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📘 Center of Mass Momentum and Collision — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Center of Mass Momentum and Collision for JEE Main with previous year questions on impulse, elastic and inelastic collisions. Each solution is explained in a simple, step-by-step manner to boost your score.
Center of Mass Momentum and Collision — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (7 April Shift 1)
A rod of length 5 L is bent right angle keeping one side length as 2 L. The position of the centre of mass of the system: (Consider L = 10 cm)
(1) \(2\hat{i} + 3\hat{j}\) (2) \(3\hat{i} + 7\hat{j}\)
(3) \(5\hat{i} + 8\hat{j}\) (4) \(4\hat{i} + 9\hat{j}\)
(3) \(5\hat{i} + 8\hat{j}\) (4) \(4\hat{i} + 9\hat{j}\)
✅ Answer & Explanation:
Correct Option: (4)
- The rod has parts 2L (along x) and 3L (along y).
- Center of mass of 2L part (mass \(2m\)) is at \((L, 0)\).
- Center of mass of 3L part (mass \(3m\)) is at \((0, 1.5L)\).
- \(X_{cm} = \frac{2m(L) + 3m(0)}{5m} = 0.4L\).
- \(Y_{cm} = \frac{2m(0) + 3m(1.5L)}{5m} = 0.9L\).
- Given \(L = 10 \, \text{cm}\), \(X_{cm} = 4\) and \(Y_{cm} = 9\).
- Position vector is \(4\hat{i} + 9\hat{j}\).
Q2. JEE Main 2025 (23 Jan Shift 1)
Consider a circular disc of radius 20 cm with centre located at the origin. A circular hole of radius 5 cm is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of centre of mass of residual or remaining disc from the origin will be :
(1) 2.0 cm (2) 1.5 cm (3) 1.0 cm (4) 0.5 cm
(1) 2.0 cm (2) 1.5 cm (3) 1.0 cm (4) 0.5 cm
✅ Answer & Explanation:
Correct Option: (3)
- Using the formula for residual center of mass: \(X = -\frac{a x}{A - a}\)
- Area of disc \(A = \pi(20)^2\), Area of hole \(a = \pi(5)^2\).
- Center of hole is at \(x = 20 - 5 = 15 \, \text{cm}\).
- \(X = -\frac{25\pi \times 15}{400\pi - 25\pi} = -\frac{375}{375} = -1.0 \, \text{cm}\).
- The distance is 1.0 cm.
Q3. JEE Main 2024 (08 Apr Shift 1)
A uniform thin metal plate of mass 10 kg with dimensions is shown. The ratio of x and y coordinates of center of mass of plate is \(n/9\). The value of n is _____ :
✅ Answer & Explanation:
Correct Answer: 15
By dividing the plate into standard rectangles and calculating the center of mass for each, the ratio \(x_{cm}/y_{cm}\) leads to the value n = 15.
By dividing the plate into standard rectangles and calculating the center of mass for each, the ratio \(x_{cm}/y_{cm}\) leads to the value n = 15.
Q4. JEE Main 2020 (08 Jan Shift 2)
As shown in figure. When a spherical cavity (centred at O) of radius 1 is cut out of a uniform sphere of radius R (centred at C), the centre of mass of remaining (shaded) part of sphere is at G, i.e., on the surface of the cavity. R can be determined by the equation :
(1) \((R^2 + R + 1)(2 - R) = 1\) (2) \((R^2 - R - 1)(2 - R) = 1\)
(3) \((R^2 - R + 1)(2 - R) = 1\) (4) \((R^2 + R - 1)(2 - R) = 1\)
(3) \((R^2 - R + 1)(2 - R) = 1\) (4) \((R^2 + R - 1)(2 - R) = 1\)
✅ Answer & Explanation:
Correct Option: (1)
- Using the center of mass balancing for volume: \(V_{\text{shaded}} \cdot CG = V_{\text{cavity}} \cdot CO\).
- Placing the geometry into this relation yields the cubic equation in Option (1).
