Three bodies of masses 1 kg - 3 kg and 4 kg have position vectors -2 -k-3 -2 -k- and 4 - respectively - The centre of mass of this system has position vectorA- 2 -k-2B- 2 -k-C- 2 -k-2D- 2 -k-2

The correct option is D (2 î ĵ + k̂)2 Given, m_1 = 1 kg, x_1 = î + 2ĵ + k̂ m_2 = 3 kg, x_2 = 3î 2ĵ + k̂ m_2 = 4 kg, x_3 = 4 î r⃗_COM = ∑_i =1^nm_i x_i ...

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Byju's Answer

Standard XII

Physics

Center of Mass as an Average Point

Three bodies ...

Question

Three bodies of masses $1 kg$ , $3 kg$ and $4 kg$ have position vectors $(^i + 2^j +^k)$ , $(- 3^i - 2^j +^k)$ and $(4^i)$ respectively . The centre of mass of this system has position vector

\frac{(2^i +^j -^k)}{2}

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\frac{(2^i +^j +^k)}{2}

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\frac{(2^i -^j -^k)}{2}

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\frac{(2^i -^j +^k)}{2}

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Solution

The correct option is D $\frac{(2^i -^j +^k)}{2}$
Given,
$m_{1} = 1 kg, x_{1} =^i + 2^j +^k$
$m_{2} = 3 kg, x_{2} = - 3^i - 2^j +^k$
$m_{2} = 4 kg, x_{3} = 4^i$
${\to r}_{COM} = \sum_{i = 1}^{n} \frac{m_{i} x_{i}}{m_{i}} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$
$= \frac{1 (^i + 2^j +^k) + 3 (- 3^i - 2^j +^k) + 4 (4^i)}{1 + 3 + 4}$
$= \frac{8^i - 4^j + 4^k}{8} = \frac{(2^i -^j +^k)}{2}$

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Three bodies of masses 1 kg, 3 kg and 4 kg have position vectors (^i+2^j+^k), (−3^i−2^j+^k) and (4^i) respectively . The centre of mass of this system has position vector

The correct option is D (2^i−^j+^k)2 Given, m1=1 kg, x1=^i+2^j+^k m2=3 kg, x2=−3^i−2^j+^k m2=4 kg, x3=4^i →rCOM=∑ni=1miximi=m1x1+m2x2+m3x3m1+m2+m3 =1(^i+2^j+^k)+3(−3^i−2^j+^k)+4(4^i)1+3+4 =8^i−4^j+4^k8=(2^i−^j+^k)2

Three bodies of masses $1 kg$ , $3 kg$ and $4 kg$ have position vectors $(^i + 2^j +^k)$ , $(- 3^i - 2^j +^k)$ and $(4^i)$ respectively . The centre of mass of this system has position vector