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A point particle of mass $0.5 \,kg$ is moving along the $X$-axis under a force described by the potential energy $V$ shown below. It is projected towards the right from the origin with a speed $v$. What is the minimum value of $v$ for which the particle will escape infinitely far away from the origin?
$2 \sqrt{2} \,ms ^{-1}$
$2 \,ms ^{-1}$
$4 \,ms ^{-1}$
The particle will never escape
Solution
(B)
When particle is projected towards right side of origin with velocity $v$, it must cross a potential barrier of $4 J$ to escape to infinity.
i.e. $\quad \frac{1}{2} m v^2 \geq 4 \Rightarrow \frac{1}{2} \times \frac{1}{2} \times v^2 \geq 4$
or $\quad v^2 \geq 16 \Rightarrow v \geq 4 \,ms ^{-1}$
But it is possible that kinetic energy of particle is less than $4 \,J$, then it is
reflected back towards left with same velocity $v$. Here, we are taking repulsion from field as elastic collision.
Then particle can escape to infinity from left side of origin when,
$\frac{1}{2} m v^2 \geq 1 \Rightarrow \frac{1}{2} \times \frac{1}{2} v^2 \geq 1 \text { or } v \geq 2 \,ms ^{-1}$
Hence, $v_{\min }$ for escape is $2 \,ms ^{-1}$.
