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 ${\nu }_1$ towards line $AB$, particle $B$ starts with a velocity ${\nu }_2$ towards line $BC$ and particle $C$ starts with velocity ${\nu }_3$ towards line $CA$. The displacement of $CM$ of three particle A, B and C after time $t$ will be (given if ${\nu }_1={\nu }_2={\nu }_3$)

• A. zero
• B. $\frac{{\nu }_1+{\nu }_2+{\nu }_3}{3}t$
• C. $\frac{{\nu }_1+\frac{\sqrt{3}}{2}{\nu }_2+\frac{{\nu }_3}{2}}{3}t$
• D. $\frac{{\nu }_1+{\nu }_2+{\nu }_3}{4}t$

Answer 1

The correct option is A zero

REF.Image.

$v_{1x}=\frac{-v_1}{2},v_{1y}=\frac{-v_1\sqrt{3}}{2}$

$v_{2x}=v_2,v_{2y}=0$

$v_{3x}=\frac{-v_3\sqrt{3}}{2},v_{3y}=v_3\frac{\sqrt{3}}{2}$

using , ${\overrightarrow{v}}_{cm}=\hat{i}{v}_x+\hat{j}{v}_y$

where,

$V_x=\frac{-m{v}_{1/2}+v_{2}m-m{v}_{3/2}}{3m}=0$

$v_y=\frac{-m{v}_1\sqrt{3}/2+m{v}_3\sqrt{3}/2}{3m}=0$

$\therefore$ Net velocity = 0 (using vector theory)

i.e. CM is at rest.

So, displacement of CM is zero.

so, (a) is correct.