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 9.
Q31 (2026)
Two cars A and B each of mass $10^{3} \text{ kg}$ are moving on parallel tracks separated by a distance of $10 \text{ m}$, in same direction with speeds $72 \text{ km/h}$ and $36 \text{ km/h}$. The magnitude of angular momentum of car A with respect to car B is ________ $\text{J}\cdot\text{s}$.
- $3.6 \times 10^{5}$
- $10^{5}$
- $3 \times 10^{5}$
- $2 \times 10^{5}$
### Related Formula
$$L_{\text{rel}} = m \cdot V_{\text{rel}} \cdot r_{\perp}$$
### Core Logic
The relative angular momentum of a particle translating uniformly with respect to another moving reference frame can be found using their relative velocity and the perpendicular distance between their lines of motion.
Relative velocity of A with respect to B:
$$V_{\text{rel}} = V_A - V_B = 72 - 36 = 36 \text{ km/h}$$
### Step 1: Conversion to SI units
$$V_{\text{rel}} = 36 \times \frac{5}{18} \text{ m/s} = 10 \text{ m/s}$$
### Step 2: Final Conclusion
Using the perpendicular distance ($r_{\perp} = 10 \text{ m}$) and mass ($m = 10^3 \text{ kg}$):
$$L = m \cdot V_{\text{rel}} \cdot r_{\perp}$$
$$L = 1000 \times 10 \times 10 = 10^{5} \text{ kg m}^2\text{/s} \text{ (or J}\cdot\text{s)}$$
### Pattern Recognition
For parallel tracks, the angular momentum of one translating body relative to another is simply $m \times v_{\text{relative}} \times \text{track separation}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Systems of Particles and Rotational Motion
Class 11 Physics: Kinematics
Q32 (2026)
The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M. Two blocks of mass of M and m are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its centre. The magnitudes of the acceleration experienced by the blocks is ________ (assume no slipping of string on pulley.)
{{IMG1}}
- $\frac{(M-m)g}{\left[\left(\frac{13}{6}\right)M+m\right]}$
- $\frac{(M-m)g}{M+m}$
- $\frac{(M-m)g}{\left[\left(\frac{8}{3}\right)M+m\right]}$
- $\frac{(M-m)g}{2M+m}$
### Related Formula
$Mg - T_2 = Ma$
$T_1 - mg = ma$
$$(T_2 - T_1)r = I \alpha = I \left(\frac{a}{r}\right)$$
### Core Logic
First, evaluate the moment of inertia of the pulley. The pulley consists of:
1. A thin rim of mass $M$ and radius $r$.
2. Two rods, each of mass $M$ and length $2r$ (diameter).
Moment of inertia of the rim: $I_{\text{rim}} = Mr^2$
Moment of inertia of two rods about the center: $I_{\text{rods}} = 2 \times \left(\frac{M(2r)^2}{12}\right) = 2 \times \frac{4Mr^2}{12} = \frac{2}{3}Mr^2$
Total $I = Mr^2 + \frac{2}{3}Mr^2 = \frac{5}{3}Mr^2$
### Step 1: Force Equations
From the free body diagrams of the descending block (mass $M$) and ascending block (mass $m$):
$$Mg - T_2 = Ma \quad \text{--- (1)}$$
$$T_1 - mg = ma \quad \text{--- (2)}$$
Torque on the pulley:
$$(T_2 - T_1)r = I\alpha = I\frac{a}{r} \implies T_2 - T_1 = \frac{I}{r^2}a \quad \text{--- (3)}$$
### Step 2: Final Conclusion
Adding equations (1), (2), and (3):
$$Mg - mg = \left(M + m + \frac{I}{r^2}\right)a$$
Substitute $I = \frac{5}{3}Mr^2$:
$$(M - m)g = \left(M + m + \frac{5M}{3}\right)a$$
$$(M - m)g = \left(\frac{8M}{3} + m\right)a$$
$$a = \frac{(M-m)g}{\left[\left(\frac{8}{3}\right)M + m\right]}$$
### Pattern Recognition
For a real pulley system, the effective mass acting against acceleration incorporates an inertia term: $m_{\text{eff}} = m_1 + m_2 + \frac{I}{R^2}$. Breaking the complex pulley into fundamental shapes (ring + rods) solves the inertia safely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Systems of Particles and Rotational Motion
Class 11 Physics: Laws of Motion
Q26 (2026)
A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is \_\_\_\_ kg.m $^{2}$
- $0.36$
- $0.72$
- $0.18$
- $0.63$
### Related Formula
$$I = \frac{7}{5}(m_1 R_1^2 + m_2 R_2^2)$$
### Core Logic
Substitute the given mass and radius values into the standard moment of inertia formula for spheres about their common tangent at the contact point:
$$I = \frac{7}{5}[5(10)^{2} + 10 \times (20)^{2}] \times 10^{-4}$$
$$I = 63 \times 10^{-2} \text{ kg m}^2 = 0.63 \text{ kg m}^2$$
### Pattern Recognition
Sees: Two touching solid spheres + moment of inertia about tangent at contact point.