Q5. JEE Main 2020 (08 Jan Shift 1)
The coordinates of the centre of mass of a uniform flag-shaped lamina (thin flat plate) of mass 4 kg. (The coordinates of the same are shown in the figure) are :
(1) (1.25 m, 1.50 m) (2) (0.75 m, 1.75 m)
(3) (0.75 m, 0.75 m) (4) (1 m, 1.75 m)
(3) (0.75 m, 0.75 m) (4) (1 m, 1.75 m)
✅ Answer & Explanation:
Correct Option: (2)
- Divide the lamina into two rectangles:
- Rectangle 1: Corners (0,0) to (1,3). Area = 3. COM = (0.5, 1.5).
- Rectangle 2: Corners (1,2) to (2,3). Area = 1. COM = (1.5, 2.5).
- \(X_{cm} = \frac{3(0.5) + 1(1.5)}{4} = \frac{3}{4} = 0.75 \, \text{m}\).
- \(Y_{cm} = \frac{3(1.5) + 1(2.5)}{4} = \frac{7}{4} = 1.75 \, \text{m}\).
Q6. JEE Main 2024 (30 Jan Shift 1)
A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m. The impulse of force imparted by the ground to the body is given by : (given \(g = 9.8 \, \text{m/s}^2\))
(1) \(4.32 \, \text{kg m s}^{-1}\) (2) \(43.2 \, \text{kg m s}^{-1}\)
(3) \(23.9 \, \text{kg m s}^{-1}\) (4) \(2.39 \, \text{kg m s}^{-1}\)
(1) \(4.32 \, \text{kg m s}^{-1}\) (2) \(43.2 \, \text{kg m s}^{-1}\)
(3) \(23.9 \, \text{kg m s}^{-1}\) (4) \(2.39 \, \text{kg m s}^{-1}\)
✅ Answer & Explanation:
Correct Option: (4)
- Velocity just before hit: \(v_1 = \sqrt{2 \times 9.8 \times 10} = 14 \, \text{m/s}\) (down).
- Velocity just after rebound: \(v_2 = \sqrt{2 \times 9.8 \times 5} = 9.9 \, \text{m/s}\) (up).
- Impulse \(J = m|v_2 - (-v_1)| = 0.1(9.9 + 14) = 0.1(23.9) = 2.39 \, \text{kg m s}^{-1}\).
Q7. JEE Main 2024 (01 Feb Shift 2)
A uniform rod AB of mass 2 kg and Length 30 cm at rest on a smooth horizontal surface. An impulse of force 0.2 N s is applied to end B. The time taken by the rod to turn through at right angles will be \(\pi/x \, \text{s}\), where \(x =\) _____.
✅ Answer & Explanation:
Correct Answer: 4
- Angular impulse \(J \cdot (\frac{L}{2}) = I \omega\).
- \(0.2 \times 0.15 = (\frac{1}{12} \times 2 \times 0.3^2) \omega \implies 0.03 = 0.015 \omega \implies \omega = 2 \, \text{rad/s}\).
- Time for \(90^\circ (\pi/2)\): \(t = \theta/\omega = (\pi/2)/2 = \pi/4 \, \text{s}\).
- Thus, \(x = 4\).
Q8. JEE Main 2023 (11 Apr Shift 2)
A nucleus disintegrates into two nuclear parts, in such a way that ratio of their nuclear sizes is \(1 : 2^{1/3}\). Their respective speed have a ratio of \(n : 1\). The value of \(n\) is _________.
✅ Answer & Explanation:
Correct Answer: 2
- Nuclear radius \(R \propto A^{\frac{1}{3}} \implies R_1/R_2 = (A_1/A_2)^{\frac{1}{3}}\).
- Given \(1/2^{\frac{1}{3}} = (A_1/A_2)^{\frac{1}{3}} \implies A_1/A_2 = \frac{1}{2}\).
- By momentum conservation: \(m_1 v_1 = m_2 v_2 \implies v_1/v_2 = m_2/m_1 = A_2/A_1 = 2\).
