Question

Particles of masses 10 g and 20 g have position vectors (5,3,0) and (2,0,3)

respectively. The position vector of their center of mass is ____.
​

Answer 1

Answer :

The position vector of their center of mass is  $\sf (3 \hat{i}+\hat{j}+2\hat{k})$

Explanation :

Given,

• Particle of mass 10 g has position vector (5,3,0)
• Particle of mass 20 g has position vector (2,0,3)

To find,

• The position vector of their center of mass

Solution,

Let

• m₁ = 10 g
• m₂ = 20 g
• x₁ = 5 , y₁ = 3 , z₁ = 0
• x₂ = 2 , y₂ = 0 , z₂ = 3

Also,

  Let the position vector of their center of mass be (x,y,z)

we know,

$\underline{\boxed{\sf x=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}}} ; \underline{\boxed{\sf y=\dfrac{m_1y_1+m_2y_2}{m_1+m_2}}} ; \underline{\boxed{\sf z=\dfrac{m_1z_1+m_2z_2}{m_1+m_2}}}$

Substitute the values,

 $\sf x=\dfrac{10(5)+20(2)}{10+20} \\\\ x=\dfrac{50+40}{30} \\\\ x=\dfrac{90}{30} \\\\ x=3$

$\sf y=\dfrac{10(3)+20(0)}{10+20} \\\\ y=\dfrac{30+0}{30} \\\\ y=\dfrac{30}{30} \\\\ y=1$

$\sf z=\dfrac{10(0)+20(3)}{10+20} \\\\ z=\dfrac{0+60}{30} \\\\ z=\dfrac{60}{30} \\\\ z=2$

⇒ (x,y,z) = (3,1,2)

Therefore,

The position vector of their center of mass is $\sf (3 \hat{i}+\hat{j}+2\hat{k})$