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Question

The mass $$100\ g$$ and $$300\ g$$ at a given time have velocities $$10\hat i-7\hat j-3\hat k$$ and $$7\hat i-9\hat j+6\hat k$$ respectively. Determine velocity of $$COM$$.


Solution

$$m_1 = 100 g = 0.1 kg$$, and 

$$v_1 = (10i - 7j -3k) m/s$$ 

$$m_2 = 300 g = 0.3 kg$$, and 

$$v_2 = (7i - 9j + 6k) m/s$$ 

Velocity of COM in x direction $$V_x =\dfrac {10\times 0.1 + 7\times 0.3}{0.1+0.3} = \dfrac {31}{4 }$$

Velocity of COM in y direction $$V_y =\dfrac {-7\times 0.1 - 9\times 0.3}{0.1+0.3} = \dfrac{-34}{4 }$$

Velocity of COM in z direction $$V_z =\dfrac {-3\times 0.1 + 6\times 0.3}{0.1+0.3} = \dfrac {15}{4 }$$

Thus, velocity of the center of mass, $$V = V_x + V_y + V_z = \dfrac {31-34+15}{4} =3m/s $$


Physics

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