- So, \(n = 2\).
Q9. JEE Main 2025 (3 April Shift 2)
Consider two blocks A and B of masses \(m_1 = 10 \, \text{kg}\) and \(m_2 = 5 \, \text{kg}\) that are placed on a frictionless table. The block A moves with a constant speed \(v = 3 \, \text{m/s}\) towards the block B kept at rest. A spring with spring constant \(k = 3000 \, \text{N/m}\) is attached with the block B as shown in the figure. After the collision, suppose that the blocks A and B, along with the spring in constant compression state, move together, then the compression in the spring is :
(1) 0.2 m (2) 0.4 m (3) 0.1 m (4) 0.3 m
✅ Answer & Explanation:
Correct Option: (3)
- Conservation of Momentum: \(10(3) = (10+5)V \implies V = 2 \, \text{m/s}\).
- Conservation of Energy at max compression:
- \(\frac{1}{2} m_1 v_1^2 = \frac{1}{2} (m_1+m_2) V^2 + \frac{1}{2} k x^2\)
- \(45 = 30 + 1500 x^2 \implies 1500 x^2 = 15 \implies x^2 = 0.01 \implies x = 0.1 \, \text{m}\).
Q10. JEE Main 2025 (29 Jan Shift 2)
Assertion (A): Three identical spheres of same mass undergo one dimensional motion as shown in figure with initial velocities \(v_A = 5 \, \text{m/s}, v_B = 2 \, \text{m/s}, v_C = 4 \, \text{m/s}\). If we wait sufficiently long for elastic collision to happen, then \(v_A = 4 \, \text{m/s}, v_B = 2 \, \text{m/s}, v_C = 5 \, \text{m/s}\) will be the final velocities.
Reason (R): In an elastic collision between identical masses, two objects exchange their velocities.
(1) (A) is false but (R) is true
(2) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
(3) Both (A) and (R) are true and (R) is the correct explanation of (A)
(4) (A) is true but (R) is false
Reason (R): In an elastic collision between identical masses, two objects exchange their velocities.
(2) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
(3) Both (A) and (R) are true and (R) is the correct explanation of (A)
(4) (A) is true but (R) is false
✅ Answer & Explanation:
Correct Option: (1)
Reason (R) is a standard law of physics. However, applying this to the sequence of collisions for A, B, and C with given velocities shows that the final state described in Assertion (A) is not the resulting sequence after all possible collisions.
Reason (R) is a standard law of physics. However, applying this to the sequence of collisions for A, B, and C with given velocities shows that the final state described in Assertion (A) is not the resulting sequence after all possible collisions.
Q11. JEE Main 2025 (29 Jan Shift 1)
As shown below, bob A of a pendulum having massless string of length 'R' is released from \(60^\circ\) to the vertical. It hits another bob B of half the mass that is at rest on a friction less table in the center. Assuming elastic collision, the magnitude of the velocity of bob A after the collision will be :
(1) \(\frac{4}{3} \sqrt{Rg}\) (2) \(\frac{2}{3} \sqrt{Rg}\)
(3) \(\sqrt{Rg}\) (4) \(\frac{1}{3} \sqrt{Rg}\)
(3) \(\sqrt{Rg}\) (4) \(\frac{1}{3} \sqrt{Rg}\)
✅ Answer & Explanation:
Correct Option: (4)
- Velocity of A just before collision: \(v = \sqrt{2gR(1-\cos 60^\circ)} = \sqrt{gR}\).
- Let \(m_A = m\) and \(m_B = \frac{m}{2}\).
- Velocity after elastic collision: \(v_A = \frac{m_A - m_B}{m_A + m_B} v = \frac{m - \frac{m}{2}}{m + \frac{m}{2}} \sqrt{gR} = \frac{\frac{1}{2}}{\frac{3}{2}} \sqrt{gR} = \frac{1}{3} \sqrt{gR}\).
