Two blocks of masses 10 kg and 20 kg are placed on the X-axis. The first mass is moved on the axis by a distance of 2 cm. By what distance should the second mass be moved to keep the position of the centre of mass unchanged?

Open in App

Solution

Let the two masses m₁ and m₂ be placed on the X-axis.
It is given that:
m₁ = 10 kg
m₂ = 20 kg

The first mass is displaced by a distance of 2 cm.
$∴ X_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} \Rightarrow X_{cm} = \frac{10 \times 2 + 20 x_{2}}{30}$

As the position of the centre of mass remains unchanged,
X_cm = 0

$\Rightarrow 0 = \frac{20 + 20 x_{2}}{30} \Rightarrow 20 + 20 x_{2} = 0 \Rightarrow 20 = - 20 x_{2} \Rightarrow x_{2} = - 1$

Therefore, to keep the position of centre of mass unchanged, the block of mass 20 kg should be moved by a distance of 1 cm, towards left.

Suggest Corrections

Similar questions

Q. Two blocks of masses 10 kg and 30 kg are placed along a vertical line. The first block is raised through a height of 7 cm. By what distance should the second mass be moved to raise the centre of mass by 1 cm?

Consider a two-particle system with the particles having masses $m_{1}$ and $m_{2}$ . If the first particle is pushed towards the centre of mass by a distance 'd', by what distance should the second particle be moved so as to keep the centre of mass at the same position?

Q. Considering a system having two masses

m_{1}

and

m_{2}

in which first mass is pushed towards centre of mass by a distance

a

. The distance the second mass must be moved to keep centre of mass at same position is :

Join BYJU'S Learning Program