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

Q: Why is energy loss maximum when particles stick together?

In a perfectly inelastic collision, the bodies lose all their relative velocity. This means the maximum possible amount of macroscopic kinetic energy is converted into internal energy, such as heat and deformation.

Q: Is momentum conserved even when kinetic energy is lost?

Yes! The law of conservation of linear momentum holds true for all types of collisions (elastic, inelastic, perfectly inelastic) as long as there are no external forces acting on the system.

Q: Why can't the particles lose 100% of their kinetic energy?

Because the system has an initial forward momentum that must be conserved. The combined mass must continue moving to carry this momentum, which means it must retain some kinetic energy.

Q: What does the fraction m/(M+m) represent if it's not the energy lost?

Interestingly, the fraction m/(M+m) represents the fraction of the initial kinetic energy that is *retained* by the system after the perfectly inelastic collision.

Q: Would the energy loss be different if the second mass M was also moving initially?

Yes. You would need to apply momentum conservation using both initial velocities to find the new final velocity, and then calculate the difference in total kinetic energy. The maximum loss still occurs when they stick together.

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 $f\left(\frac{1}{2}mv^2\right)$, 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; Statement II is the correct explanation of Statement I

(B) Statement I is true, Statement II is true; 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

The physics of collisions is a fascinating dance between two fundamental principles: the conservation of momentum and the conservation of energy. While momentum is the stubborn rule that always holds true in an isolated system, kinetic energy is more flexible—it can be conserved, or it can be lost to the universe as heat, sound, or deformation.

In this problem, we are asked to evaluate two statements regarding the maximum possible loss of kinetic energy when a moving particle strikes a stationary one. Let's dive into the mechanics of this impact and uncover the truth.

### The Condition for Maximum Energy Loss

Statement II makes a bold claim: the maximum energy loss occurs when the particles get stuck together. Is this true?

Imagine a fast-moving car crashing into a stationary truck. If they bounce off each other (an elastic collision), they retain most of their kinetic energy. But if they crumple and lock together, moving as a single mangled mass, a massive amount of energy is spent bending metal and generating heat. This is known as a **perfectly inelastic collision**, characterized by a coefficient of restitution $e = 0$.

In physics, it is a proven fact that for a given initial state, a perfectly inelastic collision dissipates the maximum possible macroscopic kinetic energy into internal energy. Therefore, **Statement II is absolutely true**.

### The Law of Unbreakable Momentum

Now, let's test Statement I, which provides a specific formula for this maximum energy loss. To find the actual energy loss, we must first figure out how fast the combined mass is moving after the collision.

Even though kinetic energy is lost, the **conservation of linear momentum** remains unbroken. The initial momentum of the system is simply the momentum of the moving particle, since the second particle is at rest.

$p_{	ext{initial}} = mv$

After the collision, the two particles stick together, forming a single body of mass $(m + M)$ moving with a new velocity, let's call it $v'$. The final momentum is:

$p_{	ext{final}} = (m + M)v'$

Equating the initial and final momentum, we can solve for the final velocity:

$mv = (m + M)v'$

$v' = \frac{mv}{m + M}$

### Calculating the Energy Deficit

With the final velocity in hand, we can now calculate the kinetic energies before and after the crash. The initial kinetic energy is just the energy of the moving particle:

$K_i = \frac{1}{2}mv^2$

The final kinetic energy belongs to the combined mass moving at velocity $v'$:

$K_f = \frac{1}{2}(m + M)v'^2$

Substituting our expression for $v'$ into this equation:

$K_f = \frac{1}{2}(m + M)\left(\frac{mv}{m + M}ight)^2$

$K_f = \frac{m^2v^2}{2(m + M)}$

The energy lost, $\Delta E$, is the difference between the initial and final kinetic energies:

$\Delta E = K_i - K_f$

$\Delta E = \frac{1}{2}mv^2 - \frac{m^2v^2}{2(m + M)}$

To simplify this, we can factor out the initial kinetic energy, $\frac{1}{2}mv^2$:

$\Delta E = \frac{1}{2}mv^2 \left[1 - \frac{m}{m + M}ight]$

Finding a common denominator for the terms inside the bracket:

$\Delta E = \frac{1}{2}mv^2 \left[\frac{m + M - m}{m + M}ight]$

$\Delta E = \frac{1}{2}mv^2 \left[\frac{M}{m + M}ight]$

### The Final Verdict

We have successfully derived the expression for the maximum energy loss. The problem states that this loss can be written as $f\left(\frac{1}{2}mv^2ight)$. By comparing our derived formula with the given expression, we can clearly see the value of the fraction $f$:

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

However, Statement I claims that $f = \frac{m}{M + m}$. Notice the subtle trick? The numerator in the statement is $m$, but our derivation proves it must be $M$. Because of this incorrect fraction, **Statement I is false**.

It is interesting to note that the fraction $\frac{m}{m+M}$ actually represents the fraction of kinetic energy that is *retained* by the system, not the energy lost. The system can never lose 100% of its kinetic energy because it must keep moving to conserve the initial forward momentum!

Select Answer:

Visualized Solution

$C o n d i t i o n f or M a x im u m E n er g y L oss$

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

$K in e t i c E n er g y S e t u p$

$C a l c u l a t in g E n er g y L oss$

$E v a l u a t in g S t a t e m e n t I$

$T h e W a y F or w a r d$

The Sigma Insight: Head-on Collision

The Condition for Maximum Energy Loss

The Law of Unbreakable Momentum

Calculating the Energy Deficit

The Final Verdict

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