Question

A uniform 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 the rod at point $\text{B}$ making an 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. None of these

Answer 1

Answer: (A), (B), (C)

impulse of the impact force is $\frac{24mv}{55}$

The ball has $V^{\prime }$ component of its velocity perpendicular to the length of rod immediately after the collision. $u$ is velocity of COM of the rod and $\omega$ is angular velocity of the rod, just after collision. The ball strikes the rod with speed $vcos{53}^{\circ }$ in perpendicular direction and its component along the length of the rod after the collision is unchanged.

Using for the point of collision:

Velocity of separation = Velocity of approach
$\Rightarrow \frac{3v}{5}=(\frac{\omega l}{4}+u)+V^{\prime }....\left(1\right)$
Conserving linear momentum (of rod + particle), in the direction $\perp$ to the rod.
$mv\cdot \frac{3}{5}=mu-{mV}^{\prime }....\left(2\right)$
Conserving angular moment about point 'D' as shown in the figure
$0=0+[mu\frac{l}{4}-\frac{m{l}^2}{12}\omega ]\Rightarrow u=\frac{\omega l}{3}....\left(3\right)$
By solving
$u=\frac{24v}{55},w=\frac{72v}{55l}$
Time taken to rotate by $\pi$ angle t = $\frac{\pi }{\omega }$
$\int N\cdot dt\cdot \frac{l}{4}=\frac{m{l}^2}{12}\frac{72v}{55l}$
$\Rightarrow \int N\cdot dt=\frac{24mv}{55}\}$
or, using impulse momentum equation on rod
$\int Ndt=mu=\frac{24mv}{55}$