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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 $A: V>E \quad$ Region $B: V < E$
Region $C: K < E \quad$ Region $D: V>E$
State with reason in each case whether a particle can be found in the given region or not.
Region $A: V>E \quad$ Region $B: V < E$
Region $C: K < E \quad$ Region $D: V>E$
State with reason in each case whether a particle can be found in the given region or not.
Solution:
2449 Upvotes
Verified Answer
As we know that,
$$
\begin{array}{lr}
\text { Total energy } E=\mathrm{PE}+\mathrm{KE} & \because \mathrm{PE}=(V) \\
E=V+K, K=(E-V) & \because \mathrm{KE}=(K) \\
&
\end{array}
$$
(i) For region $A$. Given, $V>E$, From eq. (i), $K=E-V$
as $V>E \Rightarrow E-V < 0$
So, $K < 0$ or $\mathrm{KE}$ is negative this is not possible.
(ii) For region $B$. Given, $V < E, K=E-V, K>0$
This case is possible because both energy are greater than zero PE $(V)$.
(iii) For region $C$. Given, $K>E \Rightarrow K-E>0$
From Eq. (i), $\mathrm{PE}(V)=E-K, V < 0$, (PE is negative)
This is also possible, because PE can be negative.
(iv) For region $D$. Given, $V>K, K=E-V$
This is possible as PE for a system PE $(V)$ can be greater than $\mathrm{KE}(K)$.
$$
\begin{array}{lr}
\text { Total energy } E=\mathrm{PE}+\mathrm{KE} & \because \mathrm{PE}=(V) \\
E=V+K, K=(E-V) & \because \mathrm{KE}=(K) \\
&
\end{array}
$$
(i) For region $A$. Given, $V>E$, From eq. (i), $K=E-V$
as $V>E \Rightarrow E-V < 0$
So, $K < 0$ or $\mathrm{KE}$ is negative this is not possible.
(ii) For region $B$. Given, $V < E, K=E-V, K>0$
This case is possible because both energy are greater than zero PE $(V)$.
(iii) For region $C$. Given, $K>E \Rightarrow K-E>0$
From Eq. (i), $\mathrm{PE}(V)=E-K, V < 0$, (PE is negative)
This is also possible, because PE can be negative.
(iv) For region $D$. Given, $V>K, K=E-V$
This is possible as PE for a system PE $(V)$ can be greater than $\mathrm{KE}(K)$.
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