Question

particles of masses 100 and 300 gram have position vectors (2î+5j+13k) and (-6î+4j+2k) position vector of centre of mass is

Answer 1

Answer:

X = M₁X₁ + M₂X₂ / M₁ + M₂

= (100 x 2) + (300 x -6) / 100 + 300

= 200 - 1800 / 400

= -1600 / 400

= -4

Y = M₁Y₁ + M₂Y₂ / M₁ + M₂

= (100 x 5) + (300 x 4) / 400

= 500 + 1200 / 400

= 1700 / 400

= 4.25

Z = M₁Z₁ + M₂Z₂ / M₁ + M₂

= (100 x 13) + (300 x -2) / 400

= 1300 - 600 / 400

= 700 / 400

= 1.75

Position of Centre of Mass is X=-4, Y=4.25, Z=1.75

Answer 2

Answer:

$= -4i + \frac{17}{4} j + \frac{19}{4} k$

Given:

Mass of first particle, $m_1$ = 100 g = 0.1 kg

Mass of second particle, $m_2$ = 300 g = 0.3 kg

Position vector of first particle, $r_1$ = 2i + 5j + 13k

Position vector of second particle, $r_2$ = -6i + 4j + 2k

Solution:

Let the position vector of centre of mass be $r_{C.O.M.}$ and it is given by,

$r_{C.O.M.} = \frac{m_1r_1 + m_2r_2}{m_1+m_2}$

             $= \frac{0.1(2i + 5j + 13k)+0.3(-6i + 4j + 2k)}{0.1+0.3}$

            $= \frac{0.2i + 0.5j + 1.3k - 1.8i + 1.2j + 0.6k }{0.4}$

            $= \frac{-1.6i + 1.7j + 1.9k}{0.4}$

            $= \frac{-1.6i}{0.4} +\frac{1.7j}{0.4} +\frac{1.9k}{0.4}$

            $= -4i + \frac{17}{4} j + \frac{19}{4} k$

Hence, the position vector of the center of mass  $= -4i + \frac{17}{4} j + \frac{19}{4} k$

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