Q.Consider a one-dimensional motion of a particle with total energy E. There are four regions A, B, C and D in which the relation between potential energy $V$, kinetic energy $(K)$ and total energy $E$ is as given below:
Region A : $V > E$
Region B : $V < E$
Region C : $K > E$
Region D : $V > K$
State with reason in each case whether a particle can be found in the given region or not.

The key idea is that in classical mechanics, the kinetic energy ($K$) of a particle must always be non-negative, i.e., $K \ge 0$. Since the total energy $E$ is the sum of kinetic energy and potential energy $V$ ($E = K + V$), it follows that $K = E - V$. Therefore, a particle can only exist in regions where $E - V \ge 0$, or $E \ge V$.

Here's the analysis for each region:

1. Region A: $V > E$
   • If $V > E$, then $E - V < 0$.
   • Since $K = E - V$, this implies $K < 0$. …

In classical mechanics, the total energy $E$ of a particle is the sum of its kinetic energy $K$ and potential energy $V$. This is a fundamental principle known as the conservation of mechanical energy:

A crucial physical constraint is that kinetic energy, $K = \frac{1}{2}mv^2$, must always be non-negative. Since mass $m$ is always positive and the square of velocity $v^2$ is always non-negative, $K$ can never be negative.

This condition ($K \ge 0$) is the primary determinant for whether a particle can exist in a given region. If the conditions in a region imply $K < 0$, then the particle cannot be found there. Such regions are often referred to as "classically forbidden regions."

Let's analyze each region:

1. Region A: $V > E$

   • From the energy conservation equation, we can express kinetic energy as $K = E - V$.
   • Given that $V > E$, it means that $E - V$ must be a negative quantity.
   • Therefore, in Region A, $K < 0$.
   • Since kinetic energy cannot be negative, a particle cannot be found in Region A. This is a classically forbidden region.

2. Region B: $V < E$

   • Again, using $K = E - V$.
   • Given that $V < E$, it means that $E - V$ must be a positive quantity.
   • Therefore, in Region B, $K > 0$.
   • Since kinetic energy is positive, this condition is physically allowed. A particle can be found in Region B.

3. Region C: $K > E$

   • This condition directly states that $K$ is positive (since $E$ can be positive, negative, or zero, but if $K > E$, then $K$ must be positive). A positive kinetic energy is physically allowed.
   • Let's also consider the potential energy in this region. From $E = K + V$, we have $V = E - K$. …

Concept: Kinetic Energy Must Be Non-negative

Since $E = K + V$ and $K = \frac12 mv^2 \ge 0$ always, a particle can exist in a region only if $K = E - V \ge 0$ there.

Step 1: Region A ($V > E$)

Not physically possible — a particle cannot be found here (a "classically forbidden" region).

Step 2: Region B ($V < E$)

Physically fine — a particle can be found here.

Step 3: Region C ($K > E$) …