JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 28)
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 :
_8th_April_Morning_Slot_en_28_1.png)
$${a \over 5}\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right)$$
$${a \over 5}\left( {\mathop i\limits^ \wedge - \mathop j\limits^ \wedge } \right)$$
$${a }\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right)$$
Zero
Explanation
$${\overrightarrow a _A} = - a\widehat i$$; $${\overrightarrow a _B} = - a\widehat j$$
$${\overrightarrow a _C} = - a\widehat i$$; $${\overrightarrow a _D} = - a\widehat j$$
$${\overrightarrow a _{cm}} = {{{m_a}{{\overrightarrow a }_a} + {m_b}{{\overrightarrow a }_b} + {m_c}{{\overrightarrow a }_c} + {m_d}{{\overrightarrow a }_d}} \over {{m_a} + {m_b} + {m_c} + {m_d}}}$$
$${\overrightarrow a _{cm}} = {{ - ma\widehat i + 2m\widehat j + 3ma\widehat i - 4ma\widehat j} \over {10m}}$$
= $${{2ma\widehat i - 2ma\widehat j} \over {10m}} = {a \over 5}\widehat i - {a \over 5}\widehat j = {a \over 5}\left( {\widehat i - \widehat j} \right)$$
$${\overrightarrow a _C} = - a\widehat i$$; $${\overrightarrow a _D} = - a\widehat j$$
$${\overrightarrow a _{cm}} = {{{m_a}{{\overrightarrow a }_a} + {m_b}{{\overrightarrow a }_b} + {m_c}{{\overrightarrow a }_c} + {m_d}{{\overrightarrow a }_d}} \over {{m_a} + {m_b} + {m_c} + {m_d}}}$$
$${\overrightarrow a _{cm}} = {{ - ma\widehat i + 2m\widehat j + 3ma\widehat i - 4ma\widehat j} \over {10m}}$$
= $${{2ma\widehat i - 2ma\widehat j} \over {10m}} = {a \over 5}\widehat i - {a \over 5}\widehat j = {a \over 5}\left( {\widehat i - \widehat j} \right)$$
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