Statement I Two particles moving in the same direction do not lose all their energy in a completely

Q: Why is momentum conserved in an inelastic collision?

Momentum is conserved because the forces the particles exert on each other during the collision are internal to the system. As long as there is no net external force acting on the system, the total momentum remains constant, regardless of whether kinetic energy is lost.

Q: Doesn't 'inelastic' mean that energy is lost?

Yes, in an inelastic collision, some macroscopic kinetic energy is transformed into other forms of energy (like heat, sound, or deformation energy). However, losing *some* energy does not necessarily mean losing *all* of it.

Q: When would *all* kinetic energy be lost in a collision?

All kinetic energy is lost only if the final velocity of the combined mass is exactly zero. This happens in a completely inelastic collision when the two particles approach each other from opposite directions with equal and opposite momenta ($m_1 v_1 = -m_2 v_2$).

Q: Why do the particles stick together?

That is the defining characteristic of a 'completely' (or perfectly) inelastic collision. It is the scenario where the maximum possible kinetic energy is dissipated, subject to the constraint of momentum conservation.

Q: Can the final velocity be negative?

If we define the initial direction of motion as positive, and both particles are moving in that same direction, the final velocity must also be positive. It cannot be negative or zero.

Question

Statement I Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
Statement II Principle of conservation of momentum holds true for all kinds of collisions.

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 Core Concepts

When we study collisions in physics, two fundamental principles often come into play: the conservation of momentum and the conservation of kinetic energy. However, while momentum is conserved in **all** isolated collisions (whether elastic or inelastic), kinetic energy is only conserved in perfectly elastic collisions.

In a **completely inelastic collision**, the colliding bodies stick together and move as a single combined mass after the impact. This type of collision results in the maximum possible loss of kinetic energy. But does it mean *all* kinetic energy is lost? Let's find out.

### Analyzing the Collision

Imagine two particles with masses $m_1$ and $m_2$. They are moving in the same direction with velocities $v_1$ and $v_2$, respectively. Because they are moving in the same direction, both $v_1$ and $v_2$ share the same sign (let's assume they are both positive).

According to the principle of conservation of momentum (which is our Statement II), the total momentum before the collision must equal the total momentum after the collision. This principle is a direct consequence of Newton's third law and holds true regardless of the nature of the collision.

### The Mathematical Proof

Let's set up the momentum equation for this completely inelastic collision. Before the impact, the total initial momentum is:

$$p_{initial} = m_1 v_1 + m_2 v_2$$

After the collision, the particles stick together, forming a single body of mass $(m_1 + m_2)$ moving with a common final velocity $v$. The final momentum is:

$$p_{final} = (m_1 + m_2) v$$

Equating the initial and final momenta, we get:

$$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v$$

Solving for the final velocity $v$, we find:

$$v = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$$

Here is the crucial insight: Since both $v_1$ and $v_2$ are positive (moving in the same direction), the numerator is strictly positive. The denominator, being the sum of masses, is also strictly positive. Therefore, the final velocity $v$ **must be greater than zero** ($v > 0$).

### The Final Verdict

Because the final velocity $v$ is non-zero, the final kinetic energy of the combined mass is:

$$K_f = \frac{1}{2}(m_1 + m_2)v^2 > 0$$

This mathematically proves that the system retains some kinetic energy. They do not lose *all* their energy. Thus, Statement I is absolutely true.

Furthermore, we were able to prove Statement I directly by applying the principle of conservation of momentum. Therefore, Statement II is not only true but is also the correct logical explanation for Statement I.

**Final Answer:** Option (a) is correct.

Animated Solution for Physics - System of Particles: Statement I Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. Statement II Principle of conservation of momentum holds true for all kinds of collisions.

Select Answer:

Visualized Solution

Visualizing the Setup

Conservation of Momentum

Completely Inelastic Collision

Final Velocity Analysis

Final Kinetic Energy

The Way Forward

The Sigma Insight: Conservation of Linear Momentum

The Core Concepts

Analyzing the Collision

The Mathematical Proof

The Final Verdict

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