Consider the following two statements.
I. Linear momentum of a system of particles is zero.
II. Kine

Q: Why can't kinetic energy be negative to cancel out other kinetic energies?

Kinetic energy is calculated as $K = \frac{1}{2}mv^2$. Since mass ($m$) is always positive and the square of any real number ($v^2$) is always positive or zero, kinetic energy can never be negative. Therefore, individual kinetic energies can only add up, never cancel.

Q: If momentum is conserved in a collision, is kinetic energy always conserved too?

No. In inelastic collisions, the total momentum of the system is conserved, but some kinetic energy is transformed into other forms of energy like heat, sound, or deformation energy. Only in perfectly elastic collisions are both conserved.

Q: Can a single isolated particle have zero momentum but non-zero kinetic energy?

No. For a single particle, momentum is $p = mv$ and kinetic energy is $K = \frac{p^2}{2m}$. If $p = 0$, then $K$ must strictly be $0$ as well.

Q: Why does $K_{net} = 0$ strictly mean every particle is at rest?

Because $K_{net}$ is a sum of purely non-negative terms ($\frac{1}{2}mv^2$). If even one particle was moving, its kinetic energy would be greater than zero, making the total sum greater than zero. The only way the sum is zero is if every single term is zero.

Q: Is it possible for a system to have zero kinetic energy but non-zero momentum?

Absolutely not. As we proved, if the kinetic energy is zero, every particle is at rest ($v=0$). If every particle is at rest, there is no velocity to create momentum, so the total momentum must be zero.

Question

Consider the following two statements.
I. Linear momentum of a system of particles is zero.
II. Kinetic energy of a system of particles is zero. Then,

Options:

(A) I does not imply II and II does not imply I

(B) I implies II but II does not imply I

(C) I does not imply II but II implies I

(D) I implies II and II implies I

Sigma Percentile · Accepted Answer

### The Dance of Momentum and Energy

When we study the mechanics of a system of particles, two fundamental quantities often take center stage: **Linear Momentum** and **Kinetic Energy**. While they are intimately related, their mathematical nature—one being a vector and the other a scalar—creates fascinating scenarios. Let's break down the two statements given in the problem to understand their true implications.

### Analyzing Statement I: Zero Momentum

Statement I claims that the linear momentum of a system of particles is zero. Mathematically, the total linear momentum $\vec{P}_{net}$ is the vector sum of the individual momenta of all particles:

$$\vec{P}_{net} = \sum m_i \vec{v}_i = 0$$

Does this mean the system has no kinetic energy? Imagine a thought experiment: Two identical cars, each of mass $m$, are driving directly towards each other at the exact same speed $v$. Because velocity is a vector, one car has velocity $\vec{v}$ and the other has $-\vec{v}$.

Their total momentum is:
$$\vec{P}_{net} = m\vec{v} + m(-\vec{v}) = 0$$

The total momentum is perfectly zero! However, both cars are clearly moving. Kinetic energy is a scalar quantity, meaning it doesn't care about direction. It only cares about speed. The total kinetic energy of this system is:

$$K_{net} = \frac{1}{2}mv^2 + \frac{1}{2}m(-v)^2 = mv^2$$

Since $v 
eq 0$, the kinetic energy $K_{net}$ is strictly greater than zero. This simple counterexample proves that **Statement I does not imply Statement II**.

### Analyzing Statement II: Zero Kinetic Energy

Now, let's flip the script and look at Statement II, which claims the kinetic energy of the system is zero. The total kinetic energy is the sum of the individual kinetic energies:

$$K_{net} = \sum \frac{1}{2}m_i v_i^2 = 0$$

Here is the crucial catch: Mass $m_i$ is always positive, and the square of any real speed $v_i^2$ is always non-negative (zero or positive). Therefore, every single term in this summation is non-negative.

The only mathematical way for a sum of non-negative numbers to equal exactly zero is if **every single term is individually zero**.

$$\frac{1}{2}m_i v_i^2 = 0 \implies v_i = 0 	ext{ for all } i$$

This means every single particle in the system is completely at rest. If every particle is at rest, their individual momenta are all zero ($\vec{p}_i = m_i(0) = 0$). Consequently, the vector sum of their momenta must also be zero:

$$\vec{P}_{net} = \sum \vec{p}_i = 0$$

Thus, if the kinetic energy is zero, the linear momentum is guaranteed to be zero. This proves that **Statement II implies Statement I**.

### The Verdict

By carefully distinguishing between the vector nature of momentum (which allows for cancellation) and the scalar nature of kinetic energy (which strictly accumulates), we arrive at our final conclusion: **I does not imply II, but II implies I**. This perfectly aligns with option (c).

Animated Solution for Physics - System of Particles: Consider the following two statements. I. Linear momentum of a system of particles is zero. II. Kinetic energy of a system of particles is zero. Then,

Select Answer:

Visualized Solution

$Analyzing Statement I$

$Counterexample for Statement I$

$Analyzing Statement II$

$Implication of K = 0$

$Final Conclusion$

The Sigma Insight: Conservation of Linear Momentum

The Dance of Momentum and Energy

Analyzing Statement I

Analyzing Statement II

The Verdict

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