In figure, the variation of potential energy of a particle of mass `m=2kg` 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?

Options

A. (a) If it is released at the origin, it will move in negative x-axis.
B. (b) If it is released at `x=2+Delta`, where `Deltararr0`, then its maximum speed will be `5ms^-1` and it will perform oscillatory motion.
C. (c) If initially `x=-10` and `vecu=sqrt6hati`, then it will cross `x=10`.
D. (d) `x=-5` and `x=+5` are unstable equilibrium positions of the particle.

Answer

D

Explanation

If the particle is released at the origin, it will try to go in the direction of force. Hence `dU//dx` 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+Detla`, it will reach at point of least possible potential energy `(-15J)` where it will have maximum kinetic energy.
`1/2mv_(max)^2=25`
`v_(max)=5ms^-1`
The particle will now perform oscillatory motion will `x=5` as mean position.
In (c), `E_i=u_i+k_i=15+6=21J`
At `x=10, U_f=20`
`K_f=1!=0`
So the particle crosses `x=10`.