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QJEEM19Ph2eec: JEE Main 2019 Physics - Class 11 JEEM19Ph2eec Solution: Four particles A, B, C and D with masses m A = m, m B = 2m, m C …

Four particles A, B, C and D with masses mA = m, mB = 2m, mC = 3m and mD = 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 : JEE Main 2019 (Online) 8th April Morning Slot Physics - Center of Mass and Collision Question 92 English

Multiple Choice Options

  • A a5(i∧+j∧){a \over 5}\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right)
  • B a5(i∧−j∧){a \over 5}\left( {\mathop i\limits^ \wedge - \mathop j\limits^ \wedge } \right)
  • C a(i∧+j∧){a }\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right)
  • D Zero

Step-by-Step Verified Solution

a→A=−ai^{\overrightarrow a _A} = - a\widehat i ;    a→B=−aj^{\overrightarrow a _B} = - a\widehat j

a→C=−ai^{\overrightarrow a _C} = - a\widehat i ;    a→D=−aj^{\overrightarrow a _D} = - a\widehat j

a→cm=maa→a+mba→b+mca→c+mda→dma+mb+mc+md{\overrightarrow a _{cm}} = {{{m_a}{{\overrightarrow a }_a} + {m_b}{{\overrightarrow a }_b} + {m_c}{{\overrightarrow a }_c} + {m_d}{{\overrightarrow a }_d}} \over {{m_a} + {m_b} + {m_c} + {m_d}}}

a→cm=−mai^+2mj^+3mai^−4maj^10m{\overrightarrow a _{cm}} = {{ - ma\widehat i + 2m\widehat j + 3ma\widehat i - 4ma\widehat j} \over {10m}}

= 2mai^−2maj^10m=a5i^−a5j^=a5(i^−j^){{2ma\widehat i - 2ma\widehat j} \over {10m}} = {a \over 5}\widehat i - {a \over 5}\widehat j = {a \over 5}\left( {\widehat i - \widehat j} \right)

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