Consider a uniform rod of mass M = 4m and length pivoted about its centre. A mass m moving with velocity v making angle θ = π/4 to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is: (a) (3√(2))/(7)v/ (b) 3/7v/ (c) (3)/(7√(2))v/ (d) 4/7v/

Answer: (a). Angular momentum about the pivot: L = mvsin 45°·/2 = (mv)/(2√(2)). I = 1/12(4m)² + m(/2)² = (m²)/(3) + (m²)/(4) = (7m²)/(12) ω = (mv)/(2√(2))·(12)/(7m²) = (6v)/(7√(2)) = (3√(2))/(7)v/ So the answer is (a) (3√(2))/(7)v/.

Consider a uniform rod of mass M = 4m and length pivoted about its… | JEE Main 2020 (8 Jan, Shift 1)

Q57JEE Main 2020 · 8 JAN · SHIFT 1Difficult

Angular Momentum & its Conservation

Consider a uniform rod of mass M = 4m and length $\ell$ pivoted about its centre. A mass m moving with velocity v making angle $\theta = \frac{\pi}{4}$ to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is:

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Answer: (a)

Angular momentum about the pivot: $L = mv\sin 45^\circ\cdot\frac{\ell}{2} = \frac{mv\ell}{2\sqrt{2}}$ .
$I = \frac{1}{12}(4m)\ell^2 + m\left(\frac{\ell}{2}\right)^2 = \frac{m\ell^2}{3} + \frac{m\ell^2}{4} = \frac{7m\ell^2}{12}$
$\omega = \frac{mv\ell}{2\sqrt{2}}\cdot\frac{12}{7m\ell^2} = \frac{6v}{7\sqrt{2}\,\ell} = \frac{3\sqrt{2}}{7}\frac{v}{\ell}$
So the answer is (a) $\frac{3\sqrt{2}}{7}\frac{v}{\ell}$ .

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