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JEE Main 2023 · Physics · Kinematics and Laws of Motion · hard

A small elastic ball is dropped from rest from a height of 20 m above the point where it strikes a smooth fixed plane inclined at 30^° to the…

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A small elastic ball is dropped from rest from a height of 20 m20\ \mathrm{m} above the point where it strikes a smooth fixed plane inclined at 30∘30^\circ to the horizontal. It rebounds elastically and, after one flight, lands on the same incline again. The distance between the first and second impact points, measured along the incline, is (g=10 m s−2g=10\ \mathrm{m\,s^{-2}})
  1. A. 40 m40\ \mathrm{m}
  2. B. 60 m60\ \mathrm{m}
  3. C. 100 m100\ \mathrm{m}
  4. D. 80 m80\ \mathrm{m}
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Answer: (D) 80 m80\ \mathrm{m}

Speed at impact v=2gh=20 m s−1v=\sqrt{2gh}=20\ \mathrm{m\,s^{-1}} vertically down. Resolve along and perpendicular to the incline: v⊥=vcos⁡30∘v_\perp=v\cos30^\circ, v∥=vsin⁡30∘=10v_\parallel=v\sin30^\circ=10 (down the slope). The collision is elastic and the plane is smooth, so v⊥v_\perp reverses and v∥v_\parallel is unchanged. Perpendicular motion has acceleration gcos⁡30∘g\cos30^\circ toward the plane, so flight time t=2vcos⁡30∘gcos⁡30∘=2vg=4 st=\dfrac{2v\cos30^\circ}{g\cos30^\circ}=\dfrac{2v}{g}=4\ \mathrm{s}. Along the slope the ball has 10 m s−110\ \mathrm{m\,s^{-1}} and acceleration gsin⁡30∘=5g\sin30^\circ=5: s=10(4)+12(5)(16)=80 ms=10(4)+\tfrac12(5)(16)=80\ \mathrm{m}.
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Original question written in the JEE Main 2023 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake