A tennis ball (treated as hollow spherical shell) starting from O… | 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 = $\frac{2}{3}mR^{2}$ )

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Answer: (a)

From the figure, O is $2.0 - 0.2 = 1.8$ m above A.
Rolling shell ( $k = \frac{2}{3}$ ): $mgh = \frac{1}{2}mv^2\left(1+\frac{2}{3}\right) \Rightarrow v^2 = \dfrac{6gh}{5} = \dfrac{6 \times 10 \times 1.8}{5} = 21.6$ .
Projectile range on level ground: $AB = \dfrac{v^2\sin 60^\circ}{g} = \dfrac{21.6 \times 0.866}{10}$ .
$AB \approx 1.87$ m, so the answer is (a).

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