Center of Mass Momentum and Collision JEE Main previous year questions with solutions

5 solved JEE Main questions on Center of Mass Momentum and Collision, free to read — no sign-in needed. The full chapter has 130 questions; sign in to attempt the remaining 125 in the exam simulator.

1. Q1JEE Main 2026 (23 Jan, Shift 1)Collisions
   In a perfectly inelastic collision, two spheres made of the same material with masses 15 kg and 25 kg, moving in opposite directions with speeds of $10 \mathrm{~m} / \mathrm{s}$ and $30 \mathrm{~m} / \mathrm{s}$, respectively, strike each other and stick together. The rise in temperature (in ${ }^{\circ} \mathrm{C}$), if all the heat produced during the collision is retained by these spheres, is : (specific heat of sphere material $31 \mathrm{cal} / \mathrm{kg}.{ }^{\circ} \mathrm{C}$ and $1 \mathrm{cal}=4.2 \mathrm{~J}$)
   1. A.$1.15$
   2. B.$1.75$
   3. C.1.44
   4. D.1.95
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   Answer: (C)

   Using conservation of momentum: $m_1 v_1 + m_2 v_2 = (m_1 + m_2)v_f$ where $v_2 = -30$ m/s (opposite direction). $15(10) + 25(-30) = 40v_f \Rightarrow v_f = -15$ m/s Initial kinetic energy: $KE_i = \frac{1}{2}(15)(10)^2 + \frac{1}{2}(25)(30)^2 = 750 + 11250 = 12000$ J Final kinetic energy: $KE_f = \frac{1}{2}(40)(15)^2 = 4500$ J Heat produced: $Q = 12000 - 4500 = 7500$ J $= \frac{7500}{4.2} = 1785.71$ cal Temperature rise: $\Delta T = \frac{Q}{m_{total} \cdot c} = \frac{1785.71}{40 \times 31} = 1.44°C$
2. Q2JEE Main 2024 (04 Apr, Shift 2)Centre of mass of discrete particles
   In a system two particles of masses $m_1=3 \mathrm{~kg}$ and $m_2=2 \mathrm{~kg}$ are placed at certain distance from each other. The particle of mass $m_1$ is moved towards the center of mass of the system through a distance $2 \mathrm{~cm}$. In order to keep the center of mass of the system at the original position, the particle of mass $m_2$ should move towards the center of mass by the distance _____ $\mathrm{cm}$.
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   Answer: 3

   $\begin{aligned} & \Delta \mathrm{X}_{\text {C.O.M. }}=\frac{\mathrm{m}_1 \Delta \mathrm{x}_1+\mathrm{m}_2 \Delta \mathrm{x}_2}{\mathrm{~m}_1+\mathrm{m}_2} \\ & \Rightarrow 0=\frac{3 \times 2+2(-\mathrm{x})}{3+2} \\ & \Rightarrow \mathrm{x}=3 \mathrm{~cm}\end{aligned}$
3. Q3JEE Main 2023 (13 Apr, Shift 1)Impulse
   A bullet of $10g$ leaves the barrel of gun with a velocity of $600m{s}^{-1}$. If the barrel of gun is $50cm$ long and mass of gun is $3kg,$ then value of impulse supplied to the gun will be:
   1. A.$6Ns$
   2. B.$3Ns$
   3. C.$36Ns$
   4. D.$12Ns$
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   Answer: (A)

   The formula to calculate the impulse $\left(J\right)$ on the gun is given by $J=m\left(v-u\right)...\left(1\right)$ where, $m$ is the mass, $u$ is the initial velocity and $v$ is the final velocity of the bullet. Substitute the values of the known parameters into equation (1) to calculate the required impulse. $\begin{matrix}J & = & 10g\times \dfrac{1kg}{1000g}\times \left(600-0\right)m{s}^{-1} \\ & = & 6Ns\end{matrix}$
4. Q4JEE Main 2022 (26 Jul, Shift 2)momentum
   A body of mass $8kg$ and another of mass $2kg$ are moving with equal kinetic energy. The ratio of their respective momenta will be
   1. A.$1:1$
   2. B.$2:1$
   3. C.$1:4$
   4. D.$4:1$
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   Answer: (B)

   Given: The mass of the objects ${m}_{1}=8kg$ and ${m}_{2}=2kg$. Let the momentum of the $8kg$ and $2kg$ masses is ${p}_{1}$ and ${p}_{2}$, respectively, then according to question, their kinetic energy are equal, i.e., ${K}_{1}={K}_{2}$ $\dfrac{{p}_{1}^{2}}{2{m}_{1}}=\dfrac{{p}_{2}^{2}}{2{m}_{2}}$ $\dfrac{{p}_{1}}{{p}_{2}}=\sqrt{\dfrac{{m}_{1}}{{m}_{2}}}=\sqrt{\dfrac{4}{1}}=\dfrac{2}{1}$
5. Q5JEE Main 2020 (06 Sep, Shift 2)Centre of mass of continuous mass distribution
   The centre of mass of a solid hemisphere of radius $8cm$ is $xcm$ from the centre of the flat surface. Then value of $x$ is
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   Answer: 3

   Centre of Mass of a solid hemisphere from centre of the flat surface is given by,${h}_{cm}=\dfrac{3R}{8}$ It is given that $R=8cm$ Therefore, $x={h}_{cm} =\dfrac{3R}{8} =\dfrac{3\times 8}{8} x=3cm$