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Question #43776

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass mm (in kg\text{kg}) moves in a horizontal xx-yy plane under the influence of a conservative central attractive force F⃗c=−f(r)r^\vec{F}_c = -f(r) \hat{r} (in N\text{N}) directed towards the origin OO. The region is filled with a viscous medium that exerts a drag force on the particle given by F⃗d=−βr2v⃗\vec{F}_d = -\frac{\beta}{r^2} \vec{v}, where β\beta is a positive constant (in kg⋅m2⋅s−1\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-1}), rr is the distance from the origin (in m\text{m}), and v⃗\vec{v} is the velocity vector of the particle (in m/s\text{m/s}). At time t=0t=0, the particle is at a distance r0r_0 (in m\text{m}) from the origin and is projected with a velocity v0v_0 (in m/s\text{m/s}) perpendicular to its position vector. The particle eventually loses all its transverse velocity. Assuming the particle does not reach the origin before its transverse velocity vanishes, what is the total angle Δθ\Delta \theta (in rad\text{rad}) swept by the position vector of the particle from t=0t=0 until its motion becomes purely radial?

Options

  • A.

    mr0v0β\frac{m r_0 v_0}{\beta}

  • B.

    mr0v02β\frac{m r_0 v_0}{2\beta}

  • C.

    2mr0v0β\frac{2 m r_0 v_0}{\beta}

  • D.

    ln⁡(1+mr0v0β)\ln\left(1 + \frac{m r_0 v_0}{\beta}\right)

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Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Impact Parameter Geometry|Projectile Path Equation|Damped Oscillation Equation

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to analyze non-conservative forces in central force setups.

The primary trap is the apparent unsolvability of the trajectory.

Students typically attempt to write coupled differential equations for r(t)r(t) and θ(t)\theta(t) using Newton's second law, which leads to a dead end because the central force function f(r)f(r) is unknown.

Another common mistake is blindly applying the conservation of angular momentum, ignoring the torque produced by the velocity-dependent drag force.

The correct logical path requires recognizing that while angular momentum is not conserved, the specific 1/r21/r^2 spatial dependence of the drag force allows the torque equation to be decoupled from the radial coordinate.

By decomposing the velocity into radial and transverse components, students can prove that the radial drag produces no torque, and the transverse drag produces a torque that perfectly cancels the radial dependence, yielding a direct, integrable relationship between the change in angular momentum and the swept angle.

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Question Information

SubjectPhysics
ChapterAtoms - Class 12
ConceptImpact Parameter Geometry
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree