A stone is projected vertically upwards with speed ' v '. Another… | MHT CET 2023 (9 May, Shift 1)

Q18MHT CET 2023 · 9 MAY · SHIFT 1Medium

Conservation of Energy

A stone is projected vertically upwards with speed ' $v$ '. Another stone of same mass is projected at an angle of $60^\circ$ with the vertical with the same speed ' $v$ '. The ratio of their potential energies at the highest points of their journey is $\left[\sin 30^\circ = \cos 60^\circ = 0.5, \cos 30^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}\right]$

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Answer: (a)

PE at the top $= mgH$ with $H = \frac{u_y^2}{2g}$ , so $U \propto u_y^2$ .
Vertical stone: $u_y = v$ . Other stone (at $60^\circ$ with vertical): $u_y = v\cos 60^\circ = \frac{v}{2}$ .
$\frac{U_1}{U_2} = \frac{v^2}{v^2/4} = 4$
So the answer is (a) $4 : 1$ .

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Conservation of Energy question from MHT CET 2023 (9 May, Shift 1)

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