Question

Two identical balls are interconnected with a massless and inextensible thread. The system is in gravity free space with the thread just taut. Each ball is imparted a velocity $v$, one towards the other ball and the other perpendicular to the first, at $t=0$. Then :

• A. the thread will become taut at $t=\left(L/v\right)$
• B. the thread will become taut at $t<\left(L/v\right)$
• C. the thread will always remain taut for $t>\left(L/v\right)$
• D. the kinetic energy of the system will always $m{v}^2$

Answer 1

Answer: (A), (C)

The correct options are
A the thread will become taut at $t=\left(L/v\right)$
C the thread will always remain taut for $t>\left(L/v\right)$

As shown in figure $A$ and $B$ are two identical mass. Suppose $A$ move $x$ and $B$ move $x$ because velocity is same. Length taught when $y>L$

$L=\sqrt{x^2+(L-x{)}^2}{L}^2=x^2+L^2+x^2-2lx{2x}^2-2Lx=02x(x-L)=0x=L$

time taken to travel $x=t=\frac{x}{V}=\left(\frac{L}{V}\right)$

thread will taught at $x=L$ and time $t=\frac{1}{2}$

and after that thread always taught because

$y=\sqrt{x^2-(L-x^2)}$

$y>Ldfx>L$