Question

How soon will the frame come to the orientation shown in figure (b) after collision?

• A. $\frac{\pi l}{4v}$
• B. $\frac{7\pi l}{8v}$
• C. $\frac{\pi l}{v}$
• D. $\frac{7\pi l}{4v}$

Answer 1

Answer: (D)

$\frac{7\pi l}{4v}$

Consider the L to be a single system. Its COM lies in midway of the diagonal.
The initial velocity of ball be $u$ and final velocity (in same direction) be $v$.
Linear momentum is always conserved. Momentum imparted to the L-system after collision is = $m(u-v)=2m\left(v_{Lcom}\right)$
$v_{Lcom}=(u-v)/2$
About COM of L-system(just consider an axis about it) angular momentum of the 3 body system is conserved.
$L_i=mu\left(l/2\right)$. $L_f=mv\left(l/2\right)+2m\left(v_{com}\right)\left(0\right)+I_L{\omega }_L$
$I_L=2\times m(l/\sqrt{2}{)}^2=m{l}^2$
$=>(u-v)/2=l\omega$
velocity of separation of the balls is $v_{Lcom}+\left(l/2\right)\omega -v$
$l\omega /2$ is horizontal component of velocity obtained by angular motion.
In elastic collision, velocity of approach = velocity of separation.
Therefore, $u=v_{Lcom}+(u-v)/4-v=3(u-4)/4-v$
$4(u+v)=3(u-v)=>v=-u/7$
$\omega =(u-v)/2l=4u/7l$
time taken to reach orientation = angle/$\omega$
$t=\pi /\omega =7\pi l/4u$