The $x$ location and mass of two masses are given as $(1 m, 3 k g)$ and $(- 2 m, 2 k g)$ . A third mass weighs $1 k g$ .
Given that the center of mass of the system lies at $x = - 1 m$ , what must be the $x$ position of the third mass?

x = - 5 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x = - 3 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = - 2.5 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = - 2 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = 2 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $x = - 5 m$
Given : $x_{1} = 1$ m $m_{1} = 3$ kg $m_{2} = 2 k g$ $x_{2} = - 2$ m $m_{3} = 1$ kg
Position of centre of mass $X = - 1$ m
Using $X = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$

$∴$ $- 1 = \frac{3 \times 1 + 2 \times (- 2) + 1 \times x_{3}}{3 + 2 + 1}$ $⟹ x_{3} = - 5$ m

Suggest Corrections

Similar questions

Four particles of masses 1 kg, 2 kg, 3 kg and 4 kg are placed at the corners of a square of side 2 m in the x-y plane as shown in the figure. If the origin of the coordinate system is taken at the mass of 1 kg, the (x, y) co-ordinates of the centre of mass are