First slide

Centre of mass(com)

Question

Three particles of equal masses are placed at the corners of an equilateral triangle as shown in the figure. Now particle A starts with a velocity v₁ towards line AB, particle. B starts with a velocity v₂ towards line BC and particle C starts with velocity v₃ towards line CA. The displacement of CM of three particle A, B and C after time t will be (given if v₁ = v₂ = v₃)

Moderate

A

zero
B

$\frac{v_{1} + v_{2} + v_{3}}{3} t$
C

$\frac{v_{1} + \frac{\sqrt{3}}{2} v_{2} + \frac{v_{3}}{2}}{3} t$
D

$\frac{v_{1} + v_{2} + v_{3}}{4} t$

Solution

$\begin{array}{l} Using {\vec{v}}_{CM} = \hat{i} v_{x} + \hat{j} v_{y} \\ where v_{x} = \frac{m v_{x_{1}} + m v_{x_{2}} + m v_{x_{3}}}{3 m} = 0 \\ v_{y} = \frac{m v_{y_{1}} + m v_{y_{2}} + m v_{y_{3}}}{3 m} = 0 \end{array}$
Net velocity is zero (using vector theory) i.e., CM is at rest. So, displacement of C.M. is zero. So choices (b), (c) and (d) are wrong.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App