Question

Two bodies of mass 1 kg and 3 kg have position vector i + 2j + k and -3i - 2j + k , respectively.

Answer 1

question given is not full
so i am finding the centre of mass
x=(1×1-3×3)/1+3 y=(1×2-2×3)/4 z =(1×1+3×1)/4
= -2 = -1 = 1

Centre of mass is (-2,-1,1).

Answer 2

The center of mass of the position vector of this system is $(-2i-j+k)$

Explanation:

Given that,

Mass of first body = 1 kg

Mass of second body = 3 kg

Position vector of first body = i+2j+k

Position vector of second body = -3i-2j+k

Suppose we need to calculate the center of mass of this system has a position vector.

We need to calculate the center of mass of the position vector of this system

Using formula of center of mass

$r_{cm}=\dfrac{m_{1}r_{1}+m_{2}r_{2}}{m_{1}+m_{2}}$

Put the value into the formula

$r_{cm}=\dfrac{1\times( i+2j+k)+3\times(-3i-2j+k)}{1+3}$

$r_{cm}=\dfrac{(-8i-4j+4k)}{4}$

$r_{cm}=(-2i-j+k)$

Hence, The center of mass of the position vector of this system is $(-2i-j+k)$

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Topic : center of mass

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