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title: "Two isolated point particles of masses \\(m_1\\) and \\(m_2\\) carry opposite charges \\(+Q\\) and \\(-q\\), where \\(Q > q\\). The particles are released from an initial separation \\(d\\) with non-zero initial velocities directed straight toward each other, such that the system’s total mechanical energy is negative (\\(E < 0\\), taking \\(U(\\infty) = 0\\)). A student predicts that because the system is in a bound state, the particles will reach a non-zero distance of closest approach and turn around before colliding. Which of the following statements provides the correct physical explanation for why this prediction is incorrect?"
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url: "https://nerd-notes.com/ubq/124668/"
date_modified: "2026-09-28T14:09:00+00:00"
---

# Two isolated point particles of masses \(m_1\) and \(m_2\) carry opposite charges \(+Q\) and \(-q\), where \(Q > q\). The particles are released from an initial separation \(d\) with non-zero initial velocities directed straight toward each other, such that the system’s total mechanical energy is negative (\(E < 0\), taking \(U(\infty) = 0\)). A student predicts that because the system is in a bound state, the particles will reach a non-zero distance of closest approach and turn around before colliding. Which of the following statements provides the correct physical explanation for why this prediction is incorrect?

Two isolated point particles of masses \(m_1\) and \(m_2\) carry opposite charges \(+Q\) and \(-q\), where \(Q > q\). The particles are released from an initial separation \(d\) with non-zero initial velocities directed straight toward each other, such that the system's total mechanical energy is negative (\(E < 0\), taking \(U(\infty) = 0\)). A student predicts that because the system is in a bound state, the particles will reach a non-zero distance of closest approach and turn around before colliding. Which of the following statements provides the correct physical explanation for why this prediction is incorrect?

![A horizontal line represents the axis of motion. On the left side of the line, a solid black circle is labeled \(+Q\) and has a horizontal arrow pointing to the right labeled \(\vec{v}_1\). On the right side of the line, a solid gray circle is labeled \(-q\) and has a horizontal arrow pointing to the left labeled \(\vec{v}_2\). A horizontal dimension line with arrows at both ends spans between the vertical centers of the two circles and is labeled \(d\). No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790604540-ixu22n.jpg)

- **A.** The electric potential \(V(r) = \dfrac{kQ}{r}\) established by the positive charge diverges to positive infinity as separation approaches zero, forming an infinite potential barrier that inevitably reduces the approaching negative charge's kinetic energy to zero and reflects it backward before a physical collision can occur.
- **B.** Because the electric potential energy \(U(r) = -\dfrac{kQq}{r}\) becomes increasingly negative as separation decreases, the system's kinetic energy \(K(r) = E - U(r)\) strictly increases for all \(r < d\); thus, the turning-point condition \(E = U(r)\) corresponds to a maximum separation rather than a distance of closest approach, and the particles accelerate continuously until they collide.
- **C.** A negative total mechanical energy guarantees that the two-particle system is bound, so conservation of angular momentum automatically converts the particles' initial radial approach into a stable circular orbit at a characteristic radius where the attractive Coulomb force balances the required centripetal acceleration.
- **D.** Because the two charges have opposite signs and unequal magnitudes, there exists a point along the line between them where their net electric potential is zero; upon reaching this location, the system's electric potential energy vanishes, which forces the kinetic energy to zero and causes the particles to turn around.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/124668/*