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Question: Two particles of mass 1 kg and 3 kg have position vectors \(2\widehat{i} + 3\widehat{j} + 4\widehat{...

Two particles of mass 1 kg and 3 kg have position vectors 2i^+3j^+4k^2\widehat{i} + 3\widehat{j} + 4\widehat{k}and −2i^+3j^−4k^- 2\widehat{i} + 3\widehat{j} - 4\widehat{k}respectively. The centre of mass has a position vector

A

i^+3j^−2k^\widehat{i} + 3\widehat{j} - 2\widehat{k}

B

−i^−3j^−2k^- \widehat{i} - 3\widehat{j} - 2\widehat{k}

C

−i^+3j^+2k^- \widehat{i} + 3\widehat{j} + 2\widehat{k}

D

−i^+3j^−2k^- \widehat{i} + 3\widehat{j} - 2\widehat{k}

Answer

−i^+3j^−2k^- \widehat{i} + 3\widehat{j} - 2\widehat{k}

Explanation

Solution

Here, m1=1kg,m2=3kgm_{1} = 1kg,m_{2} = 3kg

r→1=2i^+3j^+4k^,r2=−2i^+3j^−4k^{\overset{\rightarrow}{r}}_{1} = 2\widehat{i} + 3\widehat{j} + 4\widehat{k},r_{2} = - 2\widehat{i} + 3\widehat{j} - 4\widehat{k}

The position vector of the centre of mass is

r→CM=m1r→1+m2r→2m1+m2=(1)(2i^+3j^+4k^)+(3)(−2i^+3j^−4k^)1+3=2i^+3j^+4k^−6i^+9j^−12k^4=−4i^+12j^−8k^4=−i^+3j^−2k^{\overset{\rightarrow}{r}}_{CM} = \frac{m_{1}{\overset{\rightarrow}{r}}_{1} + m_{2}{\overset{\rightarrow}{r}}_{2}}{m_{1} + m_{2}} = \frac{(1)(2\widehat{i} + 3\widehat{j} + 4\widehat{k}) + (3)( - 2\widehat{i} + 3\widehat{j} - 4\widehat{k})}{1 + 3} = \frac{2\widehat{i} + 3\widehat{j} + 4\widehat{k} - 6\widehat{i} + 9\widehat{j} - 12\widehat{k}}{4} = \frac{- 4\widehat{i} + 12\widehat{j} - 8\widehat{k}}{4} = - \widehat{i} + 3\widehat{j} - 2\widehat{k}