JEE Advanced 2026 Paper 2, Physics Section 4 Q3: Angular Momentum, Its Conservation and Angular Impulse

JEE Advanced2026Paper 2Physics Section 4

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 $^{- 2}$ ]

Q3. After the collision the disk starts to rotate around point $C$ in the $X Y$ plane. The maximum change in the height (in m) of its center $O$ is:

Your answer

Solution

Geometry of P. With $C$ at the top, the centre $O$ is at distance $R = 0.2$ m below $C$ . Since $O P$ makes $4 5^{\circ}$ with the vertical, $P$ lies at horizontal distance $R / 2$ from the vertical through $C$ and at vertical distance $R + R / 2$ below $C$ (using the geometry shown). The perpendicular distance of the incoming horizontal velocity (along $- \hat{i}$ ) from $C$ equals the vertical drop of $P$ below $C$ , which is $R + R / 2 = R (1 + \frac{1}{2})$ . The perpendicular distance for the outgoing velocity (along $- \hat{j}$ ) equals the horizontal distance of $P$ from $C$ , which is $R / 2$ .

Angular momentum about $C$ (conserved during the instantaneous collision).

Moment of inertia of the disk about $C$ (using parallel axis): $I_{C} = \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2} = \frac{3}{2} (1) (0.2)^{2} = 0.06$ kg m $^{2}$ .

$m v_{i} R (1 + \frac{1}{2}) = I_{C} ω + m v_{f} \frac{R}{2}$

$(0.02) (100) (0.2) (1 + 0.707) = 0.06 ω + (0.02) (90) (0.2) (0.707)$

$0.6828 = 0.06 ω + 0.2546 \Rightarrow ω \approx 7.14$ rad/s.

Maximum rise of $O$ from energy conservation. After collision the disk rotates about $C$ , and gravity acts on its centre $O$ :

$\frac{1}{2} I_{C} ω^{2} = M g R (1 - cos θ_{m a x})$

$\frac{1}{2} (0.06) (7.14)^{2} = (1) (10) (0.2) (1 - cos θ_{m a x})$

$1.529 = 2 (1 - cos θ_{m a x})$ , but the relevant quantity is the vertical rise $Δ h = R (1 - cos θ_{m a x})$ :

$Δ h = \frac{I _{C} ω ^{2}}{2 M g} = \frac{( 0.06 ) ( 7.14 ) ^{2}}{2 ( 1 ) ( 10 )} \approx 0.153$ m $\approx 0.15$ m.

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JEE Advanced 2026 Paper 2, Physics Section 4 Q3: Angular Momentum, Its Conservation and Angular Impulse

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