Q12. JEE Main 2021 (18 Mar Shift 1)
A ball of mass 10 kg moving with a velocity \(10\sqrt{3} \, \text{m/s}\) along the x-axis, hits another ball of mass 20 kg which is at rest. After the collision, first ball comes to rest while the second ball disintegrates into two equal pieces. One piece starts moving along y-axis with a speed of \(10 \, \text{m/s}\). The second piece starts moving at an angle of \(30^\circ\) with respect to the x-axis. The velocity of the ball moving at \(30^\circ\) with x-axis is \(x \, \text{m/s}\). The value of \(x\) to the nearest integer is ________.
✅ Answer & Explanation:
Correct Answer: 20
- Initial momentum: \(P_x = 10 \times 10\sqrt{3} = 100\sqrt{3}\).
- After collision, 2nd ball (20kg) splits into two 10kg pieces.
- Piece 1: \(\vec{p}_1 = 10 \times 10 \hat{j} = 100 \hat{j}\).
- Piece 2: \(\vec{p}_2 = 10x(\cos 30^\circ \hat{i} - \sin 30^\circ \hat{j})\).
- Momentum conservation along X: \(100\sqrt{3} = 10x \cdot \frac{\sqrt{3}}{2} \implies x = 20\).
Q13. JEE Main 2020 (09 Jan Shift 2)
A particle of mass m is projected with a speed u from the ground at an angle \(\theta = \pi/3\) w.r.t. horizontal (x-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity \(u\hat{i}\). The horizontal distance covered by the combined mass before reaching the ground is :
(1) \(\frac{3\sqrt{3}}{8} \frac{u^2}{g}\) (2) \(\frac{3\sqrt{2}}{4} \frac{u^2}{g}\)
(3) \(\frac{5}{8} \frac{u^2}{g}\) (4) \(2\sqrt{2} \frac{u^2}{g}\)
(1) \(\frac{3\sqrt{3}}{8} \frac{u^2}{g}\) (2) \(\frac{3\sqrt{2}}{4} \frac{u^2}{g}\)
(3) \(\frac{5}{8} \frac{u^2}{g}\) (4) \(2\sqrt{2} \frac{u^2}{g}\)
✅ Answer & Explanation:
Correct Option: (1)
- Horizontal velocity at max height \(v_1 = u \cos 60^\circ = \frac{u}{2}\).
- Combined mass velocity after inelastic collision: \(V = \frac{m(\frac{u}{2}) + m(u)}{2m} = \frac{3u}{4}\).
- Time to reach ground from max height: \(t = \sqrt{2H/g}\).
- Horizontal distance after collision: \(R' = V \times t\).
- Substituting values leads to the expression in Option (1).
📘 Exam Preparation Tip for Center of Mass, Momentum & Collision
- Center of Mass: For composite bodies, treat each part as a point mass at its own COM. Use \(X_{cm} = \frac{\sum m_i x_i}{\sum m_i}\). For bodies with holes, use the concept of negative mass.
- Conservation of Momentum: \( \sum \vec{p}_{initial} = \sum \vec{p}_{final} \) is valid for all collisions (elastic and inelastic) when no external force acts. This is the most powerful tool for collision problems.
- Impulse: \( \vec{J} = \Delta \vec{p} = \vec{F}_{avg} \Delta t \). Remember it's a vector quantity. For impacts, impulse is the change in momentum.
- Elastic Collisions: Both momentum and kinetic energy are conserved. For 1D elastic collision of two masses: \(v_1' = \frac{m_1 - m_2}{m_1 + m_2}v_1 + \frac{2m_2}{m_1 + m_2}v_2\). For identical masses, they simply exchange velocities.
- Inelastic Collisions: Momentum is conserved but kinetic energy is not. In a perfectly inelastic collision, the bodies stick together and move with a common velocity.
- Coefficient of Restitution: \(e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}}\). For elastic, \(e = 1\). For perfectly inelastic, \(e = 0\).
- Center of Mass Frame: Analyzing collisions in the COM frame often simplifies the problem, as the total momentum is zero in this frame.