JEE Advanced 2023 Paper 1 Q04 Physics Rotational Mechanics Angular Momentum Medium

JEE Advanced 2023 Paper 1 · Q04 · Angular Momentum

A bar of mass $M = 1.00$ kg and length $L = 0.20$ m is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass $m = 0.10$ kg is moving on the same horizontal surface with $5.00$ m·s$^{-1}$ speed on a path perpendicular to the bar. It hits the bar at a distance $L/2$ from the pivoted end and returns back on the same path with speed $v$. After this elastic collision, the bar rotates with an angular velocity $\omega$. Which of the following statements is correct?

  1. A. $\omega = 6.98$ rad·s$^{-1}$ and $v = 4.30$ m·s$^{-1}$
  2. B. $\omega = 3.75$ rad·s$^{-1}$ and $v = 4.30$ m·s$^{-1}$
  3. C. $\omega = 3.75$ rad·s$^{-1}$ and $v = 10.0$ m·s$^{-1}$
  4. D. $\omega = 6.80$ rad·s$^{-1}$ and $v = 4.10$ m·s$^{-1}$
Reveal answer + step-by-step solution

Correct answer:A

Solution

Moment of inertia of bar about pivot: $I = ML^2/3 = (1)(0.2)^2/3$. Conservation of angular momentum about pivot: $m v_0 (L/2) = I\omega - m v (L/2)$, i.e. $(0.1)(5)(0.1) = (1)(0.04)/3 \cdot \omega - (0.1)(v)(0.1)$. Elastic collision: relative velocity of approach = velocity of separation at the contact point, so $v_0 = \omega(L/2) + v$, giving $5 = 0.1\omega + v$. Solving the two equations: $\omega = 300/43 \approx 6.98$ rad/s and $v = 4.30$ m/s.

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