JEE Advanced 2026 — Paper 2 — PHY-IV solution

JEE Advanced2026Paper 2PHY-IV

Question Stem (Q3-Q4).
A uniform circular disk of radius $0.2$ m and mass $1$ kg is pivoted at its top point $C$ such that it can rotate freely around $C$ in the $X Y$ plane, as shown in the figure. Initially, when the disk is at rest, a particle of mass $20$ g, travelling along negative $x$ direction in the $X Y$ plane with speed $100$ ms $^{- 1}$ , hits the circumference of the disk at a point $P$ . After collision the particle moves along negative $y$ direction at a speed of $90$ ms $^{- 1}$ . [Given: the acceleration due to gravity $g = - 10 \hat{j}$ ms $^{- 2}$ ]

Q4. Amount of energy loss (in J) in the collision is:

Your answer

Solution

KE loss during collision.

$Δ K E = [\frac{1}{2} m v_{i}^{2}] - [\frac{1}{2} m v_{f}^{2} + \frac{1}{2} I_{C} ω^{2}]$ .

$\frac{1}{2} m v_{i}^{2} = \frac{1}{2} (0.02) (100)^{2} = 100$ J.

$\frac{1}{2} m v_{f}^{2} = \frac{1}{2} (0.02) (90)^{2} = 81$ J.

$\frac{1}{2} I_{C} ω^{2} = \frac{1}{2} (0.06) (7.138)^{2} \approx 1.529$ J.

$Δ K E = 100 - (81 + 1.529) \approx 17.47$ J.

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