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JEE Advanced 2018 Paper 1 · Physics · Rotational Motion · medium

JEE Advanced 2018 Paper 1 Physics: Paragraph 1

A uniform rod of mass M=6mM=6m and length L=1.2 mL=1.2\ \mathrm{m} is hinged at one end OO on a smooth horizontal axis. It is held horizontal and released from rest. When it becomes vertical, its free end strikes a small stationary ball of mass mm resting on a smooth horizontal table, elastically and instantaneously, and the ball moves off horizontally. The hinge is frictionless, and the ball is at the level of the free end of the rod in its vertical position. Take g=10 m s−2g=10\ \mathrm{m\,s^{-2}}.

Question 1

Single correct option

The speed of the ball just after the collision is
  1. A. 6 m s−16\ \mathrm{m\,s^{-1}}
  2. B. 8 m s−18\ \mathrm{m\,s^{-1}}
  3. C. 12 m s−112\ \mathrm{m\,s^{-1}}
  4. D. 4 m s−14\ \mathrm{m\,s^{-1}}
Show answer and solution

Answer: (B) 8 m s−18\ \mathrm{m\,s^{-1}}

Falling through the rod's centre-of-mass height L/2L/2: 12(ML23)ω02=MgL2\tfrac12\left(\tfrac{ML^2}{3}\right)\omega_0^2=Mg\dfrac L2, so ω0=3g/L=5 rad s−1\omega_0=\sqrt{3g/L}=5\ \mathrm{rad\,s^{-1}} and the free-end speed is u=ω0L=6 m s−1u=\omega_0L=6\ \mathrm{m\,s^{-1}}. For an impact at the tip, the rod has effective mass Meff=I/L2=M/3=2mM_{eff}=I/L^2=M/3=2m. An elastic collision with a ball at rest gives ball speed v=2MeffMeff+mu=4m3m×6=8 m s−1v=\dfrac{2M_{eff}}{M_{eff}+m}u=\dfrac{4m}{3m}\times6=8\ \mathrm{m\,s^{-1}}. The result 6 m s−16\ \mathrm{m\,s^{-1}} corresponds to ignoring the rod's inertia, and 12 m s−112\ \mathrm{m\,s^{-1}} to treating the rod as an infinitely heavy wall.

Question 2

Single correct option

After the collision the rod swings on past the vertical to a maximum angle φ\varphi from the vertical. Then cos⁡φ\cos\varphi equals
  1. A. 89\dfrac{8}{9}
  2. B. 19\dfrac{1}{9}
  3. C. 59\dfrac{5}{9}
  4. D. 1718\dfrac{17}{18}
Show answer and solution

Answer: (A) 89\dfrac{8}{9}

With effective mass Meff=2mM_{eff}=2m and ball mass mm, the rod's tip velocity after an elastic collision is u′=Meff−mMeff+mu=u3=2 m s−1u'=\dfrac{M_{eff}-m}{M_{eff}+m}u=\dfrac u3=2\ \mathrm{m\,s^{-1}}, in the same direction as before, because the rod is heavier. This gives ω′=u′/L=53 rad s−1\omega'=u'/L=\tfrac53\ \mathrm{rad\,s^{-1}}. (Angular momentum and energy checks agree: ball takes 89\tfrac89 of the rod's kinetic energy.) Energy conservation after the collision: 12ML23ω′2=MgL2(1−cos⁡φ)\tfrac12\dfrac{ML^2}{3}\omega'^2=Mg\dfrac L2(1-\cos\varphi), so 1−cos⁡φ=Lω′23g=1.2×25/930=191-\cos\varphi=\dfrac{L\omega'^2}{3g}=\dfrac{1.2\times25/9}{30}=\dfrac19. Hence cos⁡φ=89\cos\varphi=\dfrac89.
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