Question

Find the velocity of $C$ relative to ${}^{\prime }A+B^{\prime }$, immediately after the impact.

Answer 1

REF.Image.

final com of system $=m\left(\frac{l}{2}\right)+m\left(\frac{l}{2}\right)+m(-\frac{l}{2})$

$=\frac{l}{6}$ above center of rod.

an serving angular momentum about point at final com

$mu(\frac{l}{2}-\frac{l}{6})+m\left(0\right)(\frac{l}{2}-\frac{l}{6})+m\left(0\right)(\frac{l}{6}+\frac{l}{2})$

about final com

$=Iw$ (1)

$I=m{(\frac{l}{2}-\frac{l}{6})}^2+m(\frac{l}{2}-\frac{l}{6})+m(\frac{l}{2}+\frac{1}{6})$

$=m\frac{l^2}{9}+m\frac{l^2}{9}+\frac{4l^2}{9}$

$=\frac{2}{3}m{l}^2$ (2)

From (1) and (2)

$m/mu\frac{l}{3}=\frac{2}{3}m{l}^{2}w$

$w=\frac{\mu }{2l}$

$\overline{V_{A+B}}=\overline{V_{com}}+(\frac{l}{2}-\frac{l}{6})w$

$\overline{V_C}=\overline{V_{com}}-(\frac{l}{2}+\frac{l}{6})w$

$\overline{V_C}w.r.t(A+B)=\overline{V_C}-\overline{V_{A+B}}$

$=(-\frac{l}{2}-\frac{l}{6}-\frac{l}{2}+\frac{l}{6})w$

$=-wl$

$=-\frac{\mu }{2l}l=-\frac{\mu }{2}$