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Last updated date: 07th Dec 2023
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# Two particles whose masses are $10\,kg$ and $30\,kg$ and their position vectors are $\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge$ and $- \mathop i\limits^ \wedge - \mathop j\limits^ \wedge - \mathop k\limits^ \wedge$ respectively would have the center of mass atA. $- \dfrac{{\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right)}}{2}$B. $\dfrac{{\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right)}}{2}$C. $- \dfrac{{\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right)}}{4}$D. $\dfrac{{\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right)}}{4}$

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Hint: The center of mass is a position defined relative to an object or a system of objects. It is the average position of all parts of the given system, weighted according to their masses. There are situations in which the center of mass does not fall anywhere on the object.

Given, Mass of first particle is: ${m_1} = 10\,kg$
And the position vector is $\mathop {{r_1}}\limits^ \to = \mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge$
Mass of second particle is: ${m_2} = 30\,kg$
And the position vector is $\mathop {{r_2}}\limits^ \to = - \mathop i\limits^ \wedge - \mathop j\limits^ \wedge - \mathop k\limits^ \wedge$
$\text{Center of mass}= \dfrac{{{m_1}\mathop {{r_1}}\limits^ \to + {m_2}\mathop {{r_1}}\limits^ \to }}{{{m_1} + {m_2}}}$
$\Rightarrow \text{Center of mass} = \dfrac{{10\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right) + 30\left( { - \mathop i\limits^ \wedge - \mathop j\limits^ \wedge - \mathop k\limits^ \wedge } \right)}}{{10 + 30}}$
$\therefore \text{Center of mass} = - \left( {\dfrac{{\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge }}{2}} \right)$