Shortcut: Apply parallel/perpendicular axis theorem adjustments directly via standard formula summation.
Check: Calculations yield $0.63 \text{ kg m}^2$, matching option (4). ✓
### Chapter Mix
Class 11 Physics: Rotational Motion
Q49 (2026)
A circular disc has radius $R_{1}$ and thickness $T_{1}$. Another circular disc made of the same material has radius $R_{2}$ and thickness $T_{2}$. If the moment of inertia of both discs are same and $\frac{R_{1}}{R_{2}} = 2$ then $\frac{T_{1}}{T_{2}} = \frac{1}{\alpha}$. The value of $\alpha$ is \_\_\_\_.
### Related Formula
$$I = \frac{mR^2}{2}, \quad m = \pi R^2 T \rho$$
### Core Logic
{{SOL_IMG1}}
Mass of discs:
$$m_1 = \pi R_1^2 T_1 \rho, \quad m_2 = \pi R_2^2 T_2 \rho$$
Moments of inertia:
$$I_1 = \frac{m_1 R_1^2}{2}, \quad I_2 = \frac{m_2 R_2^2}{2}$$
Equating $I_1 = I_2$:
$$\frac{\pi R_1^2 T_1 \rho R_1^2}{2} = \frac{\pi R_2^2 T_2 \rho R_2^2}{2} \implies \frac{T_1}{T_2} = \left(\frac{R_2}{R_1}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$$
Therefore, $\alpha = 16$.
### Pattern Recognition
Sees: Equal moment of inertia for two discs of different radii and thicknesses.
Shortcut: Equate $mR^2$ expressions and substitute radius ratio $\frac{R_1}{R_2} = 2$.
Check: Numerical answer is 16. ✓
### Chapter Mix
Class 11 Physics: Rotational Motion
Q29 (2026)
A uniform bar of length 12 cm and mass 20 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in opposite directions with same speed of v and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency $\omega$. The ratio of v and $\omega$ is :
{{IMG1}}
- $33$
- $2\sqrt{88}$
- $66$
- $32$
### Related Formula
$L_i = L_f$
$$I = \frac{M L_{bar}^2}{12} + \sum m_i r_i^2$$
### Core Logic
Applying angular momentum conservation about the center of mass of the rod:
$$L_i = m \cdot v \cdot 4 + 2m \cdot v \cdot 2$$
Calculating total moment of inertia $I_{final}$ after collision:
$$I_{final} = \left( \frac{20m(12)^2}{12} + m(4)^2 + 2m(2)^2 \right)$$
$$I_{final} = (240m + 16m + 8m) = 264m$$
Equating initial and final angular momentum:
$$4mv + 4mv = 264m \cdot \omega$$
$$8v = 264 \omega \implies \frac{v}{\omega} = \frac{264}{8} = 33$$
{{SOL_IMG1}}
### Step 1: Final Conclusion
The ratio $\frac{v}{\omega}$ is 33.
### Pattern Recognition
Collision on rotating bar: Conserve angular momentum about COM of rod since net external torque about COM is zero during collision.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
