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JEE Main 2013 (22 Apr, Shift 1) · Rolling Motion

Rolling Motion question from JEE Main 2013 (22 Apr, Shift 1)

Q80JEE Main 2013 · 22 APR · SHIFT 1Difficult

Rolling Motion

A tennis ball (treated as hollow spherical shell) starting from O rolls down a hill. At point A, the ball becomes air borne leaving at an angle of 30° with the horizontal. The ball strikes the ground at B. What is the value of the distance AB? (Moment of inertia of a spherical shell of mass m and radius R about its diameter = 23mR2\frac{2}{3}mR^{2})

Figure

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Answer: (a)

  1. From the figure, O is 2.0−0.2=1.82.0 - 0.2 = 1.8 m above A.
  2. Rolling shell (k=23k = \frac{2}{3}): mgh=12mv2(1+23)⇒v2=6gh5=6×10×1.85=21.6mgh = \frac{1}{2}mv^2\left(1+\frac{2}{3}\right) \Rightarrow v^2 = \dfrac{6gh}{5} = \dfrac{6 \times 10 \times 1.8}{5} = 21.6.
  3. Projectile range on level ground: AB=v2sin⁡60∘g=21.6×0.86610AB = \dfrac{v^2\sin 60^\circ}{g} = \dfrac{21.6 \times 0.866}{10}.
  4. AB≈1.87AB \approx 1.87 m, so the answer is (a).
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