Center of Mass - JEE Advanced Previous Year Questions with Solutions
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JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects. Click Here for JEE main Previous Year Topic Wise Questions of Physics with Solutions Download eSaral app for free study material and video tutorials. Simulator Previous Years JEE Advanced Questions 
[IIT-JEE-2009] 
(A) 4 (B) 3 (C) 2 (D) 1 [IIT-JEE-2009] 
Particle with velocity 'v' covers and angle of $120^{\circ}$ and after collision its velocity become '2v'. It will cover angle of $$240^{\circ}$$ 
$(\mathrm{A}) \frac{\mathrm{a}}{10}$ (B) $\frac{\mathrm{a}}{8}$ (C) $\frac{\mathrm{a}}{12}$ (D) $\frac{\mathrm{a}}{3}$ [IIT-JEE-2009] 
Conservation of linear momentum (1) u = – (1) 2 + (5) v 5v –2 = u ... (i) By definition of ‘e’ $1=\frac{v+2}{u} \Rightarrow v+2=u$... (ii) By solving above equations $\mathrm{v}=1 \mathrm{ms}^{-1}$ and $\mathrm{u}=3 \mathrm{ms}^{-1}$ For $(\mathrm{A}):$ Total momentum of system $=1 \times \mathrm{u}=3 \mathrm{kg} \mathrm{ms}^{-1}$ For (B) : Momentum of 5 kg after collision $=5(1)=5 \mathrm{kg} \mathrm{ms}^{-1}$ For $(\mathrm{C}): \mathrm{K}_{\mathrm{cm}}=\frac{1}{2}(1+5)\left(\frac{1 \times 3+0}{1+5}\right)=0.75 \mathrm{J}$ For (D) : Total kinetic energy of the system $=\frac{1}{2}(1)(3)^{2}=4.5 \mathrm{J}$ 
(A) 4.50 J (B) 7.50 J (C) 5.06 J (D) 14.06 J [IIT-JEE-2010] 
[IIT-JEE 2011] (A) $250 \mathrm{m} / \mathrm{s}$ (B) $250 \sqrt{2} \mathrm{m} / \mathrm{s}$ (C) $400 \mathrm{m} / \mathrm{s}$ (D) $500 \mathrm{m} / \mathrm{s}$ 
(A) $\theta=45^{\circ}$ (B) $\theta>45^{\circ}$ and a frictional force acts on the block towards P (C) $\theta>45^{\circ}$ and a frictional force acts on the block towards Q (D) $\theta>45^{\circ}$ and a frictional force acts on the block towards Q [IIT-JEE 2012] 
(1) If $\sin \theta=\cos \theta \Rightarrow \theta=45^{\circ} \Rightarrow$ no friction will act and the block will remain at rest. (2) If $\sin \theta>\cos \theta \Rightarrow \theta>45^{\circ} \Rightarrow$ friction will act towards $\mathrm{Q}$ (3) If $\sin \theta<\cos \theta \Rightarrow \theta<45^{\circ} \Rightarrow$ friction will act towards $\mathrm{P}$ 
$\sqrt{9 \ell_{1}}=\sqrt{5 \mathrm{g} \ell_{2}} \Rightarrow \frac{\ell_{1}}{\ell_{2}}=5$ 
[JEE Advanced-2014] 
(A) The x component of displacement of the centre of mass of the block M is : $-\frac{\mathrm{mR}}{\mathrm{M}+\mathrm{m}}$ (B) The position of the point mass is : $\mathrm{x}=-\sqrt{2} \frac{\mathrm{mR}}{\mathrm{M}+\mathrm{m}}$ (C) The velocity of the point mass m is : $\mathrm{v}=\sqrt{\frac{2 \mathrm{g} \mathrm{R}}{1+\frac{\mathrm{m}}{\mathrm{M}}}}$ (D) The velocity of the block M is : $\mathrm{V}=-\frac{\mathrm{m}}{\mathrm{M}} \sqrt{2 \mathrm{gR}}$ [JEE Advanced-2017] 
$\mathrm{M}_{\mathrm{s}} \Delta \overline{\mathrm{x}}_{\mathrm{cm}}=\mathrm{m}_{1} \Delta \overline{\mathrm{x}}+\mathrm{m}_{2} \Delta \overline{\mathrm{x}}_{2}$ $0=\mathrm{m}(+\mathrm{R}+\overline{\mathrm{x}})+\mathrm{m} \overline{\mathrm{x}}$ $\overline{\mathrm{x}}=\frac{-\mathrm{mR}}{\mathrm{M}+\mathrm{m}}$ (A) ans 
$0=\mathrm{m} \overline{\mathrm{v}}_{1}+\mathrm{M} \overline{\mathrm{v}}_{2}$ $\overline{\mathrm{v}}_{2}=-\frac{\mathrm{m} \overline{\mathrm{v}}_{1}}{\mathrm{M}}$ $\operatorname{mg} \mathrm{R}=\frac{1}{2} \mathrm{mv}_{1}^{2}+\frac{1}{2} \mathrm{Mv}_{2}^{2}$ $\mathrm{mgR}=\frac{1}{2} \mathrm{mv}_{1}^{2}+\frac{1}{2} \mathrm{M}\left(\frac{\mathrm{mv}_{1}}{\mathrm{M}}\right)^{2}$ $\operatorname{mg} \mathrm{R}=\frac{1}{2} \mathrm{mv}_{1}^{2}\left(1+\frac{\mathrm{m}}{\mathrm{M}}\right)$ $\sqrt{\frac{2 \mathrm{gR}}{\left(1+\frac{\mathrm{m}}{\mathrm{M}}\right)}}=\mathrm{v}_{1}$ 
