JEE Advanced 2020 · Paper 2 · Physics Q9

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PhysicsMultiple correct+4 / -2Question 9 of 54

A rod of mass $m$ and length $L$ , pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed $v$ strikes the rod horizontally at a distance $x$ from its pivoted end and gets embedded in it. The combined system now rotates with angular speed $\omega$ about the pivot. The maximum angular speed $\omega_M$ is achieved for $x = x_M$ . Then

A vertical rod of length L pivoted at its top end. A bullet moving horizontally with velocity v approaches and will strike the rod at distance x below the pivot.

\omega = \dfrac{3vx}{L^2 + 3x^2}

\omega = \dfrac{12vx}{L^2 + 12x^2}

x_M = \dfrac{L}{\sqrt3}

\omega_M = \dfrac{v}{2L}\sqrt3

Answer and worked solution

Official answer: A, C, D

(A)/(B) Angular momentum about the pivot is conserved in the impact:
$mvx=\left(\frac{mL^2}3+mx^2\right)\omega\implies\omega=\frac{3vx}{L^2+3x^2}.$
(A) is true and (B) is false.

(C) Maximise over $x$ :
$\frac{d\omega}{dx}\propto(L^2+3x^2)-6x^2=0\implies x_M=\frac L{\sqrt3}.$
(C) is true.

(D) At $x_M$ :
$\omega_M=\frac{3v\,L/\sqrt3}{2L^2}=\frac{\sqrt3\,v}{2L}.$
(D) is true.

(A), (C), (D)

Topics:Angular momentum and its conservation Moment of inertia Elastic and inelastic collisions

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JEE Advanced 2020 · Paper 2, Physics question 9