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Center of Mass question

2022 · 29 Jul · Shift 1 · Q53
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  5. /2022 · 29 Jul · Shift 1 · Q53

Center of Mass question

2022 · 29 Jul · Shift 1 · Q53

JEE MainPhysicsCenter of MassMCQ+4 / −1
Two bodies of mass 1 kg1 \mathrm{~kg}1 kg and 3 kg3 \mathrm{~kg}3 kg have position vectors i^+2j^+k^\hat{i}+2 \hat{j}+\hat{k}i^+2j^​+k^ and −3i^−2j^+k^-3 \hat{i}-2 \hat{j}+\hat{k}−3i^−2j^​+k^ respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector :
  1. A
    i^+2j^+k^\hat{i}+2 \hat{j}+\hat{k}i^+2j^​+k^
  2. B
    −3i^−2j^+k^-3 \hat{i}-2 \hat{j}+\hat{k}−3i^−2j^​+k^
  3. C
    −2i^+2k^-2 \hat{i}+2 \hat{k}−2i^+2k^
  4. D
    2i^−j^+2k^2 \hat{i}-\hat{j}+2 \hat{k}2i^−j^​+2k^
View written solutionFree

Correct answer: A

  1. Write the given data

Masses: m1=1 kg,m2=3 kgm_1=1\text{ kg},\quad m_2=3\text{ kg}m1​=1 kg,m2​=3 kg

Position vectors: r⃗1=i^+2j^+k^\vec r_1=\hat i+2\hat j+\hat kr1​=i^+2j^​+k^ r⃗2=−3i^−2j^+k^\vec r_2=-3\hat i-2\hat j+\hat kr2​=−3i^−2j^​+k^

  1. Use the centre of mass formula

The position vector of the centre of mass is R⃗=m1r⃗1+m2r⃗2m1+m2\vec R=\frac{m_1\vec r_1+m_2\vec r_2}{m_1+m_2}R=m1​+m2​m1​r1​+m2​r2​​

Substitute the values: R⃗=1(i^+2j^+k^)+3(−3i^−2j^+k^)1+3\vec R=\frac{1(\hat i+2\hat j+\hat k)+3(-3\hat i-2\hat j+\hat k)}{1+3}R=1+31(i^+2j^​+k^)+3(−3i^−2j^​+k^)​

  1. Simplify the numerator

First compute: 3(−3i^−2j^+k^)=−9i^−6j^+3k^3(-3\hat i-2\hat j+\hat k)=-9\hat i-6\hat j+3\hat k3(−3i^−2j^​+k^)=−9i^−6j^​+3k^

Now add: i^+2j^+k^+(−9i^−6j^+3k^)=−8i^−4j^+4k^\hat i+2\hat j+\hat k+(-9\hat i-6\hat j+3\hat k)=-8\hat i-4\hat j+4\hat ki^+2j^​+k^+(−9i^−6j^​+3k^)=−8i^−4j^​+4k^

So, R⃗=−8i^−4j^+4k^4=−2i^−j^+k^\vec R=\frac{-8\hat i-4\hat j+4\hat k}{4}=-2\hat i-\hat j+\hat kR=4−8i^−4j^​+4k^​=−2i^−j^​+k^

  1. Find the magnitude of the centre of mass position vector

∣R⃗∣=(−2)2+(−1)2+(1)2|\vec R|=\sqrt{(-2)^2+(-1)^2+(1)^2}∣R∣=(−2)2+(−1)2+(1)2​ ∣R⃗∣=4+1+1=6|\vec R|=\sqrt{4+1+1}=\sqrt{6}∣R∣=4+1+1​=6​

  1. Compare with the magnitudes of the given options

  • Option A: ∣i^+2j^+k^∣=12+22+12=6\left|\hat i+2\hat j+\hat k\right|=\sqrt{1^2+2^2+1^2}=\sqrt{6}​i^+2j^​+k^​=12+22+12​=6​

  • Option B: ∣−3i^−2j^+k^∣=9+4+1=14\left|-3\hat i-2\hat j+\hat k\right|=\sqrt{9+4+1}=\sqrt{14}​−3i^−2j^​+k^​=9+4+1​=14​

  • Option C: ∣−2i^+2k^∣=(−2)2+02+22=8\left|-2\hat i+2\hat k\right|=\sqrt{(-2)^2+0^2+2^2}=\sqrt{8}​−2i^+2k^​=(−2)2+02+22​=8​

  • Option D: ∣2i^−j^+2k^∣=4+1+4=9=3\left|2\hat i-\hat j+2\hat k\right|=\sqrt{4+1+4}=\sqrt{9}=3​2i^−j^​+2k^​=4+1+4​=9​=3

Only Option A has magnitude 6\sqrt{6}6​.

  1. Conclusion

The magnitude of the position vector of the centre of mass is equal to the magnitude of: i^+2j^+k^\boxed{\hat i+2\hat j+\hat k}i^+2j^​+k^​ So the correct option is A.

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