Question

A point particle of mass $0.5kg$ 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 $\upsilon$.

What is the minimum value of $v$ for which the particle will escape infinitely far away from the origin?

• A. $2\sqrt{2}m{s}^{-1}$
• B. $2m{s}^{-1}$
• C. $4m{s}^{-1}$
• D. The particle will never escape

Answer 1

Answer: (B)

When particle is projected towards right side of origin with velocity $v$, it must cross a potential barrier of $4J$ to escape to infinity.
i .e. $\frac{1}{2}m{\upsilon }^2\ge 4\Rightarrow \frac{1}{2}\times \frac{1}{2}\times {\upsilon }^2\ge 4$
or ${\upsilon }^2\ge 16\Rightarrow \upsilon \ge 4m{s}^{-1}$
But it is possible that kinetic energy of particle is less than $4J,$ 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,
$\frac{1}{2}m{\upsilon }^2\ge 1\Rightarrow \frac{1}{2}\times \frac{1}{2}\times {\upsilon }^2\ge 1$,or ${\upsilon }^2\ge 2m{s}^{-1}$
Hence, ${\upsilon }_{min}$ for escape is $2m{s}^{-1}.$