Center of Mass - JEE Main Previous Year Questions with Solutions
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Directions : Question number 9 contain Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best discribes the two statements. [AIEEE - 2009] 
Energy loss will be maximum when collision will be perfectly inelastic 
(By momentum) Maximum energy loss $=\mathrm{K}_{\mathrm{i}}-\mathrm{K}_{\mathrm{f}}$ $=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2}(\mathrm{m}+\mathrm{M}) \mathrm{v}_{\mathrm{f}}^{2}$ $=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2}(\mathrm{m}+\mathrm{M}) \frac{\mathrm{m}^{2} \mathrm{v}^{2}}{(\mathrm{m}+\mathrm{M})^{2}}$ $=\frac{1}{2} \mathrm{mv}^{2}\left[1-\frac{\mathrm{m}}{\mathrm{m}+\mathrm{M}}\right]$ $=\left(\frac{\mathrm{M}}{\mathrm{m}+\mathrm{M}}\right) \frac{1}{2} \mathrm{mv}^{2}$ statement 1 is false. 
Kinetic energy $=\frac{1}{2} \mathrm{m}(2 \mathrm{v})^{2} \times \frac{1}{2} 2 \mathrm{m}(\mathrm{v})^{2}$ $=3 \mathrm{mv}^{2}$ After collison Applying momentum conservation for inelastic collision $2 \mathrm{mv} \hat{\mathrm{j}}+\mathrm{m} 2 \mathrm{v} \hat{\mathrm{i}}=3 \mathrm{m} \overrightarrow{\mathrm{v}}_{\mathrm{f}}$ $\left|\overrightarrow{\mathrm{v}}_{\mathrm{f}}\right|=\sqrt{\frac{8}{9}} \mathrm{v}$ $\mathrm{K}_{\mathrm{f}}=\frac{1}{2} \times 3 \mathrm{m} \times\left(\mathrm{v}_{\mathrm{f}}^{2}\right)=\frac{4 \mathrm{mv}^{2}}{3}$ $\% \Delta \mathrm{K}=\frac{\mathrm{K}_{1}-\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{i}}} \times 100=\frac{5 \mathrm{mv}^{2} / 3}{3 \mathrm{mv}^{2}} \times 100=\frac{5}{9} \times 100=56 \%$ 
for solid cone c.m. is $\frac{\mathrm{h}}{4}$ from base so $\mathrm{z}_{0}=\mathrm{h}-\frac{\mathrm{h}}{4}=\frac{3 \mathrm{h}}{4}$ 
by momentum conservation $\mathrm{mv}_{0}=\mathrm{mv}_{1}+2 \mathrm{mv}_{2} \Rightarrow \mathrm{v}_{1}+2 \mathrm{v}_{2}=\mathrm{v}_{0}$ by $\mathrm{e}=1 \quad \mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{v}_{0}$ 
Frequently Asked Questions
Find answers to common questions.
Is Center of Mass important for JEE Advanced too?
Yes. Center of Mass is equally important for JEE Advanced, where questions are typically more conceptual and multi-step. JEE Advanced may combine COM with rotational motion, variable mass systems, or impulsive forces. Mastery of JEE Main–level problems is an essential prerequisite before attempting Advanced-level questions
What is the formula for maximum energy loss in a collision?
The maximum energy loss in a collision occurs during a perfectly inelastic collision and equals [M/(m+M)] × ½mv², where m is the moving particle's mass and M is the stationary particle's mass. This is derived using momentum conservation followed by subtracting final KE from initial KE.
How many questions appear from Center of Mass in JEE Main each year?
Center of Mass and collisions typically contribute 2–4 questions per JEE Main session. The topic is part of the Systems of Particles and Rotational Motion chapter, which NTA consistently tests across both January and April sessions. Collisions (elastic and inelastic) and momentum conservation are the most frequently tested sub-topics.
How do I calculate fractional energy loss in an elastic collision?
For an elastic collinear collision between a moving particle of mass m and a stationary particle of mass M, the fractional kinetic energy loss is 4mM/(m+M)². This formula is derived by applying both momentum conservation and e = 1. It is directly used in the neutron-deuterium and neutron-carbon JEE Main 2018 question on this page.
What is the difference between elastic and inelastic collisions for JEE?
In an elastic collision, both momentum and kinetic energy are conserved (coefficient of restitution e = 1). In a perfectly inelastic collision, momentum is conserved but maximum kinetic energy is lost and the objects stick together (e = 0). JEE Main tests both types, often using statement-based or percentage-loss formats.
What is the center of mass of a solid uniform cone?
The center of mass of a solid uniform cone lies on its axis at a distance of h/4 from the base or equivalently 3h/4 from the vertex, where h is the height of the cone. This result is independent of the base radius and is derived by integrating the mass distribution along the cone's axis.