Question

In the figure below, the variation of potential energy of a particle of mass $m=2\, kg$ is represented w.r.t its $x$-coordinate. The particle moves under the effect of the conservative force along the $x$-axis. Which of the following statements is incorrect about the particle?

• A. If it is released at the origin, it will move in negative $x$-axis.
• B. If it is released at $x=2+\Delta$, where $\Delta \rightarrow 0$, then its maximum speed will be $5\, ms ^{-1}$ and it will perform oscillatory motion.
• C. If initially $x=-10$ and $\vec{u}=\sqrt{6} \hat{i}$, then it will cross $x=10$.
• D. $x=-5$ and $x=+5$ are unstable equilibrium positions of the particle.

Answer 1

Answer: (D)

If the particle is released at the origin, it will try to go in the direction of force.
Here $d U / d x$ is positive and hence force is negative; as a result, it will move towards negative $x$-axis.
When the particle is released at $x=2+\Delta$, it will reach the point of least possible potential energy $(-15\, J )$ where it will have maximum kinetic energy.
$\frac{1}{2} m v_{\max }^{2}=25$
$v_{\max }=5\, m s ^{-1}$
The particle will now perform oscillatory motion with $x=5$ as mean position.
In (c), $E_{i}=u_{i}+k_{i}=15+6=21\, J$
At $x=10, U_{f}=20$
$K_{f}=1 \neq 0$
So the particle crosses $x=10$.