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MHT CET 2026 (17 Apr, Shift 2) · Velocity, Momentum & Energy in Flight

Velocity, Momentum & Energy in Flight question from MHT CET 2026 (17 Apr, Shift 2)

Q5MHT CET 2026 · 17 APR · SHIFT 2Medium

Velocity, Momentum & Energy in Flight

A ball P is projected at an angle of 60∘60^\circ with the vertical with certain initial speed. Another ball Q of the same mass as that of ball P is projected vertically upwards with the same initial speed as that of P. At the highest point, the ratio of potential energy of ball P to that of ball Q is

(sin⁡30∘=0.5)(\sin 30^\circ = 0.5)

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Answer: (a)

  1. P makes 30∘30^\circ with the horizontal: HP=u2sin⁡230∘2g=u28gH_P = \frac{u^2\sin^2 30^\circ}{2g} = \frac{u^2}{8g}
  2. Q goes straight up: HQ=u22gH_Q = \frac{u^2}{2g}
  3. PE ∝\propto height (same mass): UPUQ=1/81/2=14\frac{U_P}{U_Q} = \frac{1/8}{1/2} = \frac{1}{4}
  4. So the answer is (a) 1:41:4.
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