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In the arrangement shown in Figure, mA=2...

In the arrangement shown in Figure, `m_A=2kg` and `m_B=1kg`. String is light and inextensible. Find the acceleration of centre of mass of both the blocks. Neglect friction everywhere.

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The correct Answer is:
A, C
Net pulling force on the system is `(m_A-m_B)g`

or `(2-1)g=g`
Total mass being pulled is `m_A+m_B` or `3kg`
`:.` `a="Net pulling force"/"Total mass" =g/3`
Now, `a_(COM)=(m_Aa_A+m_Ba_B)/(m_A+m_B)=((2)(a)-(1)(a))/(1+2)=a/3=g/9` (downwards)
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DC PANDEY-CENTRE OF MASS, LINEAR MOMENTUM AND COLLISION-Level 2 Subjective
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