A body of mass 14 kg initially at rest explodes and breaks into three fragments of masses in the ratio 2 : 2 : 3. The two pieces of equal masses fly off perpendicular to each other with a speed of 18 m/s each. The velocity of the heavier fragment is ___m/s.

The total mass is 14 , kg and breaks in ratio 2:2:3. Let the masses be M1, M2, and M3. Since M1 and M2 fly off perpendicular to each other, let their velocity vectors be along the x and y axes. Because the total momentum must be zero, we align the fragments opposite to the final 3rd fragment's direction for simplicity, or just use standard axes. Let V1 = -18i and V2 = -18j. Note that the solution simplifies the mass ratio directly to 2, 2, 3 as relative masses for the momentum equation: Magnitude of V3: In a 3-part explosion from rest, the momentum of the third piece must be equal and opposite to the vector sum of the other two pieces. You can use the ratio of masses directly in the momentum conservation equation instead of absolute masses to save time. Class 11 Physics: Center of Mass and Collisions

A body of mass 14 kg initially at rest explodes and breaks into three fragments of masses in the ratio 2 : 2 : 3. The two pieces of equal masses fly off perpendicular to each other with a speed of 18

The total mass is 14 , kg and breaks in ratio 2:2:3. Let the masses be M1, M2, and M3. Since M1 and M2 fly off perpendicular to each other, let their velocity vectors be along the x and y axes. Because the total momentum must be zero, we align the fragments opposite to the final 3rd fragment's direction for simplicity, or just use standard axes. Let V1 = -18i and V2 = -18j. Note that the solution simplifies the mass ratio directly to 2, 2, 3 as relative masses for the momentum equation: Magnitud

A body of mass 14 kg initially at rest explodes and breaks into three fragments of masses in the ratio 2 : 2 : 3. The two pieces of equal masses fly off perpendicular to each other with a speed of 18 m/s each. The velocity of the heavier fragment is ___m/s.

The total mass is 14 , kg and breaks in ratio 2:2:3. Let the masses be M1, M2, and M3. Since M1 and M2 fly off perpendicular to each other, let their velocity vectors be along the x and y axes. Because the total momentum must be zero, we align the fragments opposite to the final 3rd fragment's direction for simplicity, or just use standard axes. Let V1 = -18i and V2 = -18j. Note that the solution simplifies the mass ratio directly to 2, 2, 3 as relative masses for the momentum equation: Magnitude of V3: In a 3-part explosion from rest, the momentum of the third piece must be equal and opposite to the vector sum of the other two pieces. You can use the ratio of masses directly in the momentum conservation equation instead of absolute masses to save time. Class 11 Physics: Center of Mass and Collisions

A body of mass 14 kg initially at rest explodes and breaks into three fragments of masses in the ratio 2 : 2 : 3. The two pieces of equal masses fly off perpendicular to each other with a speed of 18

The total mass is 14 , kg and breaks in ratio 2:2:3. Let the masses be M1, M2, and M3. Since M1 and M2 fly off perpendicular to each other, let their velocity vectors be along the x and y axes. Because the total momentum must be zero, we align the fragments opposite to the final 3rd fragment's direction for simplicity, or just use standard axes. Let V1 = -18i and V2 = -18j. Note that the solution simplifies the mass ratio directly to 2, 2, 3 as relative masses for the momentum equation: Magnitud

A body of mass 14 kg initially at rest explodes and breaks into three fragments of masses in the ratio 2 : 2 : 3. The…

Q49 jee_main_2026_24_january_evening Moment of Inertia of Continuous Bodies

A uniform solid cylinder of length L and radius R has moment of inertia about its axis equal to

I₁

. A small co-centric cylinder of length L/2 and radius R/3 carved from this cylinder has moment of inertia about its axis equals to

I₂

. The ratio

I₁/I₂

is

Numerical Answer. Answer: 162 to 162

Solution

Related Formula

I = (1)/(2) M R²

M = ρ \cdot V = ρ \cdot π R² L

Core Logic

Moment of Inertia of Continuous Bodies diagram for Q49 - JEE Main 2026 Evening

For the original cylinder (mass $M$ $M$):

I₁ = (1)/(2) M R²

For the carved cylinder, its mass $m$ $m$ is:

m = ρ \times π ((R)/(3))² \times (L)/(2)

Step 1: Calculate Mass of Carved Cylinder

m = (ρ π R² L)/(18) = (M)/(18)

Step 2: Calculate Inertia of Carved Cylinder

I₂ = (1)/(2) m ((R)/(3))²

I₂ = (1)/(2) ( (M)/(18) ) ( (R²)/(9) )

I₂ = (1)/(324) M R²

Step 3: Finding the Ratio

(I₁)/(I₂) = ((1)/(2) M R²)/((1)/(324) M R²) = (324)/(2) = 162

Pattern Recognition

For similar geometries, mass scales as $R² L$ $R^2 L$. Inertia scales as $M R²$ $M R^2$ which ultimately means $I \propto R⁴ L$ $I \propto R^4 L$. Here $R arrow R/3$ $R \rightarrow R/3$ (factor of $1/81$ $1/81$) and $L arrow L/2$ $L \rightarrow L/2$ (factor of $1/2$ $1/2$), so $I₂$ $I_2$ is $1/162$ $1/162$ of $I₁$ $I_1$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q29 jee_main_2026_28_january_morning Moment of Inertia and Torque

Two circular discs of radius each 10 cm are joined at their centres by a rod of length 30 cm and mass 600 gm as shown in figure.

