Question

A rod AC of length l and mass m is kept on a horizontal smooth plane. It is free to rotate and move. A particle of same mass m moving on the plane with velocity v strikes rod at point B making angle ${37}^{\circ }$ with the rod. The collision is elastic. After collision :

• A. The angular velocity of the rod will be $\frac{72}{55}\frac{v}{l}$
• B. The center of the rod will travel a distance $\frac{\pi l}{3}$ in the time
  in which it makes half rotation
• C. Impulse of the impact force is $\frac{24mV}{55}$
  
• D. The Centre of the rod will travel a distance $\frac{2\pi l}{3}$ in the time in which it makes half rotation

Answer 1

Answer: (B), (C)

The correct options are
B The center of the rod will travel a distance $\frac{\pi l}{3}$ in the time
in which it makes half rotation
C Impulse of the impact force is $\frac{24mV}{55}$


Where $l^1=[\frac{\left(2m\right)L^2}{12}+2m{\left(\frac{L}{6}\right)}^2+m{\left(\frac{L}{3}\right)}^2]$
Vcm and v' be the velocities of the CM of the rod and the particle
PCLM :- $m\frac{3v}{5}=m{v}_{cm}+m{v}^{\prime }$ (1)
PCAM about CM of the rod: $\frac{m3v_0}{5}\times \frac{l}{5}=\frac{1}{12}m{l}^2\omega +\frac{m{v}^{\prime }l}{4}$ (2)
$e=1=\frac{v_{cm}+\frac{\omega l}{4}-v^{\prime }}{\frac{4v}{4}}$ (3) Solving 1, 2 and 3 $\Rightarrow \omega =\frac{24v}{55l}=\frac{\omega l}{3}$
Distance traveled =$v_{cm}t$ where $t=\frac{\pi }{\omega }$
From impulse momentum equation on the rod $\int Ndt=mu=\frac{24mv}{55}$