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Question #42086

Difficulty: 8/10 | 🟢 Free

The Question

A point particle of mass $m$ is projected from a radial distance $r = R$ with an initial speed $v_0$ . The initial velocity vector makes an angle of $\frac{\pi}{4}$ with the inward radial vector. The particle moves under the influence of a central attractive potential $U(r) = -\frac{m R^2 v_0^2}{4 r^2}$ , which arises from a non-uniform field density. Assuming conservation of angular momentum, find the radial distance $r$ at which the velocity vector makes an angle of $\frac{\pi}{3}$ with the inward radial vector.

Options

A.
$\frac{R}{\sqrt{3}}$
B.
$\frac{R}{2}$
C.
$\frac{R}{\sqrt{2}}$
D.
$\frac{R}{3}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Velocity Vector Direction|Angular Velocity vs Linear Velocity|Integration

Why This Question Is Tricky

Common Mistakes

The question tests the interplay between the velocity vector's direction, angular momentum conservation, and energy in a non-standard central potential.

The specific trap is the assumption that radial velocity must vary in a central potential.

Here, the inverse-square attractive potential exactly cancels the centrifugal barrier, leading to a constant radial velocity.

Students who blindly apply circular motion approximations or fail to construct the effective potential will miscalculate the changing angle of the velocity vector.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Plane - Class 11 Part 1
Concept	Velocity Vector Direction
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free