Question

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 $\upsilon$.

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

• A. $2\sqrt{2}\, ms^{-1}$
• B. $ 2 \,ms^{-1}$
• C. $ 4 \,ms^{-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 $4\, J$ to escape to infinity.
i .e. $\frac{1}{2}m\upsilon^{2} \ge 4\Rightarrow \frac{1}{2}\times\frac{1}{2}\times\upsilon^{2}\ge4$
or $\upsilon^{2} \ge16\Rightarrow \upsilon \ge 4\, ms^{-1}$
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\upsilon^{2} \ge1\Rightarrow \frac{1}{2}\times\frac{1}{2}\times\upsilon^{2}\ge\,1$,or $\upsilon^{2} \ge2\, ms^{-1}$
Hence, $\upsilon_{min}$ for escape is $2 \,ms^{-1}.$