JEE Advanced 2026 · Paper 2 · Physics Q17

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Question Stem for Question Nos. 17 and 18

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 $XY$ 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 $XY$ 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) $=-10\,\hat\jmath$ ms $^{-2}$ ]

A disk with centre O hangs from its top point C. Point P is on the rim below and to the right of O, with OP at 45° to the downward vertical through O. A particle m approaches P moving in the −x direction; after the collision it moves in the −y direction. Axes x (right) and y (up) are drawn at the left.

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

Your answer

Answer and worked solution

Official answer: 0.14

Take the pivot $C$ as origin. Then $O=(0,-0.2)$ and $P=(0.2\sin45^\circ,\ -0.2-0.2\cos45^\circ)=(0.1414,\ -0.3414)$ m.

Angular momentum about $C$ is conserved: the pivot force acts at $C$ , and gravity gives no impulse during the brief collision. For the particle ( $m=0.02$ kg):

before, $\vec p=(-2,0)$ : $L_z=x p_y-y p_x=-0.3414\times2=-0.6828$ ;
after, $\vec p=(0,-1.8)$ : $L_z=0.1414\times(-1.8)=-0.2546$ .

So the disk gains $|L|=0.4283$ kg m²/s.

Disk. $I_C=\frac12MR^2+MR^2=0.06$ kg m², so $\omega=\frac{0.4283}{0.06}=7.14$ rad/s and
$K=\frac{L^2}{2I_C}=\frac{0.4283^2}{0.12}=1.529\text{ J}.$

Rise of $O$ . $Mg\,\Delta h=K$ gives $\Delta h=\dfrac{1.529}{10}\approx0.15$ m (less than $2R$ , so the disk swings back).

Topics:Angular momentum and its conservation Parallel and perpendicular axis theorems Conservation of mechanical energy

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JEE Advanced 2026 · Paper 2, Physics question 17