Moment of inertia diagram with rod and discs for Q29 - JEE Main 2026 Morning — Illustration showing two discs connected by a rod rotating about axis AB.

If the mass of each disc is 600 gm and applied torque between two discs is

43 \times 10⁵ ~dyne\cdot cm

, the angular acceleration of the discs about the given axis AB is ____

rad/s²

.

A. $22$ $22$
B. $11$ $11$
C. $100$ $100$
D. $27$ $27$

Solution

Related Formula

$τ = I α$ $\tau = I \alpha$ where $τ$ $\tau$ is the torque, $I$ $I$ is the moment of inertia about the axis of rotation, and $α$ $\alpha$ is the angular acceleration.

Core Logic

First, calculate the total moment of inertia of the system (two discs + one rod) about axis AB using the parallel axis theorem. Let $m = 600 ~gm$ $m = 600 \mathrm{~gm}$ and $R = 10 ~cm$ $R = 10 \mathrm{~cm}$.

Step 1: Moment of Inertia of Discs

For the left disc (distance $R$ $R$ from AB):

Ileft = (1)/(4)mR² + mR² = (5)/(4)mR²

For the right disc (distance $2R$ $2R$ from AB):

Iright = (1)/(4)mR² + m(2R)² = (17)/(4)mR²

Total for discs:

Idiscs = (5)/(4)mR² + (17)/(4)mR² = (22)/(4)mR² = (11)/(2)mR²

Step 2: Moment of Inertia of Rod

Rod has mass $m$ $m$ and length $3R = 30 ~cm$ $3R = 30 \mathrm{~cm}$. Its center of mass is at a distance $R/2$ $R/2$ from axis AB.

Irod = (m(3R)²)/(12) + m((R)/(2))²

= (9mR²)/(12) + (mR²)/(4) = (3)/(4)mR² + (1)/(4)mR² = mR²

Step 3: Total Moment of Inertia

Itotal = ((11)/(2) + 1) mR² = (13)/(2) mR²

Substitute $m = 600 ~g$ $m = 600 \mathrm{~g}$ and $R = 10 ~cm$ $R = 10 \mathrm{~cm}$:

Itotal = (13)/(2) \times 600 \times (10)² = 39 \times 10⁴ ~g \cdot cm²

Step 4: Calculating Angular Acceleration

α = (τ)/(I) = (43 \times 10⁵)/(39 \times 10⁴) ~rad/s²

= (430)/(39) \approx 11.02 ~rad/s²

Rounding off, $α \approx 11 ~rad/s²$ $\alpha \approx 11 \mathrm{~rad/s^2}$.

Pattern Recognition

In compound systems, decompose into basic shapes (rod, disc). Determine the parallel distance to the required axis for each center of mass. Keep everything in CGS units since torque is given in dyne-cm.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q48 jee_main_2026_28_january_morning Radius of Gyration

A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is

\sqrt(n)

cm. The value of n is ____

Numerical Answer. Answer: 265 to 265

Solution

Related Formula

I = Icm + md²

$I = mk²$ $I = mk^2$

Icm (solid sphere) = (2)/(5)mR²

Core Logic

Use the parallel axis theorem to find the moment of inertia about the new axis, then equate it to $mk²$ $mk^2$ to find the radius of gyration.

Step 1: Applying Parallel Axis Theorem

As per the given solution steps:

mk² = (2)/(3)mR² + md²

k² = (2)/(3)R² + d²

Step 2: Substituting Values

Given $R = 10 ~cm$ $R = 10 \mathrm{~cm}$ and $d = 15 ~cm$ $d = 15 \mathrm{~cm}$:

k² = (2)/(3)(10)² + (15)²

k² = (200)/(3) + 225

k² = (2)/(5)R² + d²

k² = (2)/(5)(10)² + (15)²

k² = 40 + 225 = 265

Step 3: Final Answer

We are given $k = \sqrt(n) \Rightarrow k² = n$ $k = \sqrt{n} \Rightarrow k^2 = n$. $n = 265$ $n = 265$

Pattern Recognition

Whenever radius of gyration is asked, divide out the mass immediately. $k² = Icm-factor R² + d²$ $k^2 = I_{\text{cm-factor}} R^2 + d^2$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q37 jee_main_2026_28_january_evening Cross Product and Vector Reversal

When the position vector

r=x i+y j+z k

changes sign as

- r

, which one of the following vector will not flip under sign change?

