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The position vector of three particles o...

The position vector of three particles of masses `m_1=1kg`.
`m_2=2kg` and `m_3=3kg` are `r_1=(hati+4hatj+hatk)m`, `r_2=(hati+hatj+hatk)m` and `r_3=(2hatj-hatj-2hatk)m` respectivley. Find the position vector of their centre of mass.

A

`(1.5hati+0.5hatj-0.5hatk)m`

B

`(0.5hati+1.5hatj-0.5hatk)m`

C

`(1.5hati+0.5hatj+0.5hatk)m`

D

`(hati+0.5hatj-0.5hatk)m`

Text Solution

Verified by Experts

The correct Answer is:
A
The positon vector of COM of the three particles is given by
`r_(COM)=(m_1r_1+m_2r_2+m_3r_3)/(m_1+m_2+m_3)`
Substituting the values, we get
`r_(COM)=((1)(hati+4hatj+hatk)+(2)(hati+hatj+hatk)+3(2hatj-hatj-2hatk))/(1+2+3)`
`=(9hati+3hatj-3hatk)/(6)`
`r_(COM)=1/2(3hati+hatj-hatk)m`
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