A stone is projected vertically upwards with velocity 'V'. Another stone of same mass is projected at an angle of 60° with the vertical with the same speed (V). The ratio of their potential energies at the highest points of their journey, is [sin 30° = cos 60° = 0.5, cos 30° = sin 60° = √(3)/2] (a) 1: 1 (b) 4: 1 (c) 3: 2 (d) 2: 1

Answer: (b). PE at the top = mgH with H = (uy²)/(2g), so U ∝ uy². Vertical stone: uy = V. Other stone (at 60° with vertical): uy = Vcos 60° = V/2. (U₁)/(U₂) = (V²)/(V²/4) = 4 So the answer is (b) 4: 1.

A stone is projected vertically upwards with velocity 'V'. Another… | MHT CET 2021 (24 Sep, Shift 1)

Q24MHT CET 2021 · 24 SEP · SHIFT 1Medium

Conservation of Energy

A stone is projected vertically upwards with velocity 'V'. Another stone of same mass is projected at an angle of $60^\circ$ with the vertical with the same speed $(\text{V})$ . 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: (b)

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 (b) $4 : 1$ .

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Conservation of Energy question from MHT CET 2021 (24 Sep, Shift 1)

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