Two particles of masses 1 kg and 3 kg have position vectors 2- - 3- - 4-k- and -2- - 3- - 4-k- respectively- The centre of mass has a position vector-

The correct option is D î + 3̂ĵ 2̂k̂ m_1=1 kg r⃗_⃗1⃗ =2î+3ĵ+4k̂ m_2 = 3 kg r⃗_⃗2⃗= 2î+3ĵ 4k̂ position vector of centre of mass will be ...

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Center of Mass as an Average Point

Two particles...

Question

Two particles of masses 1 kg and 3 kg have position vectors $^2 i +^3 j +^4 k$ and $-^2 i +^3 j -^4 k$ respectively. The centre of mass has a position vector:

^i +^3 j -^2 k

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-^i -^3 j -^2 k

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-^i +^3 j +^2 k

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-^i +^3 j -^2 k

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Solution

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m_{1}, m_{2}, m_{3}, m_{4}

_{1} =^2 i -^j +^k

_{2} =^3 i -^4 j -^4 k

_{3} = -^i +^j -^k

_{4} = -^i +^3 j +^k

\to a =^i +^j +^k, \to b =^i -^j +^k, \to c =^2 i +^3 j -^k

(\to a \times \to b) \times \to c =

^2 i +^3 j +^3 k

-^3 i =^4 j -^2 k

^i +^2 k

^i -^p j

\to a =^i +^j -^2 k

\to b =^2 i -^j +^k

\to c =^3 i -^k

m \to a + n \to b

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