Question

A unifonn rod of length $L$ and mass $2m$ rests on a smooth horizontal table, A point mass $m$ moving horizontally at right angles to the rod with velocity v collides with one end of the rod and sticks it Then

• A. angular velocity of the system after collision is $\frac{2}{5}\frac{v}{l}$
• B. angular velocity of the system after collision is $\frac{v}{2l}$
• C. the loss,in kinetic energy of the system as a whole as a result of the collision is $\frac{3}{10}m{v}^2$
• D. the loss,in kinetic energy of the system as a whole as a result of the collision is $\frac{7m{v}^2}{24}m{v}^2$

Answer 1

Answer: (A), (C)

The correct options are
A angular velocity of the system after collision is $\frac{2}{5}\frac{v}{l}$
C the loss,in kinetic energy of the system as a whole as a result of the collision is $\frac{3}{10}m{v}^2$

$C_1$ is centre of mass of rod
$C_2$ is centre of mass of both
$P_i=P_f$ $\therefore mv=3m{v}_0\Rightarrow v_0=\frac{v}{3}....\left(i\right)$
$L_i=L_f$ about $C_2$ we have,
$mv\frac{l}{6}=I_{C_2}\omega$
$=[\left(\frac{2m{l}^2}{12}\right)+2m{\left(\frac{l}{3}\right)}^2+m{\left(\frac{l}{6}\right)}^2]$
$\therefore \omega =\frac{2}{5}\frac{v}{l}$
$K_i=\frac{1}{2}m{v}^2$
$K_f=\frac{1}{2}\left(3m\right){\left(\frac{v}{3}\right)}^2+\frac{1}{2}{I}_{C2}{\omega }^2$
$=\frac{1}{6}m{v}^2+\frac{1}{2}\left(\frac{5}{12}m{l}^2\right){\left(\frac{2}{5}\frac{v}{l}\right)}^2$
$=\frac{1}{5}m{v}^2$
$\therefore$ Loss of kinetlc energy $=K_i-K_f$
$\frac{3}{10}m{v}^2$