Question

a a 13. A 2 kg mass and a 4 kg mass are placed along the X-axis at distances of 12 m and 4 m from the origin. Where should a third mass of 5 kg is to be placed so that the centre of mass of the system lies at the origin? a) 4 m b) -8 m d) -3 m c) 3 m​

Answer 1

Answer:

balance of the 2 kg and 4 kg at 12 m and 4 m respectively,

$2*12 + 4*4 = 24 + 16 = 40$

now, to balance the system,

$5x + 40 = 0 \\ x = - 8 \: m$

Answer 2

The distance of the 3rd Mass so that the center of mass of the system lies at the origin is -8m towards the left side. So, option 'B' is correct.

Given:-

Mass of 1st particle = 2kg

Mass of 2nd particle = 4kg

Mass of 3rd particle = 5kg

Distance of 1st Mass = 12m

Distance of 2nd Mass = 4m

To Find:-

The distance of the 3rd Mass so that the center of mass of the system lies at the origin.

Solution:-

We can simply calculate the distance of the 3rd Mass so that the center of mass of the system lies at the origin by using these simple steps.

As

Mass of 1st particle (m1) = 2kg

Mass of 2nd particle (m2) = 4kg

Mass of 3rd particle (m3) = 5kg

Distance of 1st Mass (x1) = 12m

Distance of 2nd Mass (x2) = 4m

Distance of 3rd Mass (x3) =?

COM = 0

According to the formula of the Center of Mass of a given system,

$x(com) = \frac{(m1 \times x1) + (m2 \times x2) + (m3 \times x3)}{m1 + m2 + m3}$

Now, putting the values we get,

$0 = \frac{(2 \times 12) + (4 \times 4) + (5 \times x3)}{2 + 4 + 5}$

$0 = \frac{24 + 16 + 5 \times x3}{11}$

$11 \times 0 = 40 + 5 \times x3$

$40 + 5 \times x3 = 0$

$5x3 = - 40$

$x3 = \frac{ - 40}{5}$

$x3 = - 8m$

Hence, The distance of the 3rd Mass so that the center of mass of the system lies at the origin is 8m towards the left side. So, option 'B' is correct.