Two bodies of masses 2 kg 5 kg have position vectors - - 2 - - k- and 3 - - 2 - - k- respectively- The centre of mass of the system has a position vector

Position vector of centre of mass for descrete particles, r⃗_⃗c⃗o⃗m⃗ = ∑ m_i r⃗_i∑ m_i = m_1 r⃗_⃗1⃗ + m_2 r⃗_⃗2⃗m_1 + m_2 (Given, m_1 = 2 kg , r⃗_⃗1⃗ = i⃗ 2 ĵ ...

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Byju's Answer

Standard XIII

Physics

Center of Mass as an Average Point

Two bodies of...

Question

Two bodies of masses $2 kg$ & $5 kg$ have position vectors $^i - 2^j +^k$ and $3^i + 2^j -^k$ respectively. The centre of mass of the system has a position vector

2^i + 4^j - 3^k

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(\frac{4^i + 5^j - 3^k}{7})

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17^i + 6^j - 3^k

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(\frac{17^i + 6^j - 3^k}{7})

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Solution

The correct option is D $(\frac{17^i + 6^j - 3^k}{7})$
Position vector of centre of mass for descrete particles, $_{com} = \frac{\sum m_{i} {\to r}_{i}}{\sum m_{i}} = \frac{m_{1}_{1} + m_{2}_{2}}{m_{1} + m_{2}}$
(Given, $m_{1} = 2 kg,_{1} = \to i - 2^j +^k$
$m_{2} = 5 kg,_{2} = 3^i + 2^j -^k$ )
$∵$ $_{com} = \frac{m_{1}_{1} + m_{2}_{2}}{m_{1} + m_{2}}$
$= \frac{2 (^i - 2^j +^k) + 5 (3^i + 2^j -^k)}{2 + 5}$
$= (\frac{17^i + 6^j - 3^k}{7})$

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Two bodies of masses 2 kg & 5 kg have position vectors ^i−2^j+^k and 3^i+2^j−^k respectively. The centre of mass of the system has a position vector

The correct option is D (17^i+6^j−3^k7)Position vector of centre of mass for descrete particles, →rcom=∑mi→ri∑mi=m1→r1+m2→r2m1+m2 (Given, m1=2 kg,→r1=→i−2^j+^k m2=5 kg, →r2=3^i+2^j−^k) ∵ →rcom=m1→r1+m2→r2m1+m2 =2(^i−2^j+^k)+5(3^i+2^j−^k)2+5 =(17^i+6^j−3^k7)

Two bodies of masses $2 kg$ & $5 kg$ have position vectors $^i - 2^j +^k$ and $3^i + 2^j -^k$ respectively. The centre of mass of the system has a position vector