Q 11-06-214JEE MainJEE Main 2019 (8 Apr, Shift 1)Easy
Four particles A, B, C and D with masses $m_A = m$, $m_B = 2m$, $m_C = 3m$ and $m_D = 4m$ are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is
Answer: (B) $\dfrac a5(\hat i - \hat j)$
From the figure: A accelerates along $-\hat i$, B along $+\hat j$, C along $+\hat i$ and D along $-\hat j$.
$$\vec a_{cm} = \frac{\sum m\vec a}{\sum m} = \frac{m(-a\hat i) + 2m(a\hat j) + 3m(a\hat i) + 4m(-a\hat j)}{10m} = \frac{2a\hat i - 2a\hat j}{10}$$
$$\vec a_{cm} = \frac a5(\hat i - \hat j)$$
Solution by Sreeraj P, M.Sc Physics