System of Particles and Rotational Motion Previous Year Questions — JEE Main Physics
58 past-year System of Particles and Rotational Motion questions from JEE Main (Physics) — page 10.
Q45 (2026)
Given below are two statements :
Statement I : For a mechanical system of many particles total kinetic energy is the sum of kinetic energies of all the particles.
Statement II : The total kinetic energy can be the sum of kinetic energy of the center of mass w.r.t. to the origin and the kinetic energy of all the particles w.r.t. the center of mass as the reference.
In the light of the above statements, choose the correct answer from the options given below :
- Both Statement I and Statement II are true
- Statement I is true but Statement II is false
- Statement I is false but Statement II is true
- Both Statement I and Statement II are false
### Related Formula
$$K_{\text{total}} = \sum \frac{1}{2} m_i v_i^2$$
$$K_{\text{total}} = \frac{1}{2} M_{\text{total}} v_{\text{cm}}^2 + K_{\text{rel, cm}}$$
### Core Logic
Evaluating Statement I:
By definition, total kinetic energy of a discrete system of particles is scalar sum of kinetic energies of individual particles: $KE = \sum \frac{1}{2}m_i v_i^2$. Statement I is true.
Evaluating Statement II:
According to König's theorem, total kinetic energy breaks down into kinetic energy of COM translation plus internal kinetic energy relative to COM:
$$KE = \frac{1}{2}(m_1+m_2) v_{\text{cm}}^2 + \frac{1}{2} \frac{m_1 m_2}{m_1+m_2} |\vec{v}_1 - \vec{v}_2|^2$$
Statement II is also true.
### Step 1: Final Conclusion
Both Statement I and Statement II are true.
### Pattern Recognition
König's Theorem for system of particles: $K_{\text{total}} = K_{\text{cm}} + K_{\text{w.r.t cm}}$. Both definitions represent valid expressions for mechanical energy.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q47 (2026)
Two masses m and 2m are connected by a light string going over a pulley (disc) of mass 30m with radius r = 0.1 m. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2m mass is released from rest and its speed when it has descended through a height of 3.6 m is ____ m/s. (Assume string does not slip and g = 10 m/s$^2$)
### Related Formula
$$\Delta U + \Delta K = 0$$
$$I_{\text{pulley}} = \frac{1}{2} M R^2 = \frac{1}{2} (30m) R^2 = 15m R^2$$
$$\omega = \frac{v}{R}$$
### Core Logic
Using law of conservation of mechanical energy:
Loss in potential energy = Gain in kinetic energy
$$2m g h - m g h = \frac{1}{2} m v^2 + \frac{1}{2} (2m) v^2 + \frac{1}{2} I \omega^2$$
$$m g h = \frac{3}{2} m v^2 + \frac{1}{2} \left(15 m R^2\right) \left(\frac{v^2}{R^2}\right)$$
$$m g h = \frac{3}{2} m v^2 + \frac{15}{2} m v^2 = 9 m v^2$$
$$v = \sqrt{\frac{g h}{9}}$$
Substituting $g = 10 \mathrm{~m/s}^2$ and $h = 3.6 \mathrm{~m}$:
$$v = \sqrt{\frac{10 \times 3.6}{9}} = \sqrt{\frac{36}{9}} = \sqrt{4} = 2 \mathrm{~m/s}$$
{{SOL_IMG1}}
### Step 1: Final Conclusion
The speed of the mass is $2 \mathrm{~m/s}$.
### Pattern Recognition
Energy conservation on pulley system: Net potential loss $mgh$ = sum of linear and rotational kinetic energies $9mv^2 \implies v = \sqrt{gh/9}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q33 (2026)
The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length L (R < L) about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as M):
- $\frac{3}{8}MR^2 +\frac{7}{12}ML^2$
- $\frac{3}{4}MR^2 +\frac{1}{6}ML^2$
- $\frac{3}{4}MR^2 +\frac{7}{12}ML^2$
- $\frac{3}{8}MR^2 +\frac{1}{6}ML^2$
### Related Formula
For a solid cylinder:
Parallel to its length passing through center: $I = \frac{M'R^{2}}{2}$
Perpendicular to length through center: $I = \frac{M'R^{2}}{4} + \frac{M'L^{2}}{12}$
Parallel Axis Theorem: $I = I_{\text{cm}} + M'd^{2}$
### Core Logic
The square loop is formed by four identical solid cylinders, each of mass $M' = \frac{M}{4}$. The given axis passes through the mid-points of two opposite cylinders. This means two cylinders have the axis passing perpendicularly through their centers, and the other two cylinders are parallel to the axis at a distance of $L/2$.
