JEE Main · Physics · Chapter 6

Centre of Mass JEE Main PYQs (2020–2026)

89 questions from Centre of Mass have appeared in JEE Main between 2020 and 2026, including 52 in the last five papers (2022–2026). The most-asked topic is Linear Momentum, Impulse, and System Dynamics (31 questions). Practise them all with instant answers and step-by-step solutions.

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Solved Centre of Mass PYQs

3 recent questions with answers. The full set of 89 is free to practise.

JEE Main 2024MediumCenter of Mass of Discrete Particle Systems
Q1
Three identical spheres each of mass $2 M$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $4 m$ each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is $\frac{4 2}{x}$ , where the value of $x$ is ___________ .
Numerical answer type
Show answer & solution
Correct answer: 3
Center of Mass Coordinates:

The center of mass (COM) of a system of particles is calculated as:

$X_{COM} = \frac{\sum _{i} m _{i} x _{i}}{\sum _{i} m _{i}}$

$Y_{COM} = \frac{\sum _{i} m _{i} y _{i}}{\sum _{i} m _{i}}$

where $m_{i}$ is the mass of the $i$ -th particle, and $(x_{i}, y_{i})$ are its coordinates.

Coordinate Setup:

Let’s place the origin at the right angle of the triangle and align the sides along the x and y axes:

Sphere 1: $(4, 0)$

Sphere 2: $(0, 4)$

Sphere 3: $(0, 0)$

Calculations:

Since all spheres have mass $2 M$ , we can simplify the COM calculations:

$X_{COM} = \frac{2 M \cdot 4 + 2 M \cdot 0 + 2 M \cdot 0}{2 M + 2 M + 2 M} = \frac{4}{3}$

$Y_{COM} = \frac{2 M \cdot 0 + 2 M \cdot 4 + 2 M \cdot 0}{2 M + 2 M + 2 M} = \frac{4}{3}$

Magnitude of Position Vector:

The position vector of the COM is $(\frac{4}{3}, \frac{4}{3})$ . Its magnitude is:

$∣ r_{COM} ∣ = (\frac{4}{3})^{2} + (\frac{4}{3})^{2} = \frac{4 2}{3}$

Finding x:

We are given that the magnitude of the position vector is of the form $\frac{4 2}{x}$ . Comparing this to our result, we find that $x = 3$ .

Answer:

The value of $x$ is 3.
JEE Main 2024MediumMotion of Center of Mass
Q2
A solid circular disc of mass $50 k g$ rolls along a horizontal floor so that its center of mass has a speed of $0.4 m / s$ . The absolute value of work done on the disc to stop it is ________ J.
Numerical answer type
Show answer & solution
Correct answer: 6
Using work energy theorem

$W = Δ K E = 0 - (\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}) W = 0 - \frac{1}{2} m v^{2} (1 + \frac{K ^{2}}{R ^{2}}) = - \frac{1}{2} \times 50 \times 0.4^{2} (1 + \frac{1}{2}) = - 6 J$

Absolute work $= + 6 J$

$W = - 6 J ∣ W ∣ = 6 J$
JEE Main 2024EasyLinear Momentum, Impulse, and System Dynamics
Q3
A stationary particle breaks into two parts of masses $m_{A}$ and $m_{B}$ which move with velocities $v_{A}$ and $v_{B}$ respectively. The ratio of their kinetic energies $(K_{B} : K_{A})$ is :
- A $v_{B} : v_{A}$
- B $1 : 1$
- C $m_{B} v_{B} : m_{A} v_{A}$
- D $m_{B} : m_{A}$
Show answer & solution
Correct answer: (A)
A stationary particle breaks into two parts with masses $m_{A}$ and $m_{B}$ , which then move with velocities $v_{A}$ and $v_{B}$ , respectively. We need to determine the ratio of their kinetic energies $K_{A}$ and $K_{B}$ .

Since the initial momentum of the particle is zero, the momentum of the two parts must be equal and opposite to conserve momentum:

$m_{A} v_{A} = m_{B} v_{B}$

Here, the ratio of kinetic energies is given by:

$\frac{K _{A}}{K _{B}} = \frac{\frac{1}{2} m _{A} v _{A}^{2}}{\frac{1}{2} m _{B} v _{B}^{2}} = \frac{m _{A} v _{A}^{2}}{m _{B} v _{B}^{2}}$

However, using the momentum relationship, we can substitute $m_{A} v_{A} = m_{B} v_{B}$ into the kinetic energy ratio, leading us to simplify:

$\frac{K _{A}}{K _{B}} = \frac{v _{A}}{v _{B}}$

Therefore, the ratio of their kinetic energies $(K_{B} : K_{A})$ is:

$\frac{K _{B}}{K _{A}} = \frac{v _{B}}{v _{A}}$