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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

BITSAT 2024 · Paper 2 · Physics · Center of Mass, Momentum and Collision

Two bodies of mass 1 kg and 3 kg have position vectors i +2 j +k...

Two bodies of mass 1 kg and 3 kg have position vectors i^+2j^+k\hat{i}+2 \hat{j}+k and −3i^−2j^+k^-3 \hat{i}-2 \hat{j}+\hat{k} respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector :

Options

  1. Ai^−2j^+k^\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}
  2. B−3i^−2j^+k^-3 \hat{i}-2 \hat{j}+\hat{k}
  3. C−2i^+2k^-2 \hat{i}+2 \hat{k}
  4. D−2i^−j^+2k^-2 \hat{i}-\hat{j}+2 \hat{k}

Correct answer

A. i^−2j^+k^\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}

Step-by-step solution

Position of COM of a mass - system is given as, r⃗com=m1r⃗1+m2r⃗2m1+m2=1(i^+2j^+k^)+3(−3i^−2j^+k^1+3=−2i^−j^+k^∣r⃗cm∣=∣−2i^−j^+k^∣=(2)2+(1)2+(1)2=6\vec{r}_{\mathrm{com}}=\frac{m_1 \vec{r}_1+m_2 \vec{r}_2}{m_1+m_2}=\frac{1(\hat{i}+2 \hat{j}+\hat{k})+3(-3 \hat{i}-2 \hat{j}+\hat{k}}{1+3}=-2 \hat{i}-\hat{j}+\hat{k}\left|\vec{r}_{c m}\right|=|-2 \hat{i}-\hat{j}+\hat{k}|=\sqrt{(2)^2+(1)^2+(1)^2}=\sqrt{6} Only option (1) magnitude is 12+(−2)2+12=6\sqrt{1^2+(-2)^2+1^2}=\sqrt{6} So option (1) is correct.
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