(A) $\Delta=\operatorname{hsin}^{2}\left(\frac{\pi}{\mathrm{n}}\right)$ (B) $\Delta=\mathrm{h} \sin \left(\frac{2 \pi}{\mathrm{n}}\right)$ (C) $\Delta=\mathrm{h}\left(\frac{1}{\cos \left(\frac{\pi}{\mathrm{n}}\right)}-1\right)$ (D) $\Delta=h \tan ^{2}\left(\frac{\pi}{2 n}\right)$ [JEE Advanced-2017] 
OA = h $\mathrm{OB}=\frac{\mathrm{h}}{\cos \frac{\pi}{\mathrm{n}}}$ Initial height of COM = h Final height of $\mathrm{COM}=\frac{\mathrm{h}}{\cos \left(\frac{\pi}{\mathrm{n}}\right)}$ 
[JEE Advanced-2018] 
$\mathrm{d}=\frac{2}{3}(\pi)=\frac{2}{3}(3.14)=2.0933 \mathrm{m}$ d = 2.09 m 
$\mathbf{V}=\mathbf{V}_{0} \mathbf{e}^{-\mathbf{t} / \tau}$ $\mathrm{v}_{0}=\frac{\mathrm{J}}{\mathrm{m}}=2.5 \mathrm{m} / \mathrm{s}$ $\mathbf{V}=\mathbf{V}_{0} \mathbf{e}^{-\mathbf{t} / \tau}$ $\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}_{0} \mathrm{e}^{-\mathrm{t} / \tau}$ $\int_{0}^{\mathrm{x}} \mathrm{dx}=\mathrm{v}_{0} \int_{0}^{\mathrm{r}} \mathrm{e}^{-t / \tau} \mathrm{dt} \quad \int \mathrm{e}^{-\mathrm{x}} \mathrm{d} \mathrm{x}=\frac{\mathrm{e}^{-\mathrm{x}}}{-1}$ 
Frequently Asked Questions
Find answers to common questions.
What is the difference between elastic and inelastic collision in JEE problems?
In an elastic collision, both kinetic energy and linear momentum are conserved. In a completely inelastic collision, only linear momentum is conserved — the objects stick together and kinetic energy is lost. JEE Advanced frequently tests problems where the first collision is elastic and the second is inelastic (as in Q1 of 2009), requiring students to apply different formulas in sequence.
Is Center of Mass in JEE Main syllabus as well?
Yes. Center of Mass is part of the JEE Main syllabus under the unit "Laws of Motion and Work, Energy, Power." JEE Main questions are generally single-step and formula-based, whereas JEE Advanced questions combine COM with projectile motion, rotational mechanics, or spring-mass systems in multi-step problems.
How many questions on Center of Mass appear in JEE Advanced each year?
Typically 1–2 questions on Center of Mass appear in JEE Advanced per year, sometimes bundled within a larger paragraph-type or multi-correct set. Over the period 2009–2023, the topic has contributed an average of 8–12 marks per paper cycle, making it one of the most consistently tested mechanics sub-topics.
Why do JEE Advanced COM problems often involve circular motion or projectile motion together?
JEE Advanced is designed to test multi-concept application, not isolated formula recall. The exam paper is set by IIT faculty who intentionally combine topics — for example, the 2013 question links elastic collision with the minimum speed condition for circular motion. Practising topic-wise questions (like this set) and then full mixed-topic papers is the most efficient two-phase preparation strategy.
Which chapters from Class 11 NCERT should I study before attempting these questions?
You should be confident in Chapter 5 (Laws of Motion), Chapter 6 (Work, Energy and Power), and Chapter 7 (System of Particles and Rotational Motion) from Class 11 Physics. These three chapters together form the complete theory base for all COM and collision problems in JEE Advanced. Detailed solutions are available at NCERT Solutions for Class 11 Physics.
How do you find the velocity of COM in a collision problem?
The velocity of COM is given by $v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$. Crucially, $v_{cm}$ does not change during a collision if no external force acts on the system. This is a key property used in multi-correct questions where the KE of COM is asked — as seen in the 2010 question above.