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MHT CET 2024 (3 May, Shift 1) · Conservation of Energy

Conservation of Energy question from MHT CET 2024 (3 May, Shift 1)

Q10MHT CET 2024 · 3 MAY · SHIFT 1Medium

Conservation of Energy

A ball ' AA ' is projected vertically upwards with certain initial speed. Another ball 'B' of same mass is projected at an angle of 30∘30^\circ with vertical with the same initial speed. At the highest point, the ratio of potential energy of ball A to that of ball B will be

(sin⁡90∘=1,sin⁡60∘=cos⁡30∘=32,sin⁡30∘=cos⁡60∘=12)\left(\sin 90^\circ = 1, \sin 60^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}, \sin 30^\circ = \cos 60^\circ = \frac{1}{2}\right)

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Answer: (a)

  1. PE at the top =mgH= mgH with H=uy22gH = \frac{u_y^2}{2g}, so U∝uy2U \propto u_y^2.
  2. Ball A: uy=uu_y = u. Ball B (at 30∘30^\circ with vertical): uy=ucos⁡30∘=32uu_y = u\cos 30^\circ = \frac{\sqrt{3}}{2}u.
  3. UAUB=u234u2=43\frac{U_A}{U_B} = \frac{u^2}{\frac{3}{4}u^2} = \frac{4}{3}
  4. So the answer is (a) 4:34 : 3.
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