Center of Mass, Momentum & Collision

JEE Main level, numerical answer, medium

38 of 130

Question 38 of 130

In a system two particles of masses m1=3 kgm_1=3 \mathrm{~kg} and m2=2 kgm_2=2 \mathrm{~kg} are placed at certain distance from each other. The particle of mass m1m_1 is moved towards the center of mass of the system through a distance 2 cm2 \mathrm{~cm}. In order to keep the center of mass of the system at the original position, the particle of mass m2m_2 should move towards the center of mass by the distance _____ cm\mathrm{cm}.

Show answer and solution

Answer: 3

ΔXC.O.M. =m1Δx1+m2Δx2 m1+m2⇒0=3×2+2(−x)3+2⇒x=3 cm\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}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
Revise Rotational Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Center of Mass, Momentum & Collision (Centre of mass of discrete particles), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.