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Question #42307

Difficulty: 8/10 | 🟢 Free

The Question

A small particle of mass mm moving with a uniform velocity v0i^v_0 \hat{i} collides elastically with a smooth, free sphere of mass M=3mM = 3m and radius RR initially at rest at the origin. The impact parameter for the collision is b=R2b = \frac{R}{2}, and the particle's initial y-coordinate is positive. What is the final velocity vector of the particle immediately after the collision?

Options

  • A.

    −18v0i^+338v0j^-\frac{1}{8} v_0 \hat{i} + \frac{3\sqrt{3}}{8} v_0 \hat{j}

  • B.

    18v0i^+338v0j^\frac{1}{8} v_0 \hat{i} + \frac{3\sqrt{3}}{8} v_0 \hat{j}

  • C.

    −14v0i^+34v0j^-\frac{1}{4} v_0 \hat{i} + \frac{\sqrt{3}}{4} v_0 \hat{j}

  • D.

    12v0i^+32v0j^\frac{1}{2} v_0 \hat{i} + \frac{\sqrt{3}}{2} v_0 \hat{j}

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Correct Answer

Step-by-Step Solution

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

Dot Product Geometry|Impact Parameter Geometry

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to apply the Impulse-Momentum Theorem in 2D scattering by utilizing Dot Product Geometry.

A common trap is assuming the impulse acts purely perpendicular to the initial velocity or misidentifying the line of impact.

Students must recognize that for a frictionless collision, the impulse is strictly along the normal vector at the point of contact.

By projecting the initial velocity onto this normal vector, the normal component of momentum change is isolated, while the tangential velocity remains invariant.

Failure to correctly decompose the vectors or account for the reduced mass of the two-body system leads to incorrect final velocity components.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWork, Energy, and Power - Class 11 Part 1
ConceptDot Product Geometry
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree