ was moving, the relative velocity $v_{rel}$ would be the difference between their velocities (e.g., $v_1 - v_2$). The formula for energy loss would still be exactly the same, but you would substitute the new $v_{rel}$. The condition for maximum loss ($e=0$) remains unchanged.

Question

This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements.

**Statement I** A point particle of mass $m$ moving with speed $v$ collides with stationary point particle of mass $M$. If the maximum energy loss possible is given as $f \left( \frac{1}{2} m v^2 \right)$, then $f = \left( \frac{m}{M + m} \right)$.

**Statement II** Maximum energy loss occurs when the particles get stuck together as a result of the collision.

Options:

(A) Statement I is true, Statement II is true, and Statement II is the correct explanation of Statement I

(B) Statement I is true, Statement II is true, but Statement II is not the correct explanation of Statement I

(C) Statement I is true, Statement II is false

(D) Statement I is false, Statement II is true

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing a classic physics experiment. A small particle of mass $m$ is zooming along a straight line with a velocity $v$.

Ahead of it lies a larger, stationary particle of mass $M$. They are on a collision course.

Our objective is to determine the exact conditions under which this collision results in the **maximum possible loss of kinetic energy**, and to calculate what that maximum loss is.

### The Master Equation

To solve this, we need to pull out our master equation for the loss of kinetic energy in a one-dimensional collision.

The energy loss, denoted by $\Delta K$, is given by the formula:

$$ \Delta K = \frac{1}{2} \frac{mM}{m+M} v_{rel}^2 (1 - e^2) $$

Here, the term $\frac{mM}{m+M}$ is known as the **reduced mass** of the system.

The term $v_{rel}$ is the relative velocity of approach, which in our case is simply $v$, since the second mass is at rest.

Finally, $e$ is the **coefficient of restitution**, a number between $0$ and $1$ that tells us how "bouncy" the collision is.

### The Condition for Maximum Loss

Now, let's look closely at the equation. We want to maximize $\Delta K$.

The masses and the initial velocity are fixed. The only variable that depends on the *nature* of the collision is $e$.

To make $\Delta K$ as large as possible, the term $(1 - e^2)$ must be maximized.

Since $e$ ranges from $0$ to $1$, the maximum value of $(1 - e^2)$ is exactly $1$, which occurs when $e = 0$.

What does $e = 0$ mean physically? It means the collision is **perfectly inelastic**. The two particles do not bounce off each other at all; instead, they stick together and move as a single combined mass.

This perfectly aligns with **Statement II**, which claims that maximum energy loss occurs when the particles get stuck together. Therefore, Statement II is absolutely true!

### Final Calculation

Now that we know $e = 0$ gives the maximum energy loss, let's substitute it back into our master equation to find the exact value.

$$ \Delta K_{max} = \frac{1}{2} \frac{mM}{m+M} v^2 (1 - 0^2) $$

$$ \Delta K_{max} = \frac{1}{2} \frac{mM}{m+M} v^2 $$

The problem asks us to express this loss as a fraction $f$ of the initial kinetic energy of the moving particle, which is $\frac{1}{2} m v^2$. Let's rearrange our result to match this format:

$$ \Delta K_{max} = \left( \frac{M}{m+M} \right) \left( \frac{1}{2} m v^2 \right) $$

By comparing this with the expression given in the problem, $\Delta K_{max} = f \left( \frac{1}{2} m v^2 \right)$, we can clearly see that the fraction $f$ is:

$$ f = \frac{M}{m+M} $$

### The Verdict

Let's evaluate **Statement I**. It claims that the fraction $f$ is equal to $\frac{m}{M+m}$.

However, our rigorous derivation shows that the numerator must be $M$ (the mass of the stationary particle), not $m$.

Therefore, Statement I is definitively false.

With Statement I being false and Statement II being true, the correct choice is option (d).

Select Answer:

Visualized Solution

Visualizing the Collision

Formula for Energy Loss

Maximizing Energy Loss

Evaluating Statement II

Calculating Maximum Energy Loss

Evaluating Statement I

Final Conclusion

The Sigma Insight: Head-on Collision

Analyzing the Setup

The Master Equation

The Condition for Maximum Loss

Final Calculation

The Verdict

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