### Step 1: Moment of Inertia for cylinders bisected perpendicularly
For the two cylinders perpendicular to the axis of rotation:
$$I_{1} = 2 \times \left(\frac{M'R^{2}}{4} + \frac{M'L^{2}}{12}\right)$$
### Step 2: Moment of Inertia for cylinders parallel to axis
For the two cylinders parallel to the axis, distance $d = L/2$. Apply the parallel axis theorem:
$$I_{2} = 2 \times \left[\frac{M'R^{2}}{2} + M'\left(\frac{L}{2}\right)^{2}\right]$$
### Step 3: Total Moment of Inertia
$$I_{\text{net}} = I_{1} + I_{2} = 2\left(\frac{M'R^{2}}{4} + \frac{M'L^{2}}{12}\right) + 2\left(\frac{M'R^{2}}{2} + \frac{M'L^{2}}{4}\right)$$
$$I_{\text{net}} = \frac{M'R^{2}}{2} + \frac{M'L^{2}}{6} + M'R^{2} + \frac{M'L^{2}}{2}$$
$$I_{\text{net}} = \frac{3M'R^{2}}{2} + \frac{4M'L^{2}}{6} = \frac{3M'R^{2}}{2} + \frac{2M'L^{2}}{3}$$
### Step 4: Substitute Total Mass
Substitute $M' = M/4$:
$$I = \frac{3}{2}\left(\frac{M}{4}\right)R^{2} + \frac{2}{3}\left(\frac{M}{4}\right)L^{2}$$
$$I = \frac{3}{8}MR^{2} + \frac{1}{6}ML^{2}$$
### Pattern Recognition
Sees: "square loop of cylinders" + "mass M of entire loop" → Always remember $M_{i} = M/4$. Calculate individual moment of inertia carefully considering whether the cylinder is oriented parallel or perpendicular.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Systems of Particles and Rotational Motion
Q42 (2026)
Two small balls with masses m and 2m are attached to both ends of a rigid rod of length d and negligible mass. If angular momentum of this system is L about an axis (A) passing through its centre of mass and perpendicular to the rod then angular velocity of the system about A is:
- $\frac{3}{2}\frac{L}{md^{2}}$
- $\frac{2L}{md^{2}}$
- $\frac{4}{3}\frac{L}{md^{2}}$
- $\frac{2L}{5md^{2}}$
### Related Formula
$$r_{\text{cm}} = \frac{m_{1}r_{1} + m_{2}r_{2}}{m_{1} + m_{2}}$$
$$I_{\text{cm}} = \mu d^{2} = \left(\frac{m_{1}m_{2}}{m_{1} + m_{2}}\right)d^{2}$$
$L = I\omega$
### Step 1: Calculate Moment of Inertia about COM
Let mass $m$ be at the origin. Position of $2m$ is $d$.
$$X_{\text{cm}} = \frac{m(0) + 2m(d)}{m + 2m} = \frac{2d}{3}$$
Distance of mass $m$ from COM is $\frac{2d}{3}$.
Distance of mass $2m$ from COM is $d - \frac{2d}{3} = \frac{d}{3}$.
$$I = m\left(\frac{2d}{3}\right)^{2} + 2m\left(\frac{d}{3}\right)^{2}$$
$$I = \frac{4md^{2}}{9} + \frac{2md^{2}}{9} = \frac{6md^{2}}{9} = \frac{2md^{2}}{3}$$
### Step 2: Calculate Angular Velocity
$L = I\omega$
$$\omega = \frac{L}{I}$$
$$\omega = \frac{L}{\frac{2md^{2}}{3}} = \frac{3L}{2md^{2}}$$
### Pattern Recognition
Sees: "two point masses" + "rotation about COM" → Quickly use reduced mass $\mu$ moment of inertia shortcut: $I_{\text{cm}} = \mu d^{2} = (\frac{m \cdot 2m}{3m})d^2 = \frac{2}{3}md^2$ to save time.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Systems of Particles and Rotational Motion
Q27 (2026)
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.
{{IMG1}}
- $10\sqrt{2}$
- $12\sqrt{2}$
- $12$
- $24\sqrt{2}$
### Related Formula
$$\vec{p}_{\text{initial}} = \vec{p}_{\text{final}}$$
$$M_1\vec{V}_1 + M_2\vec{V}_2 + M_3\vec{V}_3 = 0$$
### Core Logic
{{SOL_IMG1}}
The total mass is $14 \, \mathrm{kg}$ and breaks in ratio $2:2:3$. Let the masses be $M_1$, $M_2$, and $M_3$.
$$M_1 = \frac{2}{7} \times 14 = 4 \, \mathrm{kg}$$
$$M_2 = \frac{2}{7} \times 14 = 4 \, \mathrm{kg}$$
$$M_3 = \frac{3}{7} \times 14 = 6 \, \mathrm{kg}$$
### Step 1: Setup Momentum Equations
Since $M_1$ and $M_2$ 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 $\vec{V}_1 = -18\hat{i}$ and $\vec{V}_2 = -18\hat{j}$. Note that the solution simplifies the mass ratio directly to $2, 2, 3$ as relative masses for the momentum equation:
$$2(-18\hat{i}) + 2(-18\hat{j}) + 3\vec{V}_3 = 0$$
### Step 2: Solve for Velocity
$$\vec{V}_3 = \frac{36\hat{i} + 36\hat{j}}{3} = 12\hat{i} + 12\hat{j}$$
Magnitude of $\vec{V}_3$:
$$|\vec{V}_3| = \sqrt{12^2 + 12^2} = 12\sqrt{2} \, \mathrm{m/s}$$
### Pattern Recognition
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.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Center of Mass and Collisions