A. $Linear momentum$ $\text{Linear momentum}$
B. $Velocity$ $\text{Velocity}$
C. $Acceleration$ $\text{Acceleration}$
D. $Angular momentum$ $\text{Angular momentum}$

Solution

Related Formula

v = d rdt

p = m v

a = d vdt

L = r \times p

Core Logic

Under a sign change of coordinates (parity transformation or spatial inversion) $r arrow - r$ $\vec{r} \rightarrow -\vec{r}$. Then, velocity $v = d rdt arrow - v$ . Linear momentum $p = m v arrow - p$ . Acceleration $a = d vdt arrow - a$ .

Angular momentum $L = r \times p$ $\vec{L} = \vec{r} \times \vec{p}$. Under the transformation: $L' = (- r) \times (- p) = r \times p = L$ .

Step 1: Final Conclusion

Since both position and momentum vectors change sign, their cross product (angular momentum) retains its original sign. It does not flip.

Pattern Recognition

Angular momentum is a pseudovector (or axial vector). True vectors (polar vectors) flip signs under spatial inversion, but pseudovectors (which are cross products of two polar vectors) do not.

Chapter Mix

Class 11 Physics: Systems of Particles and Rotational Motion

Q50 jee_main_2026_28_january_evening Rotation about Fixed Axis

A fly wheel having mass

3 kg

and radius

5 m

is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to a

3 kg

mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by

3 m

is ____ J. (

g = 10 m/s²

)

Numerical Answer. Answer: 30 to 30

Solution

Related Formula

Δ K.E. + Δ P.E. = 0

Kwheel = (1)/(2) I ω²

Idisc = (m R²)/(2)

Core Logic

By conservation of mechanical energy, the loss in gravitational potential energy of the descending mass equals the gain in kinetic energy of the block and the rotational kinetic energy of the flywheel.

mg h = (1)/(2) I ω² + (1)/(2) m v²

Step 1: Relate Velocity and Angular Velocity

Since the string does not slip, the linear velocity $v$ $v$ of the mass is related to the angular velocity $ω$ $\omega$ of the wheel by:

v = ω R \Rightarrow ω = (v)/(R)

Step 2: Energy Conservation

The flywheel is treated as a solid disc/cylinder (from the $I = mR²/2$ $I = mR^2/2$ usage in the solution):

mg h = (1)/(2) ((M R²)/(2)) ω² + (1)/(2) m v²

Given $M = 3 kg$ $M = 3 \text{ kg}$ (wheel), $m = 3 kg$ $m = 3 \text{ kg}$ (block), $h = 3 m$ $h = 3 \text{ m}$. Notice that the solution text specifies the flywheel mass as $m$ $m$ and block mass also as $m$ $m$, both being $3 kg$ $3\text{ kg}$. Let's follow the PDF exactly:

mg \times 3 = (1)/(2) ((m R²)/(2)) ω² + (1)/(2) m v²

Step 3: Solve for Velocity Squared

Substitute $ω R = v$ $\omega R = v$:

mg \times 3 = (1)/(4) m v² + (1)/(2) m v²

3mg = (3)/(4) m v²

v² = 4g = 4 \times 10 = 40 (m/s)²

Step 4: Calculate Kinetic Energy of Flywheel

K.E.wheel = (1)/(2) I ω² = (1)/(4) m v²

K.E.wheel = (1)/(4) \times 3 \times 40 = 30 J

Pattern Recognition

For a mass pulling a wheel of identical mass (disc), the total $K.E.$ $K.E.$ is split between translational ( $1/2 mv²$ $1/2 mv^2$) and rotational ( $1/4 mv²$ $1/4 mv^2$). Rotational gets exactly $1/3$ $1/3$ of the total potential energy lost, $30 J$ $30\text{ J}$ out of $90 J$ $90\text{ J}$ total.

Chapter Mix

Class 11 Physics: Systems of Particles and Rotational Motion Class 11 Physics: Work, Energy and Power

Solution & Explanation

Related Formula

Core Logic

Step 1: Setup Momentum Equations

Step 2: Solve for Velocity

Pattern Recognition

Chapter Mix

Reference Study Guides

More Center of Mass and Collisions Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Calculate Mass of Carved Cylinder

Step 2: Calculate Inertia of Carved Cylinder

Step 3: Finding the Ratio

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Moment of Inertia of Discs

Step 2: Moment of Inertia of Rod

Step 3: Total Moment of Inertia

Step 4: Calculating Angular Acceleration

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Applying Parallel Axis Theorem

Step 2: Substituting Values

Step 3: Final Answer

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Relate Velocity and Angular Velocity

Step 2: Energy Conservation

Step 3: Solve for Velocity Squared

Step 4: Calculate Kinetic Energy of Flywheel

Pattern Recognition

Chapter Mix

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