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PhyHardAdv 2026 • Paper 2

Maximum Height Change of Pivoted Disk Center After Collision

Comprehension Passage

A uniform circular disk of radius 0.2 m0.2\text{ m} and mass 1 kg1\text{ kg} is pivoted at its top point CC such that it can rotate freely around CC in the XYXY plane, as shown in the figure. Initially, when the disk is at rest, a particle of mass 20 g20\text{ g}, travelling along negative xx direction in the XYXY plane with speed 100 ms−1100\text{ ms}^{-1}, hits the circumference of the disk at a point PP. After collision the particle moves along negative yy direction at a speed of 90 ms−190\text{ ms}^{-1}.

[Given: the acceleration due to gravity (g)=−10j^ ms−2(\text{g}) = - 10 \hat{j}\text{ ms}^{-2}]

After the collision the disk starts to rotate around point CC in the XYXY plane. The maximum change in the height (in m) of its center OO is:

Question Diagram 1
Official Numerical Answer0.14 to 0.16

Topics & Concepts

System of Particles and Rotational MotionCollisionAngular MomentumMoment of Inertia

Step-by-Step Solution

To find the maximum change in the height of the disk's center OO after the collision, we analyze the system using the conservation of angular momentum about the pivot point CC during the collision, followed by the conservation of mechanical energy as the disk swings upward.

1. Geometric Setup and System Parameters

Given parameters:

  • Mass of the disk, M=1 kgM = 1 \text{ kg}
  • Radius of the disk, R=0.2 mR = 0.2 \text{ m}
  • Mass of the particle, m=20 g=0.02 kgm = 20 \text{ g} = 0.02 \text{ kg}
  • Initial velocity of particle, v⃗1=−100i^ m s−1\vec{v}_1 = -100 \hat{i} \text{ m s}^{-1}
  • Final velocity of particle, v⃗2=−90j^ m s−1\vec{v}_2 = -90 \hat{j} \text{ m s}^{-1}
  • Acceleration due to gravity, g=10 m s−2g = 10 \text{ m s}^{-2}

Let point CC be the origin (0,0)(0, 0). The center OO is at (0,−R)(0, -R). From the given geometry, the point of impact PP is at a radial distance RR from OO at an angle of 45∘45^\circ with the vertical line COCO. Thus, the position vector of point PP relative to CC is: r⃗P/C=Rsin⁡45∘i^−(R+Rcos⁡45∘)j^=R2i^−R(1+12)j^\vec{r}_{P/C} = R \sin 45^\circ \hat{i} - \left(R + R \cos 45^\circ\right) \hat{j} = \frac{R}{\sqrt{2}} \hat{i} - R\left(1 + \frac{1}{\sqrt{2}}\right) \hat{j}

2. Moment of Inertia of the Disk

By the parallel axis theorem, the moment of inertia of the disk about the pivot CC is: IC=Icm+MR2=12MR2+MR2=32MR2I_C = I_{\text{cm}} + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2

Substituting the given values: IC=32(1 kg)(0.2 m)2=0.06 kg m2I_C = \frac{3}{2} (1 \text{ kg}) (0.2 \text{ m})^2 = 0.06 \text{ kg m}^2

3. Conservation of Angular Momentum about Pivot CC

Since the reaction force during the collision acts at pivot CC, its torque about CC is zero. Therefore, angular momentum about CC is conserved: L⃗i=L⃗f\vec{L}_i = \vec{L}_f

Initial Angular Momentum (L⃗i\vec{L}_i): L⃗i=r⃗P/C×(mv⃗1)\vec{L}_i = \vec{r}_{P/C} \times (m \vec{v}_1) L⃗i=[R2i^−R(1+12)j^]×(−mv1i^)=−mv1R(1+12)k^\vec{L}_i = \left[ \frac{R}{\sqrt{2}} \hat{i} - R\left(1 + \frac{1}{\sqrt{2}}\right) \hat{j} \right] \times (-m v_1 \hat{i}) = -m v_1 R\left(1 + \frac{1}{\sqrt{2}}\right) \hat{k}

Final Angular Momentum (L⃗f\vec{L}_f): L⃗f=r⃗P/C×(mv⃗2)+ICωk^\vec{L}_f = \vec{r}_{P/C} \times (m \vec{v}_2) + I_C \omega \hat{k} L⃗f=[R2i^−R(1+12)j^]×(−mv2j^)+ICωk^=[−mv2R2+ICω]k^\vec{L}_f = \left[ \frac{R}{\sqrt{2}} \hat{i} - R\left(1 + \frac{1}{\sqrt{2}}\right) \hat{j} \right] \times (-m v_2 \hat{j}) + I_C \omega \hat{k} = \left[ -m v_2 \frac{R}{\sqrt{2}} + I_C \omega \right] \hat{k}

Equating the zz-components: −mv1R(1+12)=−mv2R2+ICω-m v_1 R\left(1 + \frac{1}{\sqrt{2}}\right) = -m v_2 \frac{R}{\sqrt{2}} + I_C \omega

Rearranging for ICωI_C \omega: ICω=mR[v22−v1(1+12)]I_C \omega = m R \left[ \frac{v_2}{\sqrt{2}} - v_1 \left(1 + \frac{1}{\sqrt{2}}\right) \right]

Substituting the numerical values: 0.06ω=(0.02)(0.2)[902−100(1+12)]0.06 \omega = (0.02)(0.2) \left[ \frac{90}{\sqrt{2}} - 100 \left(1 + \frac{1}{\sqrt{2}}\right) \right] 0.06ω=0.004(−100−102)=−0.004(100+52)0.06 \omega = 0.004 \left( -100 - \frac{10}{\sqrt{2}} \right) = -0.004 (100 + 5\sqrt{2}) ω=−0.004(100+52)0.06=−20+23 rad s−1\omega = -\frac{0.004 (100 + 5\sqrt{2})}{0.06} = -\frac{20 + \sqrt{2}}{3} \text{ rad s}^{-1}

The magnitude of the angular velocity is: ∣ω∣=20+23≈7.138 rad s−1|\omega| = \frac{20 + \sqrt{2}}{3} \approx 7.138 \text{ rad s}^{-1}

4. Conservation of Energy After Collision

Right after the collision, the kinetic energy of the disk is entirely rotational: K=12ICω2K = \frac{1}{2} I_C \omega^2 K=12(0.06)(20+23)2=0.03×400+402+29=402+402300 JK = \frac{1}{2} (0.06) \left( \frac{20 + \sqrt{2}}{3} \right)^2 = 0.03 \times \frac{400 + 40\sqrt{2} + 2}{9} = \frac{402 + 40\sqrt{2}}{300} \text{ J}

Using 2≈1.4142\sqrt{2} \approx 1.4142: K≈402+40(1.4142)300=458.568300≈1.5286 JK \approx \frac{402 + 40(1.4142)}{300} = \frac{458.568}{300} \approx 1.5286 \text{ J}

As the disk swings up to its maximum height, this rotational kinetic energy is completely converted into gravitational potential energy of the disk's center of mass OO: ΔU=Mgh\Delta U = M g h 1.5286=(1 kg)(10 m s−2)h1.5286 = (1 \text{ kg}) (10 \text{ m s}^{-2}) h h=1.528610≈0.153 mh = \frac{1.5286}{10} \approx 0.153 \text{ m}

Thus, the maximum change in height of the disk's center OO is approximately 0.15 m0.15 \text{ m}.

Maximum Height Change of Pivoted Disk Center After Collision | Physics PYQ Solution - JEE Challenger