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 138 questions; sign in to attempt the remaining 133 in the exam simulator.

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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
   Show answer & solution

   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 _____ $\mathrm{cm}$.
   Show answer & solution

   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 (06 Apr, Shift 2)Energy
   A body is dropped on ground from a height ${h}_{1}$ and after hitting the ground, it rebounds to a height ${h}_{2}$. If the ratio of velocities of the body just before and after hitting ground is $4$, then percentage loss in kinetic energy of the body is $\dfrac{x}{4}$. The value of $x$ is _____.
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   Answer: 375

   The given ratio of the velocities is $\dfrac{{v}_{1}}{{v}_{2}}=4$ The percentage loss in kinetic energy is $\dfrac{\dfrac{m{{v}^{2}}_{1}}{2}-\dfrac{m{{v}^{2}}_{2}}{2}}{\dfrac{m{{v}^{2}}_{1}}{2}}\times 100=\dfrac{x}{4} \Rightarrow \dfrac{{{v}^{2}}_{1}-{{v}^{2}}_{2}}{{{v}^{2}}_{1}}=\dfrac{x}{400} \Rightarrow 1-\dfrac{{{v}^{2}}_{2}}{{{v}^{2}}_{1}}=\dfrac{x}{400} \Rightarrow 1-\dfrac{1}{16}=\dfrac{x}{400} \Rightarrow x=375$
4. Q4JEE Main 2022 (28 Jul, Shift 1)Impulse
   In two different experiments, an object of mass $5kg$ moving with a speed of $25{ms}^{-1}$ hits two different walls and comes to rest within (i) $3$ second, (ii) $5$ seconds, respectively. Choose the correct option out of the following :
   1. A.Impulse and average force acting on the object will be same for both the cases.
   2. B.Impulse will be same for both the cases but the average force will be different.
   3. C.Average force will be same for both the cases but the impulse will be different.
   4. D.Average force and impulse will be different for both the cases.
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   Answer: (B)

   Impulse $=$ Change in momentum $I=\Delta P$ Therefore, in both the cases the impulse is same. Average force is defined as, ${F}_{avg}=\dfrac{\Delta P}{\Delta t}$ Since the time in the two cases is different, the average force will be different.
5. Q5JEE Main 2021 (27 Jul, Shift 2)Uniform Motion
   A particle of mass $M$ originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation $F={F}_{0}\left[1-{\left(\dfrac{t-T}{T}\right)}^{2}\right]$ where ${F}_{0}$ and $T$ are constants. The force acts only for the time interval $2T$. The velocity $v$ of the particle after time $2T$ is:
   1. A.$\dfrac{2{F}_{0}T}{M}$
   2. B.$\dfrac{{F}_{0}T}{2M}$
   3. C.$\dfrac{4{F}_{0}T}{3M}$
   4. D.$\dfrac{{F}_{0}T}{3M}$
   Show answer & solution

   Answer: (C)

   $t=0,u=0$ $a=\dfrac{{F}_{o}}{M}-\dfrac{{F}_{o}}{M{T}^{2}}(t-T{)}^{2}=\dfrac{dv}{dt}$ ${\int }_{0}^{v}dv={\int }_{t=0}^{2T}\left(\dfrac{{F}_{o}}{M}-\dfrac{{F}_{o}}{M{T}^{2}}(t-T{)}^{2}\right)dt$ $V={\left[\dfrac{{F}_{o}}{M}t\right]}_{o}^{2T}-\dfrac{{F}_{o}}{M{T}^{2}}{\left[\dfrac{{t}^{3}}{3}-{t}^{2}T+{T}^{2}t\right]}_{0}^{2T}$ $V=\dfrac{4{F}_{o}T}{3M}$