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 \(100 \mathrm{~ms}^{-1}\), hits the circumference of the disk at a point \(P\). After collision the particle moves along negative \(y\) direction at a speed of \(90 \mathrm{~ms}^{-1}\). [Given: the acceleration due to gravity \((\mathrm{g})=-10 \hat{j} \mathrm{~ms}^{-2}\)] Amount of energy loss (in J) in the collision is:

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Conservation of Momentum

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}

]

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

Answer: 17.47

Energy loss = Initial KE - Final KE.

Initial KE:

\frac{1}{2} m v_{i}^{2} = \frac{1}{2} \times 0.02 \times (100)^{2} = 100

J.

Final KE:

\frac{1}{2} m v_{f}^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} \times 0.02 \times (90)^{2} + \frac{1}{2} \times \frac{3}{2} \times 1 \times (0.2)^{2} \times (7.138)^{2}

= 81 + \frac{3}{4} \times 0.04 \times 50.95 = 81 + 1.53 = 82.53

J.

Energy loss =

100 - 82.53 = 17.47

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Numerical answer