In figure, 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?

Moderate

A

If it is released at the origin, it will move in negative x-axis.
B

If it is released at x = 2 + $Δ$ , where $Δ$ $\to$ 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.

Solution

If the particle is released at the origin, it will try to go in the direction of force. Here 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 + $Δ$ , it will reach the least possible potential point of energy (-15 J) where it will have maximum kinetic energy
$\begin{array}{l} \frac{1}{2} {mv}_{max}^{2} = 25 \\ v_{max} = 5 {ms}^{- 1} \end{array}$
The particle will now perform oscillatory motion with x = 5 as mean position.
In (3), $E_{i} = u_{i} + k_{i} = 15 + 6 = 21 J$
At $x = 10, U_{f} = 20$
$K_{f} = 1 \neq 0$
So the particle crosses x = 10.

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