Question

AB is a straight line of length 150 cm. Two particles of masses 1 kg and 3 kg are placed at a distance of 15 cm from A and 50 cm from B respectively. The distance of the third particle of mass 2 kg from A, so that the centre of mass of the system is at the middle point of​ AB

Answer 1

Answer:

30 CM of mass 2 kg (third particles)

Answer 2

Answer:

Position of third particle is at 67.5 cm

Explanation:

Total length of AB =150 cm

From the figure we can see the arrangement of  masses from A and B. A is our origin and all the distances are calculated from a.

The center of mass is calculated as

$X{cm}=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2_+m_3}$

center of mass is= 75cm  (it's the mid point of AB)

$m_1=1kg, m_2=3 kg, m_3=2kg$  

$x_1=15cm$(Its 15 cm from A)

$x_2= 100cm$ (50 cm from B which is 100 cm from A)

putting all the values we get

$75=\frac{15+100*3+2*x_3}{1+3+2}=\frac{315+2x_3 }{6}$

⇒$x_3=\frac{450-315}{3}=67.5 cm$

Hence the position of third particle is 67.5